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JEE-Physics
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ALTERNATING CURRENT
ALTERNATING CURRENT AND VOLTAGEVoltage or current is said to be alternating if it is change continously in magnitude and perodically in directionwith time. It can be represented by a sine curve or cosine curve
I = I0 sin t or I = I
0 cos t
where I = Instantaneous value of current at time t, I0 = Amplitude or peak value
= Angular frequency = 2
T
= 2f T = time period f = frequency
I0
–I0
T4
T2
34
TT
t
I
I as a sine function of t
I0
–I0
T4
T2
34T
T
t
I
I as a cosine function of tAMPLITUDE OF AC
The maximum value of current in either direction is called peak value or the amplitude of current. It is representedby I
0. Peak to peak value = 2I
0
PERIODIC TIMEThe time taken by alternating current to complete one cycle of variation is called periodic time or time periodof the current.
FREQUENCYThe number of cycle completed by an alternating current in one second is called the frequency of the current.UNIT : cycle/s ; (Hz)In India : f = 50 Hz , supply voltage = 220 volt In USA :f = 60 Hz ,supply voltage = 110 volt
CONDITION REQUIRED FOR CURRENT/ VOLTAGE TO BE ALTERNATING• Amplitude is constant • Alternate half cycle is positive and half negative
The alternating current continuously varies in magnitude and periodically reverses its direction.
I sinosudial AC I
t
+
–
triangular AC
+
–
I
t
square wave AC saw tooth waveI
t
mixture of AC and DC
I0
t
Not AC (direction not change)
I0 I0
t t
Not AC (not periodic)
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AVERAGE VALUE OR MEAN VALUE
The mean value of A.C over any half cycle (either positive or negative) is that value of DC which would send
same amount of charge through a circuit as is sent by the AC through same circuit in the same time.
average value of current for half cycle < I > =
T / 2
0
T / 2
0
Idt
dt
Average value of I = I0 sin t over the positive half cycle :
< sin > = < sin 2 >=0 < cos >= < cos 2 >= 0< sin cos > = 0 < sin > = < cos >=
2 2 12
T
20
0av T
2
0
I sin t dtI
dt
=
T0 2
0
2 Icos t
T
02 I
• For symmetric AC, average value over full cycle = 0,
Average value of sinusoidal AC
Full cycle (+ve) half cycle (–ve) half cycle
0 2I0
–2I0
As the average value of AC over a complete cycle is zero, it is always defined over a half cycle which must be either positive or negative
MAXIMUM VALUE
• I = a sin IMax.
= a • I = a + b sin IMax.
= a + b ( if a and b > 0)
• I = a sin + b cos IMax.
= 2 2a b • I = a sin2 IMax.
= a (a > 0)
ROOT MEAN SQUARE (rms) VALUE
It is value of DC which would produce same heat in given resistance in given time as is done by the alternating
current when passed through the same resistance for the same time.
T2
0rms T
0
I dtI
dt
rms value = Virtual value = Apparent value
rms value of I = I0 sin t :
T2
00
rms T
0
(I sin t) dtI
dt
=
2T
20
0
Isin t dt
T = 0
T
0
1 1 cos2 tI dt
T 2
T
0
0
1 t sin 2 tI
T 2 2 2
= 0I
2
• If nothing is mentioned then values printed in a.c circuit on electrical appliances, any given or unknown
values, reading of AC meters are assumed to be RMS.
Cur r en t Ave r ag e P e a k R M S Angular fequency
I1 = I
0 sin t 0 I
00I
2
I2
= I0sin t cos t =
0I sin 2 t2
00I
2
0I
2 22
I3
= I0sint + I
0cost 0
02 I I0
• For above varieties of current rms = Peak value
2
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Examp l e
If I = 2 t ampere then calculate average and rms values over t = 2 to 4 s
So lu t i on
< I > =
4
342
2 2
4 42
2
2 t.dt(t )4 2
8 2 23 3(t)
dt
and Irms
=
4 42
2 2
4
2
(2 t ) dt 4t dt
2dt
=
42
2
t2 2 3 A
2
Examp l e
Find the time required for 50Hz alternating current to change its value from zero to rms value.
So lu t i on
I = I0 sin t
00
II sin t
2
1sin t
2 t
4
2t
T 4
Tt
8
12.5
8 50
ms
Examp l e
If E = 20 sin (100 t) volt then calculate value of E at t = 1
600s
So lu t i on
At t =1
600s E = 20 Sin
1100
600
= 20 sin
6
= 20 × 1
2 = 10V
Examp l e
A periodic voltage wave form has been shown in fig.
Determine. (a) Frequency of the wave form.
(b) Average value.
So lu t i on
(a) After 100 ms wave is repeated so time period is
T = 100 ms. f = 1
T = 10 Hz
(b) Average value = Area/time period = ( / )
( )
1 2 100 10
100
= 5 volt
Ex amp l e
Explain why A.C. is more dangerous than D.C. ?
So lu t i on
There are two reasons for it :
A.C. attracts while D.C. repels.
A.C. gives a huge and sudden shock.O
-E
E
t
+E = 311.080
-E = -311.080
0.01 sec
for 220 V ac Vrms
= 220 V
Hence, V0 =
rms2.V = 12.414 × 220 = 311.08 V
Voltage change from +V0 (positive peak) to –V
0 (negative peak) = 311.08 – (–311.08) = 622.16 V
This change takes place in half cycle i.e., in 1
100 s (for a 50 Hz A.C.)
A shock of 622.16 within 0.01 s is huge and sudden, hence fatal.
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Examp l e
If a direct current of value a ampere is superimposed onan alternating current I = b sint flowing through a wire,what is the
effective value of the resulting current in the circuit ?I
t
ACb
=?It
DCa+
So lu t i on
As current at any instant in the circuit will be,
I = IDC
+ IAC
= a + b sint
T T
2 2eff
0 0
1 1I I dt (a b sin t) dt
T T
T
2 2 2
0
1(a 2ab sin t b sin t)dt
T
but asT
0
1sin tdt 0
T and
T
2
0
1 1sin tdt
T 2
2 2eff
1I a b
2
SOME IMPORTANT WAVE FORMS AND THEIR RMS AND AVER AGE VALUE
Nature of Wave– fo r m RMS Value Average or mean
wave form Va l u e
Sinusoidal0
+
– 2
0I
202I
= 0.707 I0
= 0.637 I0
Half wave
0 2
0I
20I
rectifired = 0.5 I0
= 0.318 I0
Full wave
0 2
0I
2
02I
rectifired = 0.707 I0
0.637 I0
Square or+
–
I0
I0
Rectangular
Saw Tooth wave2
00I
30I
2
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MEASUREMENT OF A.C.Alternating current and voltages are measured by a.c. ammeter and a.c. voltmeter respectively. Workingof these instruments is based on heating effect of current, hence they are also called hot wire instruments.
