ALTERNATING CURRENT - einsteinclasses.com Folder/(7)Alternating_C.pdf · the variation of current and voltage as a function of time in this circuit. Solution : Figure shows a resistor

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ALTERNATING CURRENT

7.1 Introduction :

Q. What is direct current ?

Solution : Direct current does not change direction with time.

Q. What is alternating current ?

Solution : Alternating currents and voltages change its direction with time.

Q. What is the main reason for preferring use of AC voltage over DC voltage ?

Solution : The main reason for preferring use of ac voltage over dc voltage is that ac voltages can be easilyand efficiently converted from one voltage to the other by means of transformers. Further, electrical energycan also be transmitted economically over long distances.

7.2 AC Voltage Applied to a Resistor :

Q. An ac voltage is applied to a resistor. Derive the expression for the current in the circuit.

OR

Prove that the voltage and current are in the phase with each other in pure resistive circuit. Also plotthe variation of current and voltage as a function of time in this circuit.

Solution : Figure shows a resistor connected to a source of ac voltage.

Let this potential difference, also called ac voltage, be given by v = vm sin t, where v

m is the amplitude of

the oscillating potential difference and is its angular frequency.

To find the value of current through the resistor, we apply Kirchhoff’s loop rule 0)t( , to the circuit

shown in figure to get vm sin t = iR or tsin

R

vi m .

Since R is a constant, we can write this equation as i = im sin t, where the current amplitude i

m is given by

R

vi mm .

The voltage across a pure resistor and the current through it are plotted as a function of time as shown infigure. which clearly indicate that the voltage and current are in phase with each other.

Q. Find the average value of AC current over one complete cycle ? OR Prove mathematically thatthe average value of a.c. over one complete cycle is zero.

Solution : The alternating current can be expressed as I = Im sin t.

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Then the average value of a.c. cover one complete cycle will be

T

0

T

0

av dtIdtI

T

0

m

T

0

mav

tcos

T

IdttsinI

T

1I

]0cosT[cos

T

Im

T

2

]0cos2[cosT

Im

0]11[T

Im

Q. Find the average value of AC current over one-half cycle.

Solution : The alternating current can be expressed as I = Im sin t.

Then the average value of a.c. cover one-half of cycle will be

2/T

0

2/T

0

av dtIdtI

2/T

0

m

2/T

0

mav

tcos

T

I2dttsinI

T

2I

]0cos2/T[cos

T

I2 m

T

2

]0cos[cosT

I2 m

mm I2]11[

T

I2

Q. As the average current over one complete cycle is zero, does it mean that the average powerconsumed is zero and there is no dissipation of electrical energy ?

Solution : The fact that the average current is zero, however, does not mean that the average powerconsumed is zero and that there is no dissipation of electrical energy. As you know, Joule heating is givenby i2R and depends on i2 (which is always positive whether i is positive or negative) and not on i. Thus,there is Joule heating and dissipation of electrical energy when an ac current passes through a resistor.

Q. Find the average value of power in the resistor ?

Solution : The instantaneous power dissipated in the resistor is p = i2R = i2mRsin2t

The average value of p over a cycle is tsinRiRip 22m

2

where the bar over a letter (here, p) denotes its average value and <.......> denotes taking average of the

quantity inside the bracket. Since, i2m and R are constants, tsinRip 22

m

we have, 2

1tsin2 . Thus, Ri

2

1p 2

m .

Q. Define the room mean square current or effective current for ac ? Also find the value of rmscurrent ?

Solution : To express ac power in the same form as dc power (P = I2R), a special value of current is definedand used. It is called, root mean square (rms) or effective current and is denoted by I

rms or I.

It is defined by iI 2 T

0

T

0

2 dtdti =

T

0

22m dttsini

T

1 ( i = i

m sint)

T

0

2

2

1dttsin

T

1 (by integration)

Hence, I 2

ii

2

1 m2m = 0.707 i

m

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The rms current is the equivalent dc current that would produce the same average power loss as thealternating current.

Q. What is the physical meaning of rms current ?

Solution : The rms current is the equivalent dc current that would produce the same average power loss asthe alternating current.

Q. Define the rms voltage or effective voltage for ac ? Also find the value of rms voltage ?

Solution : The rms voltage or effective voltage is given by mm v707.02

vV

It is defined by vV 2 T

0

T

0

2 dtdtv =

T

0

22m dttsinv

T

1 ( v = v

m sint)

T

0

2

2

1dttsin

T

1

Hence, V 2

vv

2

1 m2m = 0.707 v

m

Q. What is the rms voltage if peak or maximum voltage equals to 220V for an ac ?

Solution : vm = 2 V = (1.414)(220 V) = 311 V

Q. A light bulb is rated at 100W for a 220 V supply. Find (a) the resistance of the bulb; (b) the peakvoltage of the source; and (c) the rms current through the bulb. [NCERT Solved Example 7.1].

Solution : (a) 484 (b) 311 V (c) 0.450 A

7.3 Representation of AC Current and Voltage by Rotating Vectors - Phasors :

Q. What is the use of phasor diagram ?

Solution : The analysis of an ac circuit is facilitated by the use of a phasor diagram.

Q. What is phasor ?

Solution : A phasor is a vector which rotates about the origin with angular speed as shown in figure.

The vertical components of phasors V and I represent the sinusoidally varying quantities v and i. Themagnitudes of phasors V and I represent the amplitudes or the peak values v

m and i

m of these oscillating

quantities. Figure shows the voltage and current phasors and their relationship at time t1 for the case of an

ac source connected to a resistor. The projection of voltage and current phasors on vertical axis arev

m sin t and i

m sint respectively represent the value of voltage and current at that instant.

Q. Draw the phasor diagram for pure resistive ac circuit.

Solution :

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Q. What is the phase angle between the voltage and the current in pure resistive ac circuit ?

Solution : The phase angle between the voltage and the current in pure resistive ac circuit is zero.

7.4 AC Voltage Applied to an Inductor :

Q. An ac voltage is applied to an ideal inductor (inductor with zero resistance). Derive the expressionfor the current in the circuit. Hence define inductive reactance.

OR

Prove that the voltage across the inductor will lead by /2 w.r.t. current passing through it i.e.,current lags the voltage by /2.

Draw the phasor diagram for this circuit. Also plot the variation of current and voltage as a functionof time.

Solution :

Figure shows an ac source connected to an inductor with negligible resistance. Thus, the circuit is a purelyinductive ac circuit. Let the voltage across the source be v = v

m sin t. Using the Kirchhoff’s loop rule,

0)t( , 0dt

diLv , where L is the self-inductance of the inductor. We have, tsin

L

v

L

v

dt

di m .

To obtain the current, we integrate : dt)tsin(L

vdi m

. and get, constant)tcos(L

vi m

.

The integration constant has the dimension of current and is time-independent. Since the source has an emfwhich oscillates symmetrically about zero, the current it sustains also oscillates symmetrically about zero,so that no constant or time-independent component of the current exists. Therefore, the integration constantis zero.

Using

2tsin)tcos( , we have

2tsinii m , where

L

vi mm

is the amplitude of the

current. The quantitiy L is analogous to the resistance and is called inductive reactance, denoted by

XL which is given by X

L = L. The amplitude of the current is, then

L

mm

X

vi .