Te r m s D.C. meter A.C. meter
Name moving coil hot wireBased on magnetic effect of current heating effect of currentReads average value r.m.s. valueIf used in A.C. circuit then they reads zero A.C. or D.C. then meter works
average value of A.C. = zero properly as it measures rms value
Deflection deflection current deflection heat (linear)
rms (non linear)
Scale Uniform Seperation Non uniform sepration = Number - 1 2 3 4 5 - 1 2 3 4 5of divisions - 1 2 3 4 5 - 1 4 9 16 25
• D.C meter in AC circuit reads zero because < AC > = 0 ( for complete cycle)
• AC meter works in both AC and DC
PHASE AND PHASE DIFFERENCE
( a ) P h a s e
I = I0 sin (t + )
Initial phase = (it does not change with time)
Instantaneous phase = t + (it changes with time)
• Phase decides both value and sign. UNIT: radian
( b ) Phase di f ference
Voltage V = V0 sin ( t +
1) Current I I
0 sin (t +
2)
• Phase difference of I w.r.t. V = 2 –
1
• Phase difference of V w.r.t. I = 1 –
2
LAGGING AND LEADING CONCEPT
( a ) V leads I or I lags V It means, V reach maximum before II=I0 sin ( t )
tV,I
V=Vsin t0
Let if V = V0 sin t then I = I
0 sin (t –
and if V = V0 sin (t+ ) then I = I
0 sin t
( b ) V lags I or I leads V It means V reach maximum after II=I0 sin ( t + )
tV,I
V=V sin t0
Let if V = V0 sin t then I = I
0 sin (t +
and if V = V0 sin (t – ) then I = I
0 sin t
PHASOR AND PHASOR DIAGRAMA diagram representing alternating current and voltage (of same frequency) as vectors (phasor) with the phase
angle between them is called phasor diagram.
t
V
V0I0
I
V0
I0
fig (a) fig (b)
Y
X
Y
X
Let V = V0 sin t and I = I
0 sin (t +)
In figure (a) two arrows represents phasors. The length of phasorsrepresents the maximum value of quantity. The projection of a phasoron y-axis represents the instantaneous value of quantity
ADVANTAGES OF AC• A.C. is cheaper than D.C• It can be easily converted into D.C. (by rectifier)• It can be controlled easily (choke coil)• It can be transmitted over long distance at negligible power loss.• It can be stepped up or stepped down with the help of transformer.
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GOLDEN KEY POINTS • AC can't be used in
(a) Charging of battery or capacitor (as its average value = 0)(b) Electrolysis and electroplating (Due to large inertia, ions can not follow frequency of A.C)
• The rate of change of A.C. Minimum, at that instant when they are near their peak values
Maximum, at that instant when they change their direction. • For alternating current I
0 > I
rms > I
av.
• Average value over half cycle is zero if one quarter is positive +
–and the other quarter is negative.
• Average value of symmetrical AC for a cycle is zero thats why average potential difference on anyelement in A.C circuit is zero.
• The instrument based on heating effect of current are works on both A.C and D.C supply and also providessame heating for same value of A.C (rms) and D.C. that's why a bulb bright equally in D.C. and A.C. of samevalue.
• If the frequency of AC is f then it becomes zero 2f times in one second and the direction of current changes2f times in one second. Also it become maximum 2f times in one second.
Examp l e
The Equation of current in AC circuit is I = 4sin 100 t3
A. Calculate.
(i) RMS Value (ii) Peak Value (iii) Frequency (iv) Initial phase (v) Current at t = 0So lu t i on
(i) Irms
= 0I
2 =
4
2= 2 2 A (ii) Peak value I
0 = 4A
(iii) = 100 rad/s frequency f = 100
2
= 50 Hz
(iv) Initial phase = 3
(v) At t = 0, I = 4sin 100 0
3
= 4 ×
3
2= 2 3 A
Examp l e
If I = I0 sin t, E = E
0 cos t
3
. Calculate phase difference between E and I
So lu t i on
I = I0 sin t and E = E
0sin t
2 3
phase difference =
2
+
3
=
5
6
Examp l eIf E = 500 sin (100 t) volt then calculate time taken to reach from zero to maximum.
So lu t i on
= 100 T = 2
100
=
1
50s, time taken to reach from zero to maximum =
T
4=
1
200s
Examp l e
If Phase Difference between E and I is 4
and f = 50 Hz then calculate time difference.
So lu t i on
2 TPhase difference time difference
2 T
Time difference =
T
2 ×
4
=
T
8 =
1
50 8 = 2.5ms
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Examp l eShow that average heat produced during a cycle of AC is same as produced by DC with I = I
rms.
So lu t i on
For AC, I = I0 sint, the instantaneous value of heat produced (per second) in a resistance R is,
H = I2R = I0
2sin2t × R the average value of heat produced during a cycle is :
T T2 20
0 0av T T
0 0
H dt (I sin t R )dtH
dt dt
20
1I R
2
T2 2 20 0
0
1I sin t dt I T
2
2
20av rms
IH R I R
2
.....(i)
However, in case of DC, HDC
= I2 R...(ii) I = Irms
so from equation (i) and (ii) HDC
= Hav
AC produces same heating effects as DC of value I = Irms
. This is also why AC instruments which are basedon heating effect of current give rms value.
DIFFERENT TYPES OF AC CIRCUITS
In order to study the behaviour of A.C. circuits we classify them into two categories :
(a) Simple circuits containing only one basic element i.e. resistor (R) or inductor (L) or capacitor (C) only.
(b) Complicated circuit containing any two of the three circuit elements R, L and C or all of three elements.
AC CIRCUIT CONTAINING PURE RESISTANCE
Let at any instant t the current in the circuit =
s
R
V = RI
E = E sin t0
Potential difference across the resistance = R.
with the help of kirchoff’s circuital law E – R = 0
E0 sin t = R
00
Esin t sin t
R (
00
E
R
= peak or maximum value of current)
Alternating current developed in a pure resistance is also of sinusoidal
tO
I=I sin t0 E and I
/2
3 /2
2X
E = E sint0
nature. In an a.c. circuits containing pure resistance, the voltage and
current are in the same phase. The vector or phasor diagram which
represents the phase relationship between alternating current and al-
ternating e.m.f. as shown in figure.
In the a.c. circuit having R only, as current and voltage are in the same
phase, hence in fig. both phasors E0 and
0 are in the same direction,
making an angle t with OX. Their projections on Y-axis represent the
instantaneous values of alternating current and voltage.
i.e. = 0 sint and E = E
0 sint.
t
tPE0
Y
E
I
OX
I0
Since 0
0
E
R , hence
0 0E
2 R 2
rmsrms
E
R
AC CIRCUIT CONTAINING PURE INDUCTANCE
A circuit containing a pure inductance L (having zero ohmic resistance)
connected with a source of alternating emf. Let the alternating e.m.f. E = E0 sin t
s
L
E = E sin0 tWhen a.c. flows through the circuit, emf induced across inductance
dL
dt
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Negative s ign indicates that induced emf acts in opposi te direction to that of applied emf.Because there is no other circuit element present in the circuit other then inductance so with the help of
Kirchoff’s circuital law d
E L 0dt
d
E Ldt
so we get 0E
sin tL 2
Maximum current 0 00
E E1
L L
, Hence,
0 sin t2
In a pure inductive circuit current always lags behind the emf by 2
.
Ean
dI E = E sint0
/2 3 /2 t2
I = I sin ( t - /2)0
O
or alternating emf leads the a. c. by a phase angle of 2
.
Expression 00
E
L
resembles the expression
ER
.
This non-resistive opposition to the flow of A.C. in a circuit is called theinductive reactance (X
L) of the circiut.
XL = L = 2 f L where f = frequency of A.C. t
E0
X
PE
Y
O/2 – t
Q0
E = E sin t0
I=I cos t0
Unit of XL : ohm
L) = Unit of L × Unit of = henry × sec–1
1Volt Voltsec ohm
Ampere / sec Ampere
Inductive reactance XL f
Higher the frequency of A.C. higher is the inductive reactance offered by an inductorin an A.C. circuit.
For d.c. c ircui t, f = 0 XL = L = 2 f L = 0
Hence, inductor offers no opposition to the flow of d.c. whereas
a resistive path to a.c. f
XL
AC CIRCUIT CONTAINING PURE CAPACITANCE
A circuit containing an ideal capacitor of capacitance C connected with a source ofC
s
E = E sin t0
alternating emf as shown in fig. The alternating e.m.f. in the circuit E = E0 sin t
When alternating e.m.f. is applied across the capacitor a similarly varying alternating current
flows in the circuit.
The two plates of the capacitor become alternately positively and negatively charged
and the magnitude of the charge on the plates of the capacitor varies sinusoidally
with time. Also the electric field between the plates of the capacitor varies sinusoidally
with time.Let at any instant t charge on the capacitor = q
Instantaneous potential difference across the capacitor E = q
C
Ean
d E = E sint0
/2 3 /2 t2
I= I sin (t + /2)0
O
q = C E q = CE0 sin t
The instantaneous value of current 0 0
dq dCE sin t CE cos t
dt dt
0Esin t
1 / C 2
0 sin t
2
where I
0 = CV
0
t90
°
I
E0
PYE
Q
I0
OX
In a pure capacitive circuit, the current always leads the e.m.f. by a phase
angle of . The alternating emf lags behinds the alternating current by aphase angle of .