From the comparison of v = vm sin t and

2tsinii m the source voltage and the current in an

inductor shows that the current lags the voltage by /2 or one-quarter (1/4) cycle. Figure (a) shows thevoltage and the current phasors in the present case at instant t

1. The current phasor I is /2 behind the

voltage phasor V. Figure (b) shows the graph for the voltage and current versus t.

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Q. What is the unit and dimension of inductive reactance ?

Solution : The unit and dimension of inductive reactance is the same as that of resistance and its SI unit isohm. The dimension of X

L is [ML2T3A–2].

Q. What is physical meaning of inductive reactance ?

Solution : The inductive reactance limits the current in a purely inductive circuit in the same way as theresistance limits the current in a purely resistive circuit.

Q. How does the inductive reactance depend on the inductance and frequency of current ?

Solution : The inductive reactance is directly proportional to the inductance and to the frequency of thecurrent.

Q. Plot a graph of inductive reactance versus angular frequency of ac for the given inductor (pure orideal). Which quantity can be determined from the slope of this graph ?

Solution : The graph of inductive reactance versus angular frequency of ac is straight line passing throughthe origin.

The slope (tan ) of this graph will give the value of inductance L.

Q. Plot a graph of inductive reactance versus self inductance (L) for the given ac source. Whichquantity can be determined from the slope of this graph ? The inductor is ideal.

Solution : The graph of inductive reactance versus self inductance (L) for the given ac source is straightline passing through the origin.

The slope (tan ) of this graph will give the value of angular frequency of ac source.

Q. The graph of inductive reactance versus frequency of ac for the given inductor (pure or ideal) isas shown in figure.

What is the value of self inductance ?

Solution : From XL = L = (2L)f, the slope tan equals to 2L where = 450 from the graph. Hence

H2

1L

.

Q. Prove that the average power supplied to an inductor over one complete cycle is zero.

Solution : The instantaneous power supplied to the inductor is

)tsin(v2

tsiniivP mmL

= –i

mv

m cos(t)sin(t) )t2sin(

2

vi mm

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So, the average power over a complete cycle is )t2sin(2

viP mm

L = 0)t2sin(2

vi mm

since the average of sin (2t) over a complete cycle is zero.

Hence the average power supplied to an inductor over one complete cycle is zero.

Q. A pure inductor of 25.0 mH is connected to a source of 220 V. Find the inductive reactance andrms current in the circuit if the frequency of the source is 50 Hz. [NCERT Solved Example 7.2].

Solution : 7.85 , 28A

7.5 AC Voltage Applied to a Capacitor :

Q. An ac voltage is applied to a capacitor. Derive the expression for the current in the circuit. Hencedefine capacitive reactance.

OR

Prove that the voltage across the capacitor will lag by /2 w.r.t. current passing through it i.e.,current leads w.r.t. the voltage by /2 (the current is /2 ahead of voltage).

Draw the phasor diagram for this circuit. Also plot the variation of current and voltage as a functionof time.

Solution :

Figure shows an ac source generating ac voltage v = vm sin t connected to a capacitor only, a purely

capacitive ac circuit. Let q be the charge on the capacitor at any time t. The instantaneous voltage v across

the capacitor is C

qv . From the Kirchhoff’s loop rule, the voltage across the source and the capacitor are

equal, C

qtsinvm . To find the current, we use the relation

dt

dqi .

dt

di (v

m C sin t) = Cv

mcos(t). Using the relation,

2tsin)tcos( .

We have

2tsinii m , where the amplitude of the oscillating current is i

m = Cv

m. We can rewrite it

as )C/1(

vi mm

. Comparing it to i

m = v

m/R for a purely resistive circuit, we find that (1/C) plays the role

of resistance. It is called capacitive reactance and is denoted by Xc, X

c = 1/C, so that the amplitude of the

current is C

mm

X

vi .

From the comparison of v = vm sin t and

2tsinii m the source voltage and the current in an

capacitor shows that the voltage lags the voltage by /2 or one-quarter (1/4) cycle i.e., the current is /2ahead of voltage. Figure (a) shows the voltage and the current phasors in the present case at instant t

1. The

current phasor I is /2 ahead the voltage phasor V. Figure (b) shows the graph for the voltage and currentversus t.

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Q. What is the unit and dimension of capacitive reactance ?

Solution : The unit and dimension of capacitive reactance is the same as that of resistance and its SI unit isohm.The dimensions of X

C is [ML2T–3A–2]

Q. What is physical meaning of capacitive reactance ?

Solution : The capacitive reactance limits the current in a purely capacitive circuit in the same way as theresistance limits the current in a purely resistive circuit.

Q. How does the capacitive reactance depend on the capacitance and frequency of current ?

Solution : The capacitive reactance is inversely proportional to the capacitance and to the frequency of thecurrent.

Q. Plot a graph of capacitive reactance versus angular frequency of ac for the given capacitance.

Solution : The graph of capacitive reactance versus angular frequency of ac is hyperbolic.

Q. Plot a graph of capacitive reactance versus capacitance (C) for the given ac source.

Solution : The graph of capacitive reactance versus capacitance for the given ac source is hyperbolic.

Q. Plot a graph of capacitive reactance versus inverse of angular frequency of ac for the givencapacitor. Which quantity can be determined from the slope of this graph ?

Solution : The graph of capacitive reactance versus inverse of angular frequency of ac is straight linepassing through the origin.

The slope (tan ) of this graph will give the value of inverse of capacitance C.

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Q. Plot a graph of capacitive reactance versus inverse of capacitance for given ac source. Whichquantity can be determined from the slope of this graph ?

Solution : The graph of capacitive reactance versus inverse of capacitance for the given ac source isstraight line passing through the origin.

The slope (tan ) of this graph will give the value of inverse of angular frequency of ac source.

Q. The graph of capacitive reactance versus inverse of frequency of ac for the given capacitor is asshown in figure.

Q. What is the value of capacitance ?

Solution : From XC =

1

C

1, the slope tan equals to

C

1 where = 300 from the graph. Hence

unit3C .

Q. Prove that the average power supplied to a capacitor over one complete cycle is zero.

Solution : The instantaneous power supplied to the capacitor is

)tsin(v2

tsiniivP mmL

= i

mv

m cos(t)sin(t) )t2sin(

2

vi mm

So, the average power over a complete cycle is )t2sin(2

viP mm

L = 0)t2sin(2

vi mm

since the average of sin (2t) over a complete cycle is zero.

Hence the average power supplied to a capacitor over one complete cycle is zero.

Q. Capacitors block dc but allow ac to pass. Why.

Solution : When a capacitor is connected to a voltage source in a dc circuit, current will flow for the shorttime required to charge the capacitor. As charge accumilates on the capacitor plates, the voltage across themincrease, opposing the current. That is, a capacitor in a dc circuit will limit or oppose the current as itcharges. When the capacitor is fully charged, the current in the circuit falls to zero.

When the capacitor is connected to an ac source, it limits or regulates the current, but does not completelyprevent the flow of charge. The capacitor is alternately charged and discharged as the current reverses eachhalf cycle.

Q. A lamp is connected in series with a capacitor. Predict you observations for dc and ac connections.What happens in each case if the capacitance of the capacitor is reduced ?

[NCERT Solved Example 7.3]

Solution : When a dc source is connected to a capacitor, the capacitor gets charged and after charging nocurrent flows in the circuit and the lamp will not glow. There will be no change even if C is reduced. Withac source, the capacitor offers capacitative reactance (1/C) and the current flows in the circuit.Consequently, the lamp will shine. Reducing C will increase reactance and the lamp will shine less brightlythan before.

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Q. A 15.0 µF capacitor is connected to a 220 V, 50 Hz source. Find the capacitive reactance and thecurrent (rms and peak) in the circuit. If the frequency is doubled, what happens to the capacitivereactance and the current ? [NCERT Solved Example 7.4].

Solution : 212 , 1.04A, 1.47 A

Q. A light bulb and an open coil inductor are connected to an ac source through a key as shown infigure.

The switch is closed and after sometime, an iron rod is inserted into the interior of the inductor. Theglow of the light bulb (a) increases; (b) decreases; (c) is unchanged, as the iron rod is inserted. Giveyour answer with reasons. [NCERT Solved Example 7.5]

Solution : As the iron is inserted, the magnetic field inside the coil magnetizes the iron increasing themagnetic field inside it. Hence, the inductance of the coil increases. Consequently, the inductive reactanceof the coil increases. As a result, a larger fraction of the applied ac voltage appears across the inductor,leaving less voltage across the bulb. Therefore, the glow of the light bulb decreases.

7.6 AC Voltage Applied to a Series LCR Circuit :

Q. Discuss an ac LCR series circuit and find the value of current flowing in the circuit using thephasor diagram. Define the impedence. Also find the phase relationship of current to the appliedalternating voltage.

Discuss the condition for the circuit to be predominantly capacitive and predominantly inductive.

Solution :

Figure shows a series LCR circuit connected to an ac source . We take the voltage of the source to bev = v

m sin t. If q is the charge on the capacitor and i the current, at time t, we have, from Kirchhoff’s loop

rule : vC

qiR

dt

diL . We want to determine the instantaneous current i and its phase relationship to the

applied alternating voltage v.

From the circuit shown in figure, we see that the resistor, inductor and capacitor are in series. Therefore, theac current in each element is the same at any time, having the same amplitude and phase. Let it bei = i

msin(t + ), where is the phase difference between the voltage across the source and the current in the

circuit. We shall construct a phasor diagram for the present case.

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Let I be the phasor representing the current in the circuit as given by equation i = imsin(t + ). Further, let

VL, V

R, V

C and V represent the voltage across the inductor, resistor, capacitor and the source, respectively.

The length of these phasors or the amplitude of VR, V

C and V

L are v

Rm = i

mR, V

Cm = i

mX

C, v

Lm = i

mX

L

respectively.

Since VC and V

L are always along the same line and in opposite directions, they can be combined which has

a magnitude |vCm

– vLm

|. Since V is represented as the hypotenuse of a right-triangle whose sides are vRm

and

|vCm

– vLm

|, the pythagorean theorem gives : 2LmCm

2Rm

2m )vv(vv .

Substituting the values of vRm

, vCm

, and vLm

we have, vm

2 = (imR)2 + (i

mX

C – i

mX

L)2 = i

m2 [R2 + (X

C – X

L)2]

or, 2

LC2

mm

)XX(R

vi

. By analogy to the resistance in a circuit, we introduce the impendance Z in

an ac circuit : Z

vi mm , where 2

LC2 )XX(RZ . Since phasor I is always parallel to phasor VV

R,

the phase angle is the angle between VR and V and can be determine from figure.

Rm

LmCm

v

vvtan

R

XXtan LC

.

If XC > X

L, is positive and the circuit is predominantly capacitive. Consequently, the current in the circuit

leads the source voltage. If XC < X

L, is negative and the circuit is predominantly inductive. Consequently,

the current in the circuit lags the source voltage.

Q. What is impedence diagram ?

Solution : Impedence diagram is a right triangle with impedence Z as its hypotenuse and it is shown in thefigure.

Q. Draw the phasor diagram and variation of voltage and current with t for series LCR circuitwhen X

c > X

L.

Solution :

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Q. Find the expression for current flowing in the series LCR circuit using the analytical solution.Also find the phase relationship of current with the voltage.

Solution : The voltage equation for the circuit is vC

qRi

dt

diL = v

m sin t

We know that i = dq/dt. Therefore, di/dt = d2q/dt2. Thus, in terms of q, the voltage equation becomes

tsinvC

q

dt

dqR

dt

qdL m2

2

...(i)

This is like the equation for a forced, damped oscillator. Let us assume a solution

q = qm sin (t + )

so that )tcos(qdt

dqm and )tsin(q

dt

qd 2m2

2

Substituting these values in equation (i), we get

tsinvC

)tsin(q)tcos(Rq)tsin(Lq– m

mm

2m

By using the relation Xc = 1/C, X

L = L, we get

qm[Rcos(t + ) + (X

C – X

L) sin (t + )] = v

m sin t ...(ii)

Multiplying and dividing the equation (ii) by 2Lc

2 )XX(RZ , we have

tsinv)tsin(Z

)XX()tcos(

Z

RZq m

LCm

...(iii)

Now, let cosZ

R and

sin

Z

)XX( LC so that R

XXtan LC1

Substituting this in equation (iii) and simplifying, we get qmZcos(t + – ) = v

m sin t

Comparing the two sides of this equation, we see that vm = q

m Z = i

m Z

where im = q

m and

2or

2. Therefore, the current in the circuit is

)tcos(i)tcos(qdt

dqi mm or )tsin(ii m

2

where 2

LC2

mmm

)XX(R

v

Z

vi

and

R

XXtan LC1

.

Q. Discuss the phenomena of resonance in LCR series AC circuit. OR Define resonant frequencyand derive the expression for resonance frequency in LCR series AC circuit.

Solution : In the phenomena of resonance in the series LCR circuit, the current amplitude will have amaximum value at a certain frequency of AC source. This frequency is known as natural frequency orresonance frequency.

For an RLC circuit driven with voltage of amplitude vm and frequency , we found that the current ampli-

tude is given by 2

LC2

mmm

)XX(R

v

Z

vi

with X

C = 1/C and X

L = L. So if is varied, then at a

particular frequency 0, X

C = X

L, and the impedance is minimum )R0RZ( 22 . This frequency is

called the resonant frequency and it is given by : LC

1orXX 0

0Lc

or

LC

10 .

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At resonant frequency, the current amplitude is maximum; im = v

m/R.

Q. Plot the graph for the variation of current amplitude (im) with angular frequency of ac source in

RLC series ac circuit. Also mention the point on the graph at which resonant frequency will occur.

Solution : The graph for the variation of current amplitude (im) with angular frequency of ac source in RLC

series ac circuit as shown in figure.

The point on the graph at which resonant frequency will occur is given by 0 at which the current amplitude

will have maximum value vm/R.

Q. Consider the RLC series ac circuit with L = 1.00 mH, C = 1.00 nF for two values of R(i) R = 100 and R = 200. The source peak voltage is 100 V. (i) Find the resonant freqency andmaximum value of current amplitude in each case. (ii) Plot the graph for the variation of currentamplitude (i

m) with angular frequency of ac source in RLC series ac circuit.

Solution : (i) The resonant frequency is given by LC

10 and the maximum value of current amplitude

equals to vm/R. For the given values of L, C, R and v

m in the problem, the value of

0 = 1.00 × 106 rad/s for

both values of R. The maximum value of current amplitude equals to 0.5A for R = 200 and 1.0A forR = 100.