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IMPORTANT POINTS
E
is the resistance R when both E and are in phase, in present case they XC
f
differ in phase by 2
, hence
1
C is not the resistance of the capacitor,,
the capacitor offer opposition to the flow of A.C. This non-resistive opposition
to the flow of A.C. in a pure capacitive circuit is known as capacitive reactance XC. C
1 1X
C 2 fC
Unit of X
C : ohm
Capacitive reactance XC is inversely proportional to frequence of A.C. X
C decreases as the frequency increases.
This is because with an increase in frequency, the capacitor charges and discharges rapidly following the flowof current.
For d.c. c ircuit f = 0 C
1X
2 fC
but has a very small value for a.c.
This shows that capacitor blocks the flow of d.c. but provides an easy path for a.c.
INDIVIDUAL COMPONENTS (R or L or C)
T E R M R L C
Circuit
R L C
Supply Voltage V = V0sin t V = V
0 sin t V = V
0 sin t
Current I = I0 sin t I = I
0 sin (t –
2
) I = I
0 sin (t+
2
)
Peak Current0
0
VI
R 0
0
VI
L
00
VI
1 C
=V
0C
Impedance ( )0
0
VR
I 0
L
0
VL X
I 0
C
0
V 1X
I C
0 rms
0 rms
V VZ
I I R = Resistance X
L=Inductive reactance. X
C=Capacitive reactance.
Phase difference zero (in same phase)2
(V leads I)
2
(V lags I)
Phasor diagramV
I
V
I
V
I
Variation of Z with f
R
f
XL
f
X fL
V
I
X C 1f
R does not depend on G,S
L,S
CG=1/R = conductance. S
L = 1/X
LS
C = 1/X
C
(mho, seiman) Inductive susceptance Capacitive susceptance Behaviour of device Same in L passes DC easily C - blocks DC in D.C. and A.C A C and D C (because X
L = 0) while (because X
C =) while
gives a high impedance provides an easy pathfor the A.C. of high for the A.C. of high
frequency (XL f) frequency C
1X
f
Ohm's law V
R = IR V
L = IX
LV
C = IX
C
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GOLDEN KEY POINTS
• Phase diference between capacitive and inductive reactance is
• Inductor called Low pass filter because it allow to pass low frequency signal.
• Capacitor is called high pass filter because it allow to pass high frequency signal.
Ex amp l e
What is the inductive reactance of a coil if the current through it is 20 mA and voltage across it is 100 V.
So lu t i on
VL = IX
LX
L =
L
3
V 100
I 20 10
= 5 k
Examp l e
The reactance of capacitor is 20 ohm. What does it mean?
What will be its reactance if frequency of AC is doubled?
What will be its, reactance when connected in DC circuit? What is its consequence?
So lu t i on
The reactance of capacitor is 20 ohm. It means that the hindrance offered by it to the flow of AC at a specific
frequency is equivalent to a resistance of 20 ohm. The reactance of capacitance C
1 1X
C 2 fC
Therefore by doubling frequency, the reactance is halved i.e., it becomes 10 ohm. In DC circuit f = 0. Thereforereactance of capacitor = (infinite). Hence the capacitor can not be used to control DC.
Examp l e
A capacitor of 50 pF is connected to an a.c. source of frequency 1kHz Calculate its reactance.
So lu t i on
XC =
1
C=
3 12
1
2 10 50 10 =
710
Examp l e
In given circuit applied voltage V = 50 2 sin 100t volt and
L
A
Vammeter reading is 2A then calculate value of L
So lu t i on
Vrms
= Irms
XL
Reading of ammeter = Irms
XL = rms
rms
V
I= 0
rms
V
2 I =
50 2
2 2 = 25 L =
LX
=
25
100=
1
4 H
Examp l e
A 50 W, 100 V lamp is to be connected to an AC mains of 200 V, 50 Hz. What capacitance is essential tobe put in series with the lamp ?
So lu t i on
resistance of the lamp 2 2sV (100)
RW 50
= 200 and the maximum curent V 100 1
I AR 200 2
when the lamp is put in series with a capacitance and run at 200 V AC, from V = IZ
V 200Z 400
1I
2
Now as in case of C–R circuit 2
2
1Z R
( C)
,
2 2
2
1R (400)
( C )
4 2 4
2
116 10 (200) 12 10
( C)
2112 10
C
2
1C F
100 12 10
100F 9.2 F
12
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RESISTANCE AND INDUCTANCE IN SERIES (R-L CIRCUIT)A circuit containing a series combination of a resistance R and an inductance L, connected with a sourceof alternating e.m.f. E
as shown in figure.
s
R L
E = E sin t0
PHASOR DIAGRAM FOR L-R CIRCUITLet in a L-R series circuit, applied alternating emf is E = E
0 sint. As R
and L are joined in series, hence current flowing through both will be sameat each instant. Let be the current in the circuit at any instant and V
L and
VR the potential differences across L and R respectively at that instant.
Then VL = X
L and V
R = R
Now, VR is in phase with the current while V
L leads the current by
2
.
R
X
E
QY
VL
PVRSo VR and V
L are mutually perpendicular (Note : E V
R + V
L)
The vector OP represents VR
(which is in phase with ), while OQ represents VL (which leads by 90°).
The resultant of VR and V
L = the magnitude of vector OR 2 2
R LE V V
Thus E² = VR² + V
L² = ² (R² + X
L²)
2 2L
E
R X
The phasor diagram shown in fig. also shows that in L-R circuit the appliedXL
R
ZL
emf E leads the current or conversely the current lags behind the e.m.f.
E. by a phase angle L L L
R
V X X Ltan
V R R R
1 L
tanR
Inductive Impedance ZL :
In L-R circuit the maximum value of current 00
2 2 2
E
R L
Here 2 2 2R L represents the effective
opposition offered by L-R circuit to the flow of a.c. through it. It is known as impedance of L-R circuit and
is represented by ZL. 2 2 2
LZ R L 2 2R (2 fL ) The reciprocal of impedance is called admittance
L2 2 2
L
1 1Y
Z R L
RESISTANCE AND CAPACITOR IN SERIES (R-C CIRCUIT)
s
RC
E = E sin t0
A circuit containing a series combination of a resistance R and a capacitorC, connected with a source of e.m.f. of peak value E
0 as shown in fig.
PHASOR DIAGRAM FOR R-C CIRCUITCurrent through both the resistance and capacitor will be same at every instantand the instantaneous potential differences across C and R are
VC = X
C and V
R = R
where XC = capacitive reactance and = instantaneous current.
Now, VR is in phase with , while V
C lags behind by 90°.
OPVR
X
SQ
VC
E (applied emf)
E=
V+ V2
C
R
2
The phasor diagram is shown in fig.The vector OP represents V
R (which is in phase with )
and the vector OQ represents VC (which lags behind by
2
).
The vector OS represents the resultant of VR and
VC = the applied e.m.f. E.
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Hence VR
2 + VC
2 = E 2 2 2R CE V V O
PR
X
SQXC
Z=R
+X2
C2
E² = ² (R² + XC²)
2 2C
E
R X
The term 2 2C(R X ) represents the effective resistance of the R-C
circuit and called the capacitive impedance ZC of the circuit.