(ii)

Graph (i) is for 100 resistance and graph (ii) is for 200 resistance.

Q. Draw the phasor diagram for series LCR ac circuit at resonance. What is the phase differencebetween current and voltage for series LCR ac circuit at resonance ?

Solution : The phasor diagram for series LCR ac circuit at resonance is as shown in figure.

The phase difference between current and voltage for series LCR ac circuit at resonance is zero i.e., currentand voltage are in the same phase

Q. What is the application of resonant RLC circuits for ac source ? OR Explain the application ofresonant circuit in the tuning mechanism of a radio or a TV set ?

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Solution : Resonant circuits have a variety of applications, for example, in the tuning mechanism of a radioor a TV set. The antenna of a radio accepts signals from many broadcasting stations. The signals picked upin the antenna acts as a source in the tuning circuit of the radio, so the circuit can be driven at manyfrequencies. But to heat one particular radio system, we tune the radio. In tuning, we vary the capacitanceof a capacitor in the tuning circuit such that the resonant fequency of the circuit becomes nearly equal to thefrequency of the radio signal received. When this happens, the amplitude of the current with the frequencyof the signal of the particular radio station in the circuit is maximum.

Q. Why cannot have resonance in a RL or RC circuit ? OR Why resonance phenomenon is exhibitedby a circuit only if both L and C are present in the circuit ?

Solution : The resonance phenomenon is exhibited by a circuit only if both L and C are present in thecircuit. Only then do the voltages across L and C cancel each other (both being out of phase) and the currentamplitude is v

m/R, the total source voltage appearing across R. This means that we cannot have resonance

in a RL or RC circuit.

Q. What is the power dissipated by series LRC circuit if the current amplitude is 1/2 times themaximum value of current amplitude ? Let the power dissipation at the maximum value of currentamplitude is P.

Solution : The power dissipated by the circuit becomes P/2 if the current amplitude is 1/2 times themaximum value of current amplitude.

Q. (a) What is bandwidth of the circuit and how it is related with the sharpness of resonance ?

OR

What is sharpness of resonance ?

(b) Derive the expression of bandwidth of the circuit and the sharpness of resonance and hencedefine the quality factor ?

(c) What is the physical meaning of quality factor ?

Solution : (a) For values of other than 0, the amplitude of the current is less than the maximum value.

Suppose we choose a value of for which the current amplitude is 1/2 times its maximum value.

From the curve in figure, we see that there are two such values of say, and

2, one greater and the other

smaller than 0 and symmetrical about

0. We may write

1 =

0 +

2 =

0 –

The difference 1 –

2 = 2 is often called the bandwidth of the circuit. The quantity (

0 / 2) is

regarded as a measure of the sharpness of resonance. The smaller the , the sharper or narrower is theresonance.

(b) To get an expression for , we note that the current amplitude im is max

mi)2/1( (the maximum value

of im equals to v

m/R) for

1 =

0 + . Therefore, at

1,

2

11

2

mm

C

1LR

vi

= 2R

v

2

i mmaxm

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2RC

1LR

2

11

2

2

2

11

2 R2C

1LR

R

C

1L

11

which may be written as, RC)(

1L)(

00

R

1C

11L

00

00

Using LC

120 in the second term on the left hand side, we get R

1

L1L

0

0

00

We can approximate

0

1

0

1as1 since 10

.

Therefore, R1L1L0

00

0

R

2L

00

L2

R

L

R2

Hence the bandwidth of the circuit equals to L

R. The sharpness of resonance is given by,

R

L

200

The ratio R

L0 is also called the quality factor, Q of the circuit.

(c) From equation, we see that Q

2 0 . So, larger the value of Q, the smaller is the value of 2 or the

bandwidth and sharper is the resonance. Using 02 = 1/LC, equation can be equivalently expressed as

Q = 1/0CR.

If the resonance is less sharp, not only is the maximum current less, the circuit is close to resonance for alarger range of frequencies and the tuning of the circuit will not be good. So, less sharp the resonance,less is the selectivity of the circuit or vice versa. If quality factor is large, i.e., R is low or L is large, thecircuit is more selective.

Q. A resistor of 200 and a capacitor of 15.0 µF are connected in series to a 220 V, 50 Hz ac source.(a) Calculate the current in the circuit; (b) Calculate the voltage (rms) across the resistor and thecapacitor. Is the algebraic sum of these voltages more than the source voltage ? If yes, resolve theparadox. [NCERT Solved Example 7.6]

Solution : (a) 0.775 A (b) 151 V, 160.3 V

7.7 Power in AC Circuit : The Power Factor :

Q. Find the average power over a cycle in ac circuit ?

Solution : The voltage v = vm sin t is applied to ac circuit drives a current in the circuit given by

i = im sin (t + ). Therefore, the instantaneous power p supplied by the source is

p = vi = (vm sin t) × [i

m sin(t + )] )]t2cos([cos

2

iv mm .

[ 2sinAsinB = cos (A – B) – cos (A + B)]

The average power over a cycle is given by the average of the two terms in R.H.S. of above equation. It isonly the second term which is time-dependent. Its average is zero (the positive half of the cosine cancels the

negative half). Therefore, cos2

i

2

vcos

2

ivP mmmm = V I cos .

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This can also be written as, P = I2 Z cos .

Q. Define the power factor ?

Solution : The average power in an ac circuit is given by V I cos . The average power dissipated dependsnot only on the voltage and current but also on the cosine of the phase angle between them. The quantitycos is called the power factor.

Q. Find the power factor for the following circuit (i) resistive circuit (ii) purely inductive or purelycapacitive circuit (iii) LCR series circuit (iv) LCR series circuit at resonance ?

Solution : (i) Resistive circuit : If the circuit contains only pure R, it is called resistive. In that case = 0,cos = 1. There is maximum power dissipation.

(ii) Purely inductive or capacitance circuit : If the circuit contains only an inductor or capacitor, we knowthat the phase difference between voltage ande current is /2. Therefore, cos = 0, and no power isdissipated even through a current is flowing in the circuit. This current is sometimes referred to as wattlesscurrent.

(iii) LCR series circuit : In an LCR series circuit, power dissipated is given by I2 Z cos , where = tan–1

(XC – X

L)/R. So, may be non-zero in a RL or RC or RCL circuit. Even in such cases, power is dissipated

only in the resistor.

(iv) Power dissipated at resonance in LCR circuit : At resonance XC – X

L = 0, and = 0. Therefore,

cos = 1 and P = I2Z = I2R. That is, maximum power is dissipated in a circuit (through R) at resonance.

Q. How can you obtain wattless current in an AC circuit ? OR What is wattless current ?

Solution : If the circuit contains only an inductor or capacitor, we know that the phase difference betweenvoltage ande current is /2. Therefore, cos = 0, and no power is dissipated even through a current isflowing in the circuit. This current is sometimes referred to as wattless current.

Q. Name device through which power consumed an AC circuit is zero.

Solution : Pure inductor or pure capacitive circuit.

Q. What is the power dissipation in an a.c. circuit in which voltage and current are given by :

V = 300 sin (t + /2) and I = 5 sin t ?

Solution : Zero

Q. (a) For circuits used for transporting electric power, a low power factor implies large power lossin transmission. Explain. (b) Power factor can often be improved by the use of a capacitor ofappropriate capacitance in the circuit. Explain. [NCERT Solved Example 7.7]

Solution : (a) We know that P = I V cos where cos is the power factor. To supply a given power at agiven voltage. If cos is small, we have to increase current accordingly. But this will lead to large powerloss (I2r) in transmission. (b) Suppose in a circuit, current I lags the voltage by an angle . Then powerfactor cos = R/Z.

We can improve the power factor (tending to 1) by making Z tend to R. Let us understand, with the help ofa phasor diagram, how this can be achieved. Let us resolve I into two components. I

p along the applied

voltage V and Iq perpendicular to the applied voltage. I

q as you have learnt in Section 7.7, is called the

wattless component since corresponding to this component of current, there is no power loss. Ip is known as

the power component because it is in phase with the voltage and corresponds to power loss in the circuit.

It’s clear from this analysis that if we want to improve power factor, we must completely neutralize the

lagging wattless current Iq by an equal leading wattless current qI . This can be done by connecting a

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capacitor of appropriate value in parallel so that Iq and qI cancel each other and P is effectively I

p V..

Q. A sinusoidal voltage of peak value 283 V and frequency 50 Hz is applied to a series LCR circuit inwhich R = 3, L = 25.48 mH, and C = 796 µF. Find (a) the impedance of the circuit; (b) the phasedifference between the voltage across the source and the current; (c) the power dissipated in thecircuit; and (d) the power factor [NCERT Solved Example 7.8].

Solution : (a) 5 (b) –53.10. Since is negative, the current in the circuit lags the voltage across thesource. (c) 4800 W (d) 0.6

Q. Suppose the frequeny of the source in the previous example can be varied. (a) What is thefrequency of the source at which resonance occurs ? (b) Calculate the impendence, the current, andthe power dissipated at the resonant condition. [NCERT Solved Example 7.9]

Solution : (a) 35.4 Hz (b) 3, 66.7 A, 13.35 kW

Q. At an airport, a person is made to walk through the doorway of a metal detector, for securityreasons. If she/he is carrying anything made of metal, the metal detector emits a sound. On whatprinciple does this detector work ? [NCERT Solved Example 7.10]

Solution : The metal detector works on the principle of resonance in ac circuits. When you walk through ametal detector, you are, in fact, walking through a coil of many turns. The coil is connected to a capacitortuned so that the circuit is in resonance. When you walk through with metal in your pocket, the impendanceof the circuit changes – resulting in significant change in current in the circuit. This change in current isdetected and the electronic circuitry causes a sound to be emitted as an alarm.

7.8 LC Oscillations :

Q. (a) What is LC oscillations ?

(b) Derive the expression for the frequency of LC oscillations. Also find the charge on the capacitorat any time ‘t’ and current in the LC circuit ?

Solution : (a) We know that a capacitor and an inductor can store electrical and magnetic energy,respectively. When a capacitor (initially charged) is connected to an inductor, the charge on the capacitorand the current in the circuit exhibit the phenomenon of electrical oscillations similar to oscillations inmechanical systems.This is known as LC oscillation.

(b) Let a capacitor be charged qm (at t = 0) and connected an inductor as shown in figure.

The moment the circuit is completed, the charge on the capacitor starts decreasing, giving rise to current inthe circuit. Let q and i be the charge and current in the circuit at time t. According to Kirchhoff’s loop rule,

0dt

diL

C

q .

i = –(dq/dt) in the present case (as q decreases, i increases). Therefore the equation becomes :

0qLC

1

dt

qd2

2

.

This equation has the form 0xdt

xd 202

2

for a simple harmonic oscillator. The charge, therefore,

oscillates with a natural frequency LC

10 and varies sinusoidally with time as q = q

m cos (

0t + )

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where qm is the maximum value of q and is a phase constant. Since q = q

m at t = 0, we have cos = 1 or

= 0. Therefore, in the present case, q = qm cos(

0t).

The current

dt

dqi is given by i = i

m sin (

0t) where i

m =

0q

m.

Q. Show that in the free oscillations of an LC circuit, the sum of energies stored in the capacitor andthe inductor is constant in time. [NCERT Solved Example 7.11]

Solution : C2

q20 , This sum is constant in time as q

0 and C, both are time-independent. Note that it is equal to

the initial energy of the capacitor.

7.9 Transformers :

Q. (a) What is transformers ? On which principle does it work ?

(b) Explain with the help of a labelled diagram, the construction and working of a transformer(step of transformer or step down transformer). Also write down the assumptions.

(c) If the transformer is assumed to be 100% afficient then find the relation of primary andsecondary currents with number of terms in primary coil and secondary coil.

Solution : (a) For many purposes, it is necessary to change (or transform) an alternating voltage from oneto another of greater or smaller value. This is done with a device called transformer using the principle ofmutual induction.

(b) A transformer consists of two sets of coils, insulated from each other. They are wound on a soft-ironcore, either one on top of the other as in figure (a) or on separate limbs of the core as in figure (b). One ofthe coils called the primary coil has N

p turns. The other coil is called the secondary coil; it has N

s turns.

Often the primary coil is the input coil and the secondary coil is the output coil of the transformer.

When an alternating voltage is applied to the primary, the resulting current produces an alternatingmagnetic flux which links the secondary and induced an emf in it. The value of this emf depends on thenumber of turns in the secondary. We consider an ideal transformer in which the primary has negligibleresistance and all the flux in the core links both primary and secondary windings. Let be the flux in eachturn in the core at time t due to current in the primary when a voltage v

p is applied to it.

Then the induced emf or voltage s, in the secondary with N

s turns is

dt

dNss

.

The alternating flux also induces an emf, called back emf in the primary. This is dt

dNpp

.

But p = v

p. If this were not so, the primary current would be inifinite since the primary has zero resistance

(as assumed). If the secondary is an open circuit or the current taken from it is small, then to a goodapproximation

s = v

s, where v

s is the voltage across the secondary. Therefore, equations can be written as

dt

dNv ss

,

dt

dNv pp

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From above equations, we have p

s

p

s

N

N

v

v

Note that the above relation has been obtained using three assumptions : (i) the primary resistance andcurrent are small; (ii) the same flux links both the primary and the secondary as very little flux escapes fromthe core, and (iii) the secondary current is small.

(c) If the transformer is assumed to be 100% efficient (no energy losses), the power input is equal to thepower output, and since p = iv, i

pv

p = i

sv

s

Combining equations we have p

s

p

s

s

p

N

N

v

v

i

i .

Q. What is transformer ratio and hence define step up transformer and step down transformer ?What is the value of transformer ratio of step up transformer and step down transformer ?

Solution : We have,

ps

psp

p

ss I

N

NIandV

N

NV

Ns/N

p is defined as the transformer ratio.

If the secondary coil has a greater number of turns than the primary (Ns > N

p), the voltage is stepped up

(Vs > V

p). This type of arrangement is called a step-up transformer. However, in this arrangement, there is

less current in the secondary than in the primary (Np/N

s < 1 and I

s < I

p). The value of transformer ratio is

greater than one for step up transformer.

If the secondary coil has less turns than primary (Ns < N

p), we have a step-down tranformer. In this case,

Vs < V

p and I

s > I

p. That is, the voltage is stepped down, or reduced, and the current is increased. The value

of transformer ratio is less than one for step up transformer.