Hence, in C-R circuit2
2 2 2C C
1Z R X R
C
Capacit ive Impedance ZC
:
In R-C circuit the term 2 2CR X effective opposition offered by R-C circuit to the flow of a.c. through
it. It is known as impedance of R-C circuit and is represented by ZC
The phasor diagram also shows that in R-C circuit the applied e.m.f. lags behind the current (or the current leads the emf E) by a phase angle given by
C C
R
V X 1 / C 1tan
V R R CR
,
CX 1tan
R CR
1 1tan
CR
COMBINATION OF COMPONENTS (R-L or R-C or L-C)
T E R M R-L R-C L-C
Circuit
R L RC C
L
I is same in R & L I is same in R & C I is same in L & C
Phasor diagram
VL
VR
I
VR
I
VC
VI
VC
VL
V2 = 2 2R LV V V2 = 2 2
R CV V V = VL – V
C (V
L>V
C)
V = VC – V
L (V
C>V
L)
Phase difference V leads I ( = 0 to 2
) V lags I ( = –
2
to 0 ) V lags I ( =
2
,if X
C>X
L)
in between V & I V leads I ( =2
,if X
L>X
C) Impedance 2 2
LZ R X
Z = 22CR X Z = L cX X
Variation of Z as f,Z as f, as f, Z first then
with f
Z
f
R
Z
f
R
Z
f
At very low f Z ~ R (XL 0) Z ~ X
CZ ~ X
C
At very high f Z ~ XL
Z ~ R (XC 0) Z ~ X
L
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Examp l e
Calculate the impedance of the circuit shown in the figure.30 40
So lu t i on
Z = 2 2cR (X ) = 2 2(30) (40) = 2500 = 50
Examp l e
If XL = 50and X
C = 40 Calculate effective value of current in given circuit.
So lu t i on
Z = XL – X
C = 10
XL =50 X=40C
V=40sin 100 volt
=
0V
Z=
40
10 = 4 A I
rms =
4
2 = 2 2 A
Examp l e
In given circuit calculate, voltage across inductor
V=100 2 sin voltt
V=60R VL=?
So lu t i on
V V VR L2 2 2 V
L2 = V2 – V
R2
VL
2 2RV V = 2 2(100) (60) = 6400 = 80 V
Examp l e
In given circuit find out (i) impedance of circuit (ii) current in circuit 6 8
V=20sin voltt
So lu t i on
(i) Z = 2 2 2 2CR X (6) (8) = 10
(ii) V = IZ 0V 20I 2A
Z 10 so I
rms =
2
2 = 2 A
Examp l e
When 10V, DC is applied across a coil current through it is 2.5 A, if 10V, 50 Hz A.C. is applied current reducesto 2 A. Calculate reactance of the coil.
So lu t i on
For 10 V D.C. V = IR Resistance of coil R = 10
42.5
For 10 V A.C. V = I Z V 20
Z 5I 10
2 2LZ R X 5 2 2
LR X 25 X L2 = 52 – 42 X
L= 3
Examp l e
When an alternating voltage of 220V is applied across a device X, a current of 0.5 A flows through the
circuit and is in phase with the applied voltage. When the same voltage is applied across another device
Y, the same current again flows through the circuit but it leads the applied voltage by /2 radians.
(a) Name the devices X and Y.
(b) Calculate the current flowing in the circuit when same voltage is applied across the seriescombination of X and Y.
So lu t i on
(a) X is resistor and Y is a capacitor
(b) Since the current in the two devices is the same (0.5A at 220 volt)
When R and C are in series across the same voltage then
R = XC = 220
0 5. = 440 Irms =
V
R X
rms
C
2 2 =
220
440 4402 2
( ) ( ) =
220
440 2 = 0.35A
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INDUCTANCE, CAPACITANCE AND RESISTANCE IN SERIES
(L-C-R SERIES CIRCUIT)
A circuit containing a series combination of an resistance R, a coil of inductance
A.C. source
RL C
E = E sin t0
L and a capacitor of capacitance C, connected with a source of alternating e.m.f.
of peak value of E0, as shown in fig.
PHASOR DIAGRAM FOR SERIES L-C-R CIRCUIT
Let in series LCR circuit applied alternating emf is E = E0 sin t.
As L,C and R are joined in series, therefore, current at any instant
through the three elements has the same amplitude and phase.
However voltage across each element bears a di f ferent
phase relationship with the current.
Let at any instant of time t the current in the circuit is
Let at this time t the potential differences across L, C, and R
Q
Y
XVR P
VL
O
VC
VL = X
L, V
C = X
C and V
R =
R
Now, VR is in phase with current but V
L leads by 90°
While VC legs behind by 90°.
The vector OP represents VR (which is in phase with ) the vector
OQ represent VL (which leads by 90°)
and the vector OS represents VC (which legs behind by 90°)
VL and V
C are opposite to each other.
XVR P
E (app
lied em
f)
K
Q
Y
VL
( VL-V
)C
O
T
VC
If VL > V
C (as shown in figure) the their resultant will be (V
L – V
C) which
is represented by OT.
Finally, the vector OK represents the resultant of VR and (V
L – V
C), that
is, the resultant of all the three = applied e.m.f.
Thus 2 2R L CE V (V V ) = I 2 2
L CR (X X ) 2 2L C
E
R (X X )
Impedance 2 2L CZ R (X X ) =
2
2 1R L
C
XR
P
Z
K
Q
Y
XL
( XL-X
)C
O
T
XC
The phasor diagram also shown that in LCR circuit the applied e.m.f.
leads the current by a phase angle tan = L CX X
R
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SERIES LCR AND PAR ALLEL LCR COMBINATION
SERIES L-C-R CIRCUIT PARALLEL L-C-R CIRCUIT
1 . Circui t diagram
R L C
R
L
C
V same for R,L and C
I same for R, L & C V same for R, L & C
2 . Phasor diagram
I
VC
VL
VR
V
IL
IC
IR
(i) If VL > V
C then (i) if I
C > I
L then
I
VR
VL–VC V
IR
I -IC L
(ii) If VC > V
L then (ii) if I
L > I
C then
IVRV –VC LVIRI -IL C
(iii) V = 2 2R L CV (V V ) (iii) I = 2 2
R L CI (I I )
Impedance Z = 2 2L CR (X X ) Admittance Y = 2 2
L CG (S S )
tan = L CX X
R
= L C
R
V V
V
tan =
L CS S
G
=
L C
R
I I
I
(iv) Impedance triangle (iv) Admittance triangle
Z
R
X=X–XL C
S -SL C
G
Y
GOLDEN KEY POINTS
• Se r i e s Pa r a l l e l
(a) if XL > X
C then V leads I, (positive) (a) if S
L > S
C (X
L< X
C) then V leads I, (positive)
circuit nature inductive circuit nature inductive
(b) if XC> X
L then V lags I, (negative) (b) if S
C> S
L ( X
C < X
L) then V lags I, (negative)
circuit nature capacitive circuit nature capacitive
• In A.C. circuit voltage for L or C may be greater than source voltage or current but it happens
only when circuit contains L and C both and on R it never greater than source voltage or current.
• In parallel A.C.circuit phase difference between IL and I
C is
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Examp l e
Find out the impedance of given circuit. 49
6
So lu t i on
2 2L CZ R (X X ) 2 24 (9 6) = 2 24 3 25 5
( XL > X
C Inductive)
Examp l e
Find out impedance of given circuit.3
6
6
So lu t i on
Y2 = G2 + (SL – S
C)2
21 1 1
36 6 3
Y = 2
6 Z =
6
2 (capacitive, because X
L > X
C)
Ex amp l e
Find out reading of A C ammeter and also calculate the potential difference across, resistance and capacitor.
So lu t i on
2 2L CZ R (X X ) 10 2 0
0
V 100 10I A
Z 10 2 2
1020 10
A
E=100sin t100 t volt
ammeter reads RMS value, so its reading = 10
2 2 = 5A
so VR = 5 × 10 = 50 V and V
C = 5 × 10 = 50 V
Examp l e
In LCR circuit with an AC source R = 300 , C = 20 F, L = 1.0 H, Erms
= 50V and f = 50/Hz. Find RMS
current in the circuit.