Q. What are different energy losses in actual transformer and how are they minimised ?

Solution : The different energy losses in actual transformers are following :

(i) Flux Leakage : There is always some flux leakage; that is, not all of the flux due to primary passesthrough the secondary due to poor design of the core or the air gaps in the core. It can be reduced bywinding the primary and secondary coils one over the other.

(ii) Resistance of the windings : The wire used for the windings has some resistance and so, energy is lostdue to heat produced in the wire (I2R). In high current, low voltage windings, these are minimised by usingthick wire.

(iii) Eddy currents : The alternating magnetic flux induces eddy currents in the iron core and causesheating. The effect is reduced by having a laminated core.

(iv) Hysteresis : The magnetisation of the core is repeatedly reversed by the alternating magnetic field. Theresulting expenditure of energy in the core appears as heat and is kept to a minimum by using a magneticmaterial which has a low hysteresis loss.

Q. How the transformer is used for the transmission and distribution of electrical energy ?

The large scale transmission and distribution of electrical energy over long distances is done with the use oftransformers. The voltage output of the generator is stepped-up (so that current is reduced andconsequently, the I2R loss is cut down). It then transmitted over long distances to an area sub-station nearthe consumers. There the voltage is stepped down. It is further stepped down at distributing sub-stationsand unility poles before a power supply of 240 V reaches our homes.

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NCERT EXERCISE

7.1 A 100 resistor is connected to a 220 V, 50 Hz a.c. supply.

(a) What is the rms value of current in the circuit ?

(b) What is the net power consumed over a full cycle ?

7.2 (a) The peak voltage of an a.c. supply is 300 V. What is the rms voltage ?

(b) The rms value of current in an a.c. circuit is 10 A. What is the peak current ?

7.3 A 44 mH inductor is connected to 220 V, 50 Hz a.c. supply. Determine the rms value of the current inthe circuit.

7.4 A 60 µF capacitor is connected to a 110 V, 60 Hz a.c. supply. Determine the rms value of the currentin the circuit.

7.5 In exercise 7.3 and 7.4, what is the net power absorbed by each circuit over a complete cycle. Explainyour answer.

7.6 Obtain the resonant frequency r of a series LCR circuit with L = 2.0 H, C = 32 µF and R = 10 .

What is the Q-value of this circuit ?

7.7 A charged 30 µF capacitor is connected to a 27 mH inductor. What is the angular frequency of freeoscillations of the circuit ?

7.8 Suppose the initial charge on the capacitor in Exercise 7.7 is 6 mC. What is the total energy stored inthe circuit initially ? What is the total energy at later time ?

7.9 A sereis LCR circuit with R = 20 , L = 1.5 H and C = 35 µF is connected to a variable-frequency200 V a.c. supply. When the frequency of the supply equals the natural frequency of the circuit, whatis the average power transferred to the circuit in one complete cycle ?

7.10 A radio can tune over the frequency range of a portion of MW broadcast band : (800 kHz to 1200kHz). If its LC circuit has an effective inductance of 200 µH, what must be the range of its variablecapacitor ?

7.11 Figure shows a series LCR circuit connected to a variable frequency 230 V source. L = 5.0 H,C = 80µF, R = 40 .

(a) Determine the source frequency which drives the circuit in resonance.

(b) Obtain the impendance of the circuit and the amplitude of current at the resonatingfrequency.

(c) Determine the rms potential drops across the three elements of the circuit. Show that thepotential drop across the LC combination is zero at the resonating frequency.

7.12 An LC circuit contains a 20 mH inductor and a 50 µF capacitor with an initial charge of 10 mC. Theresistance of the circuit is negligible. Let the instant, the circuit is closed be t = 0.

(a) What is the total energy stored initially ? Is it conserved during LC oscillations ?

(b) What is the natural frequency of the circuit ?

(c) At what time is the energy stored

(i) completely electrical (i.e., stored in the capacitor) ?

(ii) completely magnetic (i.e., stored in the inductor) ?

(d) At what times is the total energy shared equally between the inductor and the capacitor ?

(e) If a resistor is inserted in the circuit, how many energy is eventually dissipated as heat ?

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7.13 A coil of inductance 0.50 H and resistance 100 is connected to a 240 V, 50 Hz a.c. supply.

(a) What is the maximum current in the coil ?

(b) What is the time lag between the voltage maximum and the current maximum ?

7.14 Obtain the answers (a) to (b) exercise 7.13 if the circuit is connected to a high frequency supply(240 V, 10 kHz). Hence, explain the statement that a very high frequency, an inductor in a circuitnearly amounts to an open circuit. How does an inductor bechave in a d.c. circuit after the steadystate ?

7.15 A 100 µF capacitor in series with a 40 resistance is connected to a 110 V, 60 Hz supply.

(a) What is the maximum current in the circuit ?

(b) What is the time lag between the current maximum and the voltage maximum ?

7.16 Obtain the answers to (a) and (b) above in the problem if the circuit is connected to a 110 V, 12 kHzsupply ? Hence. explain the statement that a capacitor is a conductor at very high frequencies.Compare this behaviour with that of capacitor in a d.c. circuit after the steady state.

7.17 Keeping the source frequency equal to the resonating frequency of the series LCR circuit, if thethree elements L, C and R are arranged in parallel, show that the total current in the parallel LCRcircuit is minimum at this frequency. Obtain the current rms value in each branch of the circuit forthe elements and source specified in Exercise 7.11 for this frequency.

7.18 A circuit containing a 80 mH inductor and a 60 µF capacitor in series is connected to a 230 V, 50 Hzsupply. The resistance of the circuit is negligible.

(a) Obtain the current amplitude and rms values.

(b) Obtain the rms values of potential drops across each element.

(c) What is the average power transferred to the inductor ?

(d) What is the average power transferred to the capacitor ?

(e) What is the total average power absorbed by the circuit ?

7.19 Suppose the circuit in exercise 7.18 has a resistance of 15 . Obtain the average powertransferred to each element of the circuit and the total power absorbed.

[Answers : (7.1) (a) 2.2 A (b) 484 W (7.18) (a) 11.64 A (b) 436.8 V (c) 0 (d) /2 (e) 0 (7.19) 790.6 W]

7.20 A series LCR circuit with L = 0.12 H, C = 480 nF, R = 23 is connected to a 230 V variable frequencysupply.

(a) What is the source frequency for which current amplitude is maximum ? Obtain thismaximum value.

(b) What is the source frequency for which average power absorbed by the circuit ismaximum ? Obtain the value of this maximum power.

(c) For which frequencies of the source is the power transferred to the circuit half the powerat resonant frequency ? What is the current amplitude of these frequencies ?

(d) What is the Q-factor of the given circuit ?

7.21 Obtain the resonant frequency and Q-factor of a series LCR circuit with L = 3.0 H, C = 27 µF andR = 7.4 . It is desired to improve the sharpness of the resonance of the circuit by reducing its ‘fullwidth at half maximum’ by a factor of 2. Suggest a suitable way.

7.22 Answer the following questions :

(a) In any ac circuit, is the applied instantaneous voltage equal to the angebraic sum of theinstantaneous voltages across the series elements of the circuit ? Is the same true for rmsvoltage ?