So lu t i on
rms rmsrms
2 2
2 2
6
E E 50I
Z 1 50 1R L 300 2 1
50C20 10 2
rms
232
50 50 1I 0.1A
10100 9 1610(300) 100
2
Examp l e
Calculate impedance of the given circuit :
(i) 120Z
Z=?
240
(ii)
12V4V5V
I=2A(iii)
0.2F 1H3
V=100sint volt
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So lu t i on
(i) It is parallel circuit so Y is evaluated
L C
1 1 1Y S S
120 240 240 Z = 240 (inductive)
(ii) Vs2 = 52 + 122 = 169 V
s = 13 volt therefore
sV 13Z 6.5
I 2
(iii) R = 3, XL = L = 1 as ( = 1)
XC =
1
C =
1
(0.2).1 = 5 so Z2 = R2 + (X
L – X
C)2 = 32 + (1 – 5)2 = 25 Z = 5
RESONANCE
A circuit is said to be resonant when the natural frequency of circuit is equal to frequency of the applied voltage.
For resonance both L and C must be present in circuit.
There are two types of resonance : (i) Series Resonance (ii) Parallel Resonance
SERIES RESONANCE
( a ) At Resonance
(i) XL = X
C(ii) V
L = V
C(iii) = 0 (V and I in same phase)
(iv) Zmin
= R (impedance minimum) (v) Imax
= V
R (current maximum)
( b ) Resonance frequency
XL = X
C
rL = r
1
C 2r
1
LC r
1
LC
r
1f
2 LC
( c ) Var iat ion of Z w ith f
f
Z
R
fr
(i) If f< fr then X
L < X
Ccircuit nature capacitive, (negative)
(ii) At f = fr then X
L = X
Ccircuit nature, Resistive, = zero
(iii) If f > fr then X
L > X
Ccircuit nature is inductive, (positive)
Variat ion of I with f as f increase, Z first decreases then increase
( d )
Imax=VR
Imax
2
f1 fr f2 f
If
Imax
as f increase, I first increase then decreases
• At resonance impedance of the series resonant circuit is minimum so it is called 'acceptor circuit' as it most readilyaccepts that current out of many currents whose frequency is equal to its natural frequency. In radio or TV tuningwe receive the desired station by making the frequency of the circuit equal to that of the desired station.
Half power frequencies
The frequencies at which, power become half of its maximum value called half power frequencies
Band width = f = f2 – f
1
Quality factor Q Q-factor of AC circuit basically gives an idea about stored energy & lost energy.
maximum energy stored per cycle
Q 2maximum energy loss per cycle
(i) It represents the sharpness of resonance. (ii) It is unit less and dimension less quantity
(iii) Q = L r(X )
R =
C r(X )
R =
r2 f L
R =
1
R
L
C =
rf
f =
rf
band width
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Magn i f i c a t i on
At resonance VL or V
C = QE (where E = supplied voltage) R1
R2
R3
R < R < R1 2 3
ffr
I
So at resonance Magnification factor = Q-factor
Sha rp ne s s
Sharpness Quality factor Magnification factor
R decrease Q increases Sharpness increases
PARALLEL RESONANCE
L R
C( a ) At resonance
(i) SL = S
C(ii) I
L = I
C (iii) = 0
(iv) Zmax
= R (impedance maximum)
(v) Imin
= V
R (current minimum)
( b ) Resonant frequency fr =
1
2 LC
( c )
fr
ZVariation of Z with f as f increases , Z first increases then decreases
If f < fr thenS
L > S
C, (positive), circuit nature is inductive
If f > fr then S
C > S
L, (negative), circuit nature capacitive.
( d ) Var iat ion of I w ith f as f increases , I first decreases then increases
I
fr f
RImin=V
Note : For this circuit
2
r 2
1 1 Rf
2 LC L max
LZ
RC For resonance
2
2
1 R
LC L
GOLDEN KEY POINTS
• Series resonance circuit gives voltage amplificaltion while parallel resonance circuit gives current amplification.
• At resonance current does not depend on L and C, it depends only on R and V.
• At half power frequencies : net reactance = net resistance.
• As R increases , bandwidth increases
• To obtain resonance in a circuit following parameter can be altered :
(i) L (ii) C (iii) frequency of source.
• Two series LCR circuit of same resonance frequency f are joined in series then resonance frequency of series
combination is also f
• The series resonance circuit called acceptor whereas parallel resonance circuit called rejector circuit.
• Unit of LC is second
Examp l e
For what frequency the voltage across the resistance R will be maximum.
So lu t i on
It happens at resonanceR
1 F 1
H
6
1 1f 500
2 LC 1 12 10
Hz
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Examp l e
A capacitor, a resistor and a 40 mH inductor are connected in series to an
V1 V2 V3
R=220
300V 300V XV
I
110V, 60Hz
AC source of frequency 60Hz, calculate the capacitance of the capacitor, if
the current is in phase with the voltage. Also calculate the value of X and I.
So lu t i on
At resonance
1L
C,
2
1C
L
2 2
1
4 f L
2 2 3
1176 F
4 (60) 40 10
V = VR
X = 110 V and V 110
I 0.5 AR 220
Examp l e
A coil, a capacitor and an A.C. source of rms voltage 24 V are connected in series, By varying the frequency
of the source, a maximum rms current 6 A is observed, If this coil is connected to a bettery of emf 12 V, and
internal resistance 4 , then calculate the current through the coil.
So lu t i on
At resonance current is maximum. I = V
R Resistance of coil R =
V
I =
24
6= 4
When coil is connected to battery, suppose I current flow through it then I =
E
R r =
12
4 4= 1.5 A
Examp l e
Radio receiver recives a message at 300m band, If the available inductance is 1 mH, then calculate required
capacitance
So lu t i on
Radio recives EM waves. ( velocity of EM waves c = 3 x 108 m/s)
c = f f = 83 10
300= 106 Hz Now f =
1
2 LC= 1 × 106 C =
2 12
1
4 L 10 = 25 pF
Examp l e
In a L–C circuit parallel combination of inductance of 0.01 H and a
capacitor of 1 F is connected to a variable frequency alternating current
source as shown in figure. Draw a rough sketch of the current variation
as the frequency is changed from 1kHz to 3kHz.
C
LA
So lu t i on
L and C are connected in parallel to the AC source,
so resonance frequency
4
6
1 1 10f 1.6kHz
22 LC 2 0.01 10
f
I
1.6kHzIn case of parallel resonance, current in L–C circuit at resonance is zero,
so the I-f curve will be as shown in figure.
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POWER IN AC CIRCUITThe average power diss ipation in LCR AC circui t
Let V = V0 sint and I = I
0 sin (t – )
Instantaneous power P = (V0 sint)(I
0 sin(t – ) = V
0I0 sint (sintcos – sincost)
Average power <P> = T
20 0 0 0
0
1(V I sin t cos V I sin t cos t sin )dt
T
=
T T
20 0 0 0
0 0
1 1 1V I sin t cos dt sin t cos t sin dt V I cos 0 sin
T T 2
<P> = 0 0V I cos
2 = V
rms I
rm,s cos
Instantaneous Average power/actual power/ Virtual power/ apparent Peak power
power dissipated power/power loss Power/rms Power
P = VI P = Vrms
Irms
cos P = Vrms
Irms
P = V0 I
0
• Irms
cos is known as active part of current or wattfull current, workfull current. It is in phase with voltage.
• Irms
sinis known as inactive part of current, wattless current, workless current. It is in quadrature (90) with voltage.
Power factor :
Average power rms rmsP E I cos r m s power cos
Power factor (cos ) Average power
r m s Powerand cos=
R
Z
Power factor : (i) is leading if I leads V (ii) is lagging if I lags V
GOLDEN KEY POINTS
• Pav
< Prms
.
• Power factor varies from 0 to 1
• Pure/Ideal V Power factor = cos Average power
R 0 V, I same Phase 1 (maximum) Vrms
. Irms
L
2
V leads I 0 0
C
2
V lags I 0 0
Choke coil
2
V leads I 0 0
• At resonance power factor is maximum (= 0 socos= 1) and Pav
= Vrms
Irms
Ex amp l e
A voltage of 10 V and frequency 103 Hz is applied to
1F capacitor in series with a resistor of 500. Find
the power factor of the circuit and the power dissipated.So lu t i on
C 63
1 1X 500
2 f C 102 10
Z = 2 2CR X 2 2(500) (500) 500 2
Power factor cos=R
Z=
500
500 2=
1
2 Power
dissipated =V
rms I
rms cos
2rmsV
Zcos =
2(10) 1 1
10500 2 2W
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Examp l e
If V = 100 sin 100 t volt and I = 100 sin (100 t +
3) mA for an A.C. circuit then find out
(a) phase difference between V and I (b) total impedance, reactance, resistance(c) power factor and power dissipated (d) components contains by circuits
So lu t i on
(a) Phase difference
3
(I leads V)
(b) Total impedance Z =
0
30
V 1001k
I 100 10Now resistance
1R Z cos 60 1000 500
2
reactance 3 500
X Z sin 60 10002 3
R
XZ
60°
(c) = – 60° Power factor = cos = cos (–60°) = 0.5 (leading)
Power dissipated rms rms
100 0.1 1P V I cos 2.5
22 2W
(d) Circuit must contains R as
2
and as is negative so C must be their, (L may exist but XC > X
L)
Examp l e
If power factor of a R-L series circuit is 1
2 when applied voltage is V = 100 sin 100t volt and resistance of
circuit is 200 then calculate the inductance of the circuit.So lu t i on
cos = R
Z
1
2=
R
Z Z = 2R 2 2
LR X = 2R XL 3 R
L = 3 R L =
3R =
3 200
100 =
2 3H
Examp l eA circuit consisting of an inductance and a resistacne joined to a 200 volt supply (A.C.). It draws a currentof 10 ampere. If the power used in the circuit is 1500 watt. Calculate the wattless current.
So lu t i onApparent power = 200 × 10 = 2000 W
Power factor cos =True power
Apparent power=
1500
2000 =
3
4
Wattless current = rms
sin = 10
23 10 7
1 A4 4
Examp l eA coil has a power factor of 0.866 at 60 Hz. What will be power factor at 180 Hz.
So lu t i onGiven that cos = 0.866, = 2f = 2 × 60 = 120 rad/s, ’ = 2f’ = 2 × 180 = 360 rad/sNow, cos = R/Z R = Z cos = 0.866 Z
But Z = 2 2R ( L ) L = 2 2Z R = 2 2Z (0.866 Z) = 0.5 Z L =
0.5Z =
0.5Z
120When the frequency is changed to ’ = 2 × 180 = 3 × 120 = 300 rad/s, theninductive reactance ’ L = 3 L = 3 × 0.5 Z = 1.5 Z
New impedence Z’ = 2[R ' ( 'L ) ] = 2 2(0.866 Z) (1.5 Z) = Z 2 2[(0.866 ) (1.5 ) ] = 1.732Z
New power factor = R
Z ' =
0.866 Z
1.732 Z = 0.5
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CHOKE COIL
In a direct current circuit, current is reduced with the help of a resistance.
Hence there is a loss of electrical energy I2 R per sec in the form of
heat in the resistance. But in an AC circuit the current can be reduced
by choke coil which involves very small amount of loss of energy. Choke
coil is a copper coil wound over a soft iron laminated core. This coil is
put in series with the circuit in which current is to be reduced. It also
known as ballast.
tube light rod
starter
choke coil
Circuit with a choke coil is a series L-R circuit. If resistance of choke coil = r (very small)
The current in the circuit E
IZ
with 2 2Z (R r ) ( L ) So due to large inductance L of the coil, the
current in the circuit is decreased appreciably. However, due to small resistance of the coil r,
The power loss in the choke Pav = V
rms I
rms cos 0
2 2 2
r r rcos 0
Z Lr L
GOLDEN KEY POINT
• Choke coil is a high inductance and negligible resistance coil.
• Choke coil is used to control current in A.C. circuit at negligible power loss
• Choke coil used only in A.C. and not in D.C. circuit
• Choke coil is based on the principle of wattless current.
• Iron cored choke coil is used generally at low frequency and air cored at high frequency.
• Resistance of ideal choke coil is zero
Examp l e
A choke coil and a resistance are connected in series in an a.c circuit and a potential of 130 volt is applied
to the circuit. If the potential across the resistance is 50 V. What would be the potential difference across the
choke coil.
So lu t i on
V = 2 2R LV V 2 2 2 2
L RV V V (130) (50) = 120 V
Examp l e
An electric lamp which runs at 80V DC consumes 10 A current. The lamp is connected to
100 V – 50 Hz ac source compute the inductance of the choke required.
So lu t i on
Resistance of lamp V 80
R 8I 10
Let Z be the impedance which would maintain a current of 10 A through the Lamp when it is run on
100 Volt a.c. then. Z = V
I =
100
10 = 10 but Z = 2 2R ( L )
(L)2 = Z2 – R2 = (10)2 – (8)2 = 36L = 6 L =
6 =
6
2 50 = 0.02H
Examp l e
Calculate the resistance or inductance required to operate a lamp (60V, 10W) from a source of
(100 V, 50 Hz)
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So lu t i on
(a) Maximum voltage across lamp = 60VR
100V, 50Hz
V Lamp
+ VR = 100 V
R = 40V
Now current througth Lamp is = Wattage
voltage =
10
60=
1
6A
But VR= IR 40 =
1
6(R) R = 240
(b) Now in this case (VLamp
)2 + (VL)2 = (V)2
100V, 50Hz
L
(60)2 + (VL)2 = (100)2 V
L = 80 V
Also VL = IX
L =
1
6X
Lso X
L = 80 × 6 = 480 = L (2f) L = 1.5 H
A capacitor of suitable capacitance replace a choke coil in an AC circuit, the average power consumed in
a capacitor is also zero. Hence, like a choke coil, a capacitor can reduce current in AC circuit without power
dissipation.
Cost of capacitor is much more than the cost of inductance of same reactance that's why choke coil is used.
Examp l e A choke coil of resistance R and inductance L is connected in series withCL, R
V=V sin t(volt)0 w~
a capacitor C and complete combination is connected to a.c.
voltage, Circuit resonates when angular frequency of supply is = 0.
(a) Find out relation betwen 0, L and C
(b) What is phase difference between V and I at resonance,
is it changes when resistance of choke coil is zero.
So lu t i on (a) At resonance condition XL = X
C
0L =
1
0 C
0 =
1
LC
(b) cos = R
Z =
R
R = 1 = 0° No, It is always zero.
LC OSCILLATION
The oscillation of energy between capacitor (electric field energy) and inductor (magnetic field energy) is called
LC Oscillation.
UNDAMPED OSCILLATION
When the circuit has no resistance, the energy taken once from the source and given to capacitor keeps on
oscillating between C and L then the oscillation produced will be of constant amplitude. These are called undamped
oscillation.
LC
I
t
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After switch is closed
Q di
L 0C dt
2
2
Q d QL 0
C dt
2
2
d Q 1Q 0
LCdt
By comparing with standard equation of free oscillation
22
2
d xx 0
dt
2 1
LCFrequency of oscillation
1f
2 LC
Charge varies sinusoidally with time q = qm cos t
current also varies periodically with t I = dq
dt = q
m cos
( t )
2
If initial charge on capacitor is qm then electrical energy strored in capacitor is U
E =
2mq1
2 C
At t = 0 switch is closed, capacitor is starts to discharge.
As the capacitor is fully discharged, the total electrical energy is stored in the inductor in the form of magnetic
energy.
UB =
2m
1LI
2where I
m = max. current
(Umax
)EPE
= (Umax
)MPE
2
2mm
q1 1LI
2 C 2
DAMPED OSCILLATION
Practically, a circuit can not be entirely resistanceless, so some part of energy is lost in resistance and amplitude
of oscillation goes on decreasing. These are called damped oscillation.