(b) A capacitor is used in the primary instantaneous circuit of an induction coil.

(c) An applied voltage signal consists of a superposition of a dc voltage and an ac voltage ofhigh frequency. The circuit consists of an inductor and a capacitor in series. Show that thedc signal will appear across C and the ac signal across L.

(d) A choke coil in series with a lamp is connected to a dc line. The lamp is seen to shinebrightly. Insertion of an iron core in the choke causes no change in the lamp’s brightness.Predict the corresponding observations if the connection is to an ac line.

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(e) Why is choke coil needed in the use of fluorescent tubes with ac mains ? Why can we notuse an ordinary resistor instead of the choke coil ?

7.23 A power transmission line feeds input power at 2300 V to a step-down transformer with its primarywindings having 4000 turns. What should be the number of turns in the secondary in order to getoutput power at 230 V ?

7.24 At a hydroelectric power plant, the water pressure head is at a height of 300 m and the water flowavailable is 100 m3 s–1. If the turbine generator effieiency is 60%, estimate the electric poweravailable from the plant (g = 9.8 m s–2).

7.25 A small town with a demand of 800 kW of electric power at 220 V is situated 15 km away from anelectric plant generating power at 440 V. The resistance of the two wire line carrying power is 0.5 per km. The town gets power from the line through a 4000 - 220 V step down transformer at asub-station in the town.

(i) Estimate the line power loss in the form of heat.

(ii) How much power must the plant supply, assuming there is negligible power lossdue to leakage ?

(iii) Characterise the step up transformer at the plant.

7.26 Do the same exercise as above with the replacement of the earlier transformer by a 40,000-220 Vstep down transformer (Neglect, as before, leakage losses through this may not be a goodassumption any longer because of the very high voltage transmission involved). Hence, explain, whyhigh voltage transmission is preferred ?

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ANSWERS

7.1 (a) 2.20 A (b) 484 W

7.2 (a) 212.1 V (b) 14.1 A

7.3 15.9 A

7.4 2.49 A

7.5 Zero in each case

7.6 25

7.7 1.11 × 103s–1

7.8 0.6 J, same at later times

7.9 2 kW

7.10 The variable capacitor should have a range of about 88 pF to 198 pF.

7.11 (a) 50 rad s–1 (b) 40 , 8.1 A (c) 0

7.12 (a) 1.0 J. Yes, sum of the energies stored in L and C is conserved if R = 0

(b) = 103 rad s–1, v = 159 Hz

(c) q = q0 cos t

(i) Energy stored is completely electrical at ....,2

T3,T,

2

T,0t

(ii) Energy stored is completely magnetic (i.e., electrical energy is zero) at .....4

T5,

4

T3,

4

Tt wheree

ms3.6v

1T .

(d) At .....,,.........8

T5,

8

T3,

8

Tt because

2

q

4cosq

8

Tcosqq 0

00

. Therefore, electrical energy

=

C2

q

2

1

C2

q 20

2

which is half the total energy..

(e) R damps out the LC oscillations eventually. The whole of the initial energy (= 1.0 J) is eventuallydissipiated as heat.

7.13 (a) 1.82 A (b) 3.194 ms

7.14 3.98 × 10–6 s

7.15 (a) 3.24 A (b) 1.55 × 10–3s

7.16 (a) 3.89 A

7.17 In R branch, IRrms

= 5.75 A

In L branch, ILrms

= 0.92 A

In C branch, ICrms

= 0.92 A

7.18 (a) I0 = 11.6 A, I

rms = 8.24 A (b) V

Lrms = 207 V, V

Crms = 437 V (c) zero (d) zero (e) zero

7.19 Irms

= 7.26 A

Average power to R = I2rms

R = 791 W

Average power to L = Average power to C = 0

Total power absorbed = 791 W

7.20 (a) 14.14 A (b) 2300 W (c) 10 A (d) 21.7

7.21 0 = 111 rad s–1; Q = 45

To double Q without changing 0, reduce R to 3.7 .

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7.22 (a) Yes. The same is not true for rms voltage, because voltages across different element may not be inphase. See, for example, answer to Exercise 7.18.

(b) The high induced voltage, when the circuit is broken, is used to charge the capacitor, thusavoiding sparks, etc.

(c) For dc, impedance of L is negligible and of C very high (infinite), so the dc signal appears acrossC. For high frequency ac, impendence of L is high and that of C is low. So, the ac signal sppearsacross L.

(d) For a steady state dc, L has no effect, even if it is increased by an iron core. For ac, the lamp willshine dimly because of additional impedance of the choke. It will dim further when the iron core isinserted which increases the choke’s impendance.

(e) A choke coil reduces voltage across the tube without wasting power. A resistor would waste poweras heat.

7.23 400

7.24 176 MW

7.25 (i) 600 kW (ii) 1400 kW (iii) 7000 V

7.26 (i) 6 kW (ii) 806 kW (iii) 40,300 V

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ADDITIONAL QUESTIONS AND PROBLEMS

Q. The primary coil of an ideal step up transformer has 100 turns and the transformer ratio is also 100.The input voltage and the power are 220 V and 1100 W respectively. Calculate

(i) number of turns in the secondary (ii) the current in the primary

(iii) voltage across the secondary (iv) the current in the secondary

(v) power in the secondary

A. (i) 10000 (ii) 5 A (iii) 22000 V (iv) 0.05 A (v) 1100 W

Q. What is form factor of an a.c. ?

A. Form factor of an a.c. is defined as the ratio of its rms value and average value for half a cycle.

Form factor = 11.122I

2

2/I

)I(

I

m

m

2

1av

rms

Q. Can a moving coil galvanometer measure an a.c. ?

A. No, because moving coil galvanometer will measure the average value of a.c. for whole cycle, which iszero.

Q. What value of current/voltage do you measure with an a.c. ammeter/voltmeter ?

A. Rms value of current/voltage.

Q. Can an alternating current be used for electrolysis phenomenon ?

A. No, because then the electrodes will alternately behave as anode and cathode.

Q. What are the maximum and minimum values of power factor of an a.c. circuit ?

A. Maximum value of power factor is 1 and its minimum value if zero i.e., 0 cos 1.

Q. Why is a choke coil called wattless ?

A. A choke coil has a high inductance and almost zero resistance, hence, it does not consume power and calledwattless.

Q. A choke coil and a bulb are connected in series to a d.c. source. The bulb shines brightly. How doesits brightness change when an iron core in inserted in the choke coil ?

A. There will be no effect on the brightness of bulb because choke coil does not play any role in d.c.

Q. A choke coil and a bulb are connected in series to an a.c. source. The bulb shines brightly. How doesits brightness change when an iron core is inserted in the choke coil ?

A. On inserting an iron core, reactance of choke coil increases and hence brightness of the bulb decreases.

Q. A bulb and a capacitor are connected in series to an a.c. source of variable frequency. How will thebrightness of the bulb change on increasing the frequency of the a.c. source. Give reason.

A. Brightness of the bulb increases on increasing the frequency of a.c. source because with increase infrequency capacitive reactance X

C and consequently the impendance decreases.