RL
C
Angular frequency of oscillation 2
2
1 R
LC 4L
frequency of oscillation
2
2
1 1 Rf
2 LC 4L
I
t
oscillation to be real if 2
2
1 R0
LC 4L
Hence for oscilation to be real 2
2
1 R
LC 4L
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GOLDEN KEY POINTS
• In damped oscillation amplitude of oscillation decreases exponentially with time.
• At T 3T 5T
t , , .....4 4 4
energy stored is completely magnetic.
• At T 3T 5T
t , , .....8 8 8
energy is shared equally between L and C
• Phase difference between charge and current is
2 when charge is maximum, current minimumwhen charge is minimum,current maximum
Examp l e
An LC circuit contains a 20mH inductor and a 50F capacitor with an initial charge of 10mC. The resistanceof the circuit is negligible. Let the instant the circuit is closed to be t = 0.
(a) What is the total energy stored initially.
(b) What is the natural frequency of the circuit.
(c) At what time is the energy stored is completely magnetic.
(d) At what times is the total energy shared equally between inductor and the capacitor.
So lu t i on
(a) UE =
21 q
2 C=
3 2
6
1 (10 10 )
2 50 10= 1.0J
(b) = 1
LC =
3 6
1
20 10 50 10= 103 rad/sec f = 159 Hz
(c) q = q0 cos t
Energy stored is completely magnetic (i.e. electrical energy is zero, q = 0)
a tT 3T 5T
t , , .........4 4 4
where 1
Tf
= 6.3 ms
(d) Energy is shared equally between L and C when charge on capacitor become 0q
2
so, at T 3T 5T
t , , ......8 8 8
energy is shared equally between L and C
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Examp le#1
For the given circuit
(A) The phase difference between IL & IR1
is 0°
(B) The phase difference between VC & VR2
is 90°
(C) The phase difference between IL & IR1
is 180°
(D) The phase difference between VC & VR2
is 180°
So lu t i on Ans. (B)
The phase difference between VC and
2RV is rad or 90°
Examp le#2
A periodic voltage V varies with time t as shown in the figure. T is the time period. The r.m.s. value of thevoltage is :-
V0
T/4 T
V
t
(A) 0
8
V(B)
0
2
V(C) V
0(D)
0
4
V
So lu t i on Ans. (B)
Root mean square value <V> =
T / 42 2 20 20 0
0 0 0
T
0
T TV dt V VV V4 4
T T 4 2dt
Examp le#3
The potential difference V and current I flowing through the AC circuit is given by V = 5 cos(t – /6) volt and
I = 10sint ampere. The average power dissipated in the circuit is
(A) 25 3
2 W (B) 12.5 W (C) 25 W (D) 50 W
SOME WORKED OUT EXAMPLES
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So lu t i on Ans. (B)
V = 5 cos (t - /6); i = 10 sin t = 10 cos (t - /2)
2 6 3;
VI 5 10 1P cos 12.5W
2 2 2
Examp le#4
The radius of a coil decreases steadily at the rate of 10–2 m/s. A constant and uniform magnetic field of in-
duction 10–3 Wb/m2 acts perpendicular to the plane of the coil. The radius of the coil when the induced e.m.f.
in the coil is 1V, is :-
(A)
2 cm (B)
3 cm (C)
4 cm (D)
5 cm
So lu t i on Ans. (D)
d de
dt dt (r2B) = 2rB
6
3 2
dr e 10 5r r cm
drdt 2 10 102 B
dt
Examp le#5
A time varying voltage V = 2t volt is applied across an ideal inductor of inductance
L = 2H as shown in figure. Then select incorrect statement
(A) current versus time graph is a parabola. .
V=2t
2H
(B) energy stored in magnetic field at t = 2 s is 4J
(C) potential energy at time t = 1 s in magnetic field is increasing at a rate of 1 J/s
(D) energy stored in magnetic field is zero all the time
So lu t i on Ans. (D)
V = 2t Ldi
dt = 2t 2 ×
di
dt = 2t
di
dt i =
2t
2 i – t graph parabola
U = 1
2 Li2 =
1
2 × 2 × 4 = 4J and
dU
dt=Li
di
dt = 2 ×
2t
2 × t = t3 = 1 J/s
Examp le#6
A circular coil of 500 turns encloses an area of 0.04 m2. A uniform magnetic field of induction
0.25 Wb/m2 is applied perpendicular to the plane of the coil. The coil is rotated by 90° in 0.1 second at a
constant angular velocity about one of its diameters. A galvanometer of resistance 25 was connected in series
with the the coil. The total charge that will pass through the galvanometer is -
(A) 0.4 C (B) 1 C (C) 0.2 C (D) Zero
So lu t i on Ans. (C)
Induced current
e dq e d 1 NBAI q
R dt R dt R R R
Total charge q = 500 0.25 0.04
25= 0.2 C
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Examp le#7
A condenser of capacity 6 µF is fully charged using a 6-volt battery. The battery is removed and aresistanceless 0.2 mH inductor is connected across the condenser. The current which is flowing through theinductor when one-third of the total energy is in the magnetic field of the inductor is :-
(A) 0.1 A (B) 0.2 A (C) 0.4 A (D) 0.6 A
So lu t i on Ans. (D)
Total energy = Intial energy on capacitor = 1
2CV2, Magnetic field energy =
1
2LI2
1
2LI2 =
1 1
3 2 CV2 I =
2 6
3
CV 6 10 6 6
3L 3 2.0 10 = 0.6 A
Examp le#8
For the circuit shown, which of the following statement(s) is(are) correct?
(A) Its time constants is 2 second.
(B) In steady state, current through inductance will be 1A.
(C) In steady state, current through 4 resistance will be 2/3 A.
(D) In steady state, current throgh 8 resistance will be zero.
So lu t i on Ans. (B)
Time constant L 12
1sR 12
In steady state
current through 4 resistance =
6 11 A
6 12 3
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Examp le#9
In the circuit shown the capacitor has charge Q. At t = 0 sec the key is closed. The charge on the capacitor atthe instant potential difference across the inductor L
1 is zero, is
(A) Q (B) Q
3 (C)
2Q
3(D) 0
So lu t i on Ans. (D)
When Vacross L
= 0 i = imax
qc = 0
Examp l e#10
The figure shows a rod of length with points A and B on it. The rod is moved in a uniform magnetic field
(B0) i n d i f ferent ways as shown. In wh ich case potent ial d i f ference (V
A–V
B) between
A & B is minimum?
(A)
A
BB0
(B)
A
BB0
v=2
(C) /2 /2
A B
B0
(D)
/2A
BB0
/2
S o l u t i o n Ans. (C)
For (A): 2 2
B A A B
1 1V V B V V B
2 2
For (B) : 2 2
B A A B
1 1V V Bv B B V V B
2 2 2
For (C) : VA _ V
B = 0
For (D) :
2
2 2 2
B A A B
1 1 3 3V V B B B V V B
2 2 2 8 8
Examp l e#11
A bent rod PQR (PQ = QR = ) shown here is rotating about its end P with angular speed in a region of
transverse magnetic field of strength B.
(A) e.m.f. induced across the rod is B2
(B) e.m.f. induced across the rod is B2/2
Q
P R
60°
B
(C) Potential difference between points Q and R on the rod is B2/2
(D) Potential difference between points Q and R on the rod is zeroSo lu t i on Ans. (B,D)
The rod in equivalent to a rod joining the ends P and R of the rod rotating is the same sense.
60°
RP
R
B
2
R P
BV V
2;
2
Q P
BV V
2;
2
R P
BV V
2; V
Q – V
R = 0
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Examp l e#12
x
y
BA conducting loop is kept so that its center lies at the origin of the coordinate
system. A magnetic field has the induction B pointing along Z-axis as shown
in the figure
(A) No emf and current will be induced in the loop if it rotates about Z-axis(B) emf is induced but no current flows if the loop is a fiber when it rotates about y-axis.(C) emf is induced and induced current flows in the loop if the loop is made of copper & is rotated about y-axis.(D) If the loop moves along Z-axis with constant velocity, no current flows in it.