Q. Can a transformer be used in d.c. circuits ? Why ?

A. No, because there is no change in magnetic flux in a d.c. circuit.

Q. Why is core of a transformer laminated one ?

A. In order to reduce eddy currents loss in the transformer.

Q. Can a choke coil be replaced by a capacitance of suitable capacitance ?

A. Yes, it can be replaced by a capacitor because even then the current is wattless.

Q. An ordinary moving coil ammeter used for d.c. cannot be used to measure an a.c., even if itsfrequency is low. Explain, why ?

A. An ordinary moving coil ammeter measures the average value of current. The average value of alternatingcurrent over a full cycle, even when its frequency is low, is zero. Hence, the ammeter cannot measure suchan a.c.

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Q. A radio frequency choke is air-cored coil whereas an audio frequency choke is iron cored. Givereason for this difference.

A. A choke must have high value of reactance and reactance of a coil is XL = L = L 2 v.

For a radio frequency choke v is extremely high and L need not be so high, hence, we design only an aircored solenoid coil. For an audio frequency choke v is small, hence for high value of X

L, the inductance L

should be high. Therefore, we take an iron cored coil so as to have high value of inductance L.

Q. Distinguish between reactance and impedance of an a.c. circuit.

A. (i) Reactance in an a.c. circuit is the opposite offered to the flow of a.c. by an inductor or a capacitor. On theother hand, impedance of a.c. circuit is the total opposition offered by LCR circuit for flow of a.c. throughit.

(ii) Inductive reactance XL = L and capacitive reactance

C

1XC , but impedance

2CL

2 )XX(RZ . It is self-evident that ZZ and consequently current flowing in the circuit

increases.

Q. An inductor L of reactance XL is connected in series with a bulb B to an a.c. source as shown.

Briefly explain how does the brightness of the bulb change, when (i) number of turns of the inductoris reduced ? and (ii) a capacitor of reactance X

C = X

L is included in series in the same circuit ?

Solution : If R be the resistance of the filament of bulb, then total impedance of the circuit will be

2L

2 XRZ and current Z

VI . (i) If number of turns of the inductor is reduced then its inductance L

and consequently the reactance XL and hence impedance Z decreases. As a result, current flowing through

the bulb increases and brightness of the bulb increases. (ii) When a capacitor of reactance XC = X

L is

included in series in the circuit, new impedence R)XX(R)XX(RZ 2LL

22CL

2 , i.e.,

the impedance has its minimum value. Consequently, the bulb glows with maximum brightness.

Q. An alternating voltage of frequency f is applied across a series LCR circuit. Let fr be the

resonance frequency for the circuit. Will the current in the circuit lag, lead or remain in phase withthe applied voltage when (i) f > f

r, (ii) f < f

r ? Explain your answer in each case.

Solution : When frequency of alternating voltage is more than the resonant frequency (i.e., f > fr), the

current will lag behind the voltage, because now XL (= L) will be more than

C

1XC and hence net

reactance is inductive in nature.

(ii) When frequency of alternating voltage is less than the resonant frequency (i.e., f < fr), the current will

lead the voltage, because now XL < X

C and net reactance is capacitive in nature.

Q. Mention the factors on which the resonant freqnency of a series LCR circuit depends. Plot a graphshowing variation of impedance of a series LCR circuit with the frequency of the applied a.c. source.

Solution : In the series LCR circuit the resonant frequency depends on the value of inductance L andcapacitance C present in the circuit.

A graph showing variation of impedance Z of a series LCR circuit with the freqency v of the applied a.c.source is shown below :

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Q. Can a.c., source be connected to a circuit and yet deliver no power to it ? If so under whatcircumstances ?

A. It is possible that an a.c. source be connected to a circuit and yet deliver no power to it. This is possible

where there is a phase difference of 2

between the applied voltage and the resulting current. As an

example, if we supply an alternating voltage to a pure inductor or a pure capacitor, the net power deliveredis zero.

Q. In India, domestic power supply is at 220 V, 50 Hz, while in USA, it is 110 V, 50 Hz. Give one advanceand one disadvantage of 220 V supply over 110 V supply.

A. Advantage : For a given power the current in line will be half at 220 V as compared to that at 110 V.Therefore, the line wire may be or lesser thickness thereby saving the cost.

Disadvantage : As peak voltage is 2202 = 311 V, better (thicker) insulation of the domestic wiring isneeded.

Q. A lamp is connected in series with a capacitor. Predict your observations for d.c. and a.c.connections. What happens in each case if the capacitance of the capacitor is reduced ?

A. (a) In d.c. connection, the capacitor will get charged and after charging no current flows in the circuit. So,the lamp will not glow. If we reduce the capacitance of the capacitor, it makes no difference and the lampdoes not glow at all.

(b) In a.c. connection, the capacitor offers a finite capacitive reactance

C

1XC and a current flows in

lamp-capacitor circuit. Thus, the lamp will glow on reducing the capacitance of the capacitor its reactanceincreases and consequently the glow of lamp decreases.

Q. Explain, why it is more economical to transmit electrical energy over long distance at high voltage.

A. If a given amount of power p is transmitted at a high voltage V, then the current flowing through line wire

V

PI and consequently energy loss due to joule heating 2

22

V

RPRI will be less. Moreover, as V is high

and I is comparatively less, we may ude a thin wire as transmission line and its cost will be less.

Q. Give two advantages and two disadvantages of a.c. over d.c.

A. Advantages of a.c. :

(i) The generation and transmission of a.c. is more economical than d.c.

(ii) The alternating voltage may be easily stepped up or down as per need by using suitable transformers.

Disadvantages of a.c. :

(i) The a.c. supply is more fatal and dangereous than d.c.

(ii) a.c. cannot be used to study chemical effect or magnetic effect of current.

Q. Calculate the current drawn by the primary of a transformer, which steps down 200 V to 20 V tooperate a device of resistance 20 . Assume the efficiency of the transformer to be 80%.

A. 0.125 A

Q. What is the value of current in the a.c. circuit containing R = 10 , C = 50 µF in series across 200 V,50 Hz a.c. source ?

A. 3.123 A

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Q. Derive an expression for the impedance of a coil in an a.c. circuit.

Q. An alternating source of emf is applied to an inductor and resistor in series. Investisgate the phaserelationship between the current and emf.What is the impendance of the circuit ?

Q. Derive the expression for the impendance of an a.c. circuit with a capacitor and a resistor in series.

Q. An alternating source of voltage is applied to a RC series circuit. Investigate the phase relationshipbetween current and voltage.

Q. For a given a.c. circuit, distinguish between resistance, reactance and impendance.

A. Distinguish between resistance, reactance and impedance :

Resistance :

(i) It is resistance offered to flow of an a.c. by the ohmic resistor and its value of R.

(ii) Resistance does not depend upon frequency of a.c. and has a constant value.

Reactance :

(i) It is the resistance offered to the flow of an a.c. by either an inductor L or a capacitor C. Inductive

reactance XL = L and capacitive reactance

C

1XC

(ii) Inductive reactance increases but capacitive reactance decreases with increase in a.c. frequency.

Impedance :

(i) It is the total opposition offered by the entire LCR circuit for flow of an a.c. Impedance Z

2CL

2 )XX(R 2

2

C

1LR

(ii) Impedance of an a.c. circuit depends upon the values of circuit elements L, C and R as well as thefrequency of a.c. supply.

Documents

ALTERNATING CURRENT - einsteinclasses.com Folder/(7)Alternating_C.pdf · the variation of current and voltage as a function of time in this circuit. Solution : Figure shows a resistor