S o l u t i o n Ans. (A ,C,D)If the loop rotates about Z axis, the variation of flux linkage will be zero. Therefore no emf is induced.Consequently no current flows in the loop. When it rotates about y axis, its flux linkage changes. However,in insulators there can not be motional emf. If the loop is made of copper, it is conductive thereforeinduced current is set up. If the loop moves along the Z axis variation of flux linkage is zero. Therefore theemf and current will be equal to zero.n
Examp l e#13Initially key was placed on (1) till the capacitor got fully charged. Key is placed on (2) at t=0. The timewhen the energy in both capacitor and inductor will be same-
(1) (2)
C
EL
(A) LC
4(B)
LC
2(C)
5 LC
4(D)
5 LC
2
So l u t i on . Ans. (A,C)
For given situation 2 2
2
2 2
q di d q q d qL 0 0 q 0
C dt LCdt dt q = q
0 cos t & i = –q
0sint
According to given conditions
22
2 2 2 200
q cos tq 1 1Li Lq sin t
2C 2 2C 2
cot2t = 1 t = 3 5 7
, , ,4 4 4 4
.........
LC 3 LC 5 LC 7 LC
t , , ,4 4 4 4
.....
Examp l e#14
For an LCR series circuit, phasors of current i and applied voltage V = V0 sint are shown in diagram
at t =0. Which of the following is/are CORRECT?
(A) At t =
2, instantaneous power supplied by source is negative.
(B) From 0 < t <
2
3, average power supplied by source is positive. /3
i0
V0
(C) At t =
5
6, instantaneous power supplied by source is negative.
(D) If is increased slightly, angle between the two phasors decreases.
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So lu t i on Ans. (BCD)
The graph shows V & I as function of time.
Current leads the voltage by /3
Power is positive if V & I are of same sign.
Power is negative if V & I are of opposite sign
Voltage
tcurrent
i/v
If
1
C thus angle decreases.
Example#15 to 17
Consider a conducting circular loop placed in a magnetic field as shown in
figure. When magnetic field changes with time, magnetic flux also changes
and emf
d
edt
is induced.
If resistance of loop is R then induced current is e
iR
. For current, charges must have non–zero average
velocity. Magnetic force cannot make the stationary charges to move. Actually there is an induced electric field
in the conductor caused by changing magnetic flux, which makes the charges to move,
dE d
dt.This
induced electric field is non conservative by nature.
1 5 . A cylindrical space of radius R is filled with a uniform magnetic induction B
parallel to the axis of the cylinder. If dB
dt= constant, the graph, showing thee
variation of induced electric field with distance r from the axis of cylinder, is
(A) (B) (C) (D)
1 6 . A square conducting loop is placed in the time varying magnetic field
dBve constant
dt. The centre of
square coincides with axis of cylindrical region of magnetic field. The directions of induced electric field at point
a, b and c.
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(A) (B) (C) (D)
1 7 . A line charge per unit length is pasted uniformly onto the rim of a wheel of mass m and radius R. The wheel
has light non–conducting spokes and is free to rotate about a vertical axis as shown in figure.A uniform magnetic
field B exist as shown in figure. What is the angular velocity of the wheel when the field is suddenly switched off?
(A) 22 a B
mR(B)
2a B
mR(C)
23 a B
mR(D)
2a B
2mR
So lu t i on
1 5 . Ans. (B)
2 dBFor r R; E 2 r r E r
dt 2 dB 1
For r R; E 2 r R Edt r
1 6 . Ans. (A)
Induced electric field is circular.
1 7 . Ans. (B)
2L I mR
t t t but
22 B a BR
2 R ER a Rt t
2 2 2mR a BR a B
t t mR
Example#18 to 20
A conducting square wire frame ABCD of side is pulled by horizontal force so that it moves with constant
velocity v. A uniform magnetic field of strength B is existing perpendicular to the plane of wire. The resistance
per unit length of wire is and negligible self inductance. If at t= 0, frame is just at the boundary of magnetic
field. Then
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1 8 . The emf across AB at t =
2v is
(A) Bv
4(B) zero (C) Bv (D) None of these
1 9 . Potential difference across BC at time t =
2vis
(A) Bv
4(B) Bv (C)
3Bv
4(D) None of these
2 0 . Find the applied horizontal force on BC, (F) as a function of time 't' (t <
v)
(A)
2B vt (B)
2B v
2(C)
2B v
4(D) None of these
So lu t i on
1 8 . Ans. (B)
e B . ( v ) as
e| | v = 0° e = 0
1 9 . Ans. (C)
B
C
A
D
V = E – ir = Bv –
B v 3B v[ ]
4 ( ) 4
2 0 . Ans. (C)
2 2 2B (B v) B v B vF Bi
4 ( ) 4 ( ) 4
Examp l e#21
Magnetic flux in a circular coil of resistance 10 changes with time as shown in figure. direction indicates a
direction perpendicular to paper inwards.
2 68 10 14 16
(Wb)
10
–10t(s)
× × ×
× × ×
× × ×
× × ×
Co lumn–I Column–I I
(A) At 1 second induced current is (P) Clockwise
(B) At 5 second induced current is (Q) Anticlockwise
(C) At 9 second induced current is (R) 0.5 A
(D) At 15 second induced current is (S) 5 A
(T) None of these
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So lu t i on Ans. (A) Q,R (B) T (C) P,R (D) Q,R
For (A)
d
5dt
5 1
i10 2
A, Anticlockwise For (B)
d
0dt
i=0 = zero
For (C) d
dt= –5
5 1i
10 2A, Clockwise For (D)
d
dt=5
5 1i A
10 2, Anticlockwise
Examp l e#22
In the circuit shown in figure E=25V, L=2H, C=60 F, R1 = 5 and R2 = 10. Switch S is closed at t = 0.
C
R1
R2
L
S
E
Column- I Co lumn- I I
(A) Current through R1 at t = 0 (P) 0
(B) Current through R2 at t = 0 (Q) 5A
(C) Current through R1 at t = (R) 2.5 A
(D) Current through R2 at t = (S) 7.5 A
(T) None of these
So lu t i on Ans. (A) (Q), (B) (P), (C) (P), (D) (R)
R1
ER2
At t = 0
R1R2
E
At t =
I1 = 1
E
R = 25
5= 5A I1 = 0
I2 = 0 I2 = 2
E
R = 25
10 = 2.5 A
Examp l e#23
A uniform but time-varying magnetic field B(t) exists in a circular region of radius a and is directed into
the plane of the paper, as shown. The magnitude of the induced electric field at point P at a distance
r from the centre of the circular region is proportional to 1/rn. Find the value of n.
a
x
x
x
x x
r
P
So lu t i on Ans. 1
d 1E 2 r E n 1
dt r
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Examp l e#24
An inductor (XL = 2) a capacitor (X
C = 8) and a resistance (8) is connected in series with an ac source. The
voltage output of A.C source is given by V = 10 cos 2 50t. Find the instantaneous p.d. between A and B when
the voltage output from source is half of its maximum.
8X =2L A
X = 8C
~
B
So lu t i on Ans. 3
AB
6 10V 3
10 2
2 2
AB C C C LR X X and Z X X R
Examp l e#25
Figure shows a uniform circular loop of radius ‘a’ having specific resistance placed in a uniform magnetic field
B perpendicular to plane of figure. A uniform rod of length 2a & resistance Rmoves with a velocity v as shown.
Find the current in the rod when it has moved a distance a
2 from the centre of circular loop.
[Given B = 3, a =2, v =3,
9
4, R = 2/3 all in SI units]
x xx x
x x x
x
x
x
x
x
x
x
x
a
a/2v
B
So lu t i on Ans. 6
6 33 1
a 3 × B × v = 18 V
18 18
63
i A
