Chapter 32 Fundamentals of Circuits L.A. Bumm. Schematic Symbols ideal wire with zero resistance

Chapter 32Fundamentals of Circuits

L.A. Bumm

Schematic Symbols

ideal wire with zero resistance

Basic Circuits and Kirchhoff’s Loop Law

0

;

0

;

loop i bat Ri

bat R

bat R

R

V V V V

V V IR

V V IR

IR

I VR

E

E

E

EE

Power in Electrical Circuits

;bat bat

bat

U q V V

U q

dU dqP I

dt dt

E

E

E E

Power supplied by batteriesbatP I E

Power in Electrical Circuits

Power dissipated in resistors.Electrical energy is converted in to heat.

mech

thermal R

R thermal R R

R R

W F s qEd

E K qEd

E qEL q V

d dqP E V I V

dt dtP I V

22 R

R R

VP I V I R

R

Because current and voltage difference for resistors are related by resistance (Ohm’s Law), we can eliminate I or ΔV from the power equation. But this only works for resistors!!.

P I V This is always true:

;RR R R R

VP I V I IR V V IR

R

Clicker Question

Which schematic is NOT equivalent to the cartoon picture?

A) C)B) D)

Clicker Question

Which equation is not Ohm’s Law

A) C)B) D) E)

VR

I

VI

R

RI

V

V I R

These are all statements of

Ohm’s Law

Resistors in Series and Parallel

• Resistors in series– The current is the same through each resistor.– The voltage drops across each resistor sum to the

voltage drop across all the resistors.– The equivalent resistance will be equal to or

greater than the largest resistance in the series circuit.

• Resistors in Parallel– The voltage drop is the same across each resistor.– The current through each resistor sums to the

total current supplied to the parallel circuit.– The equivalent resistance is less than or equal to

the smallest resistor in the parallel circuit.

321 RRRReq

321

1111

RRRReq

Combinations of series and parallel resistors.

Any network of resistors can be reduced to a single equivalent resistance. When the network is not a simple series or parallel combination, it can always be reduced to a single resistor by the sequential application of the series and parallel resistor combination rules. Don’t be afraid to redraw the circuit to make it look more familiar to you.

Reduce this circuit to its equivalent resistance.1

1 130

90 45

11 1

15.4 40 25

10 30 40

Example: A nichrome wire with diameter d and length L is connected to a battery of EMF E by two lengths of copper wire with the same d and L as the nichrome wire. Draw a schematic of this circuit. Do not neglect the resistance of the copper wire. What is the resistance of each segment of wire. What is the current in the circuit. How much power is supplied by the battery. How much power is dissipated in the nichrome wire. What percentage of the power is delivered to the nichrome wire? E = 1.5 V; d = 1.0 mm; L = 200 cm; ρnichrome = 1.5×10−6 Ωm; ρCu = 1.7×10−8 Ωm

LR

A

VI

R

We are ask to find RNC, RCu, I, Pbat, PNC, and PNC/Pbat.

3.8 NCR E RNC

I

RCu1

RCu2

6 2

24

1.5 10 m 200 10 m3.8197

5 10 m

NCNC

LR

A

8 2

24

1.7 10 m 200 10 m0.0433

5 10 m

CuCu

LR

A

24.3 10 CuR

2P I V I R

Example: A nichrome wire with diameter d and length L is connected to a battery of EMF E by two lengths of copper wire with the same d and L as the nichrome wire. Draw a schematic of this circuit. Do not neglect the resistance of the copper wire. What is the resistance of each segment of wire. What is the current in the circuit. How much power is supplied by the battery. How much power is dissipated in the nichrome wire. What percentage of the power is delivered to the nichrome wire?E = 1.5 V; d = 1.0 mm; L = 200 cm; ρnichrome = 1.5×10−6 Ωm; ρCu = 1.7×10−8 Ωm

VI

R


2

2

2 4.3 10 3.8 3.9063

Cu NC Cu Cu NCR R R R R R

1.5 V0.3840 A

3.9063 mI

R

E

0.38 AI

E RNC

I

RCu1

RCu2

2P I V I R

3.9 R

Example: A nichrome wire with diameter d and length L is connected to a battery of EMF E by two lengths of copper wire with the same d and L as the nichrome wire. Draw a schematic of this circuit. Do not neglect the resistance of the copper wire. What is the resistance of each segment of wire. What is the current in the circuit. How much power is supplied by the battery. How much power is dissipated in the nichrome wire. What percentage of the power is delivered to the nichrome wire?E = 1.5 V; d = 1.0 mm; L = 200 cm; ρnichrome = 1.5×10−6 Ωm; ρCu = 1.7×10−8 Ωm


0.3840 A 1.5 V 0.5760 WbatP I E

0.5630 W

0.9778360.5760 W

NC

bat

P

P

E RNC

I

RCu1

RCu2

2P I V I R

22 0.3840 A 3.8197 0.5630 WNC NCP I R

0.58 WbatP

0.56 WNCP

98 %NC

bat

P

P

We could have ignored the resistance of the copper wire, depending on our application.

Kirchhoff’s Laws

Kirchhoff’s Junction Law (Conservation of charge)At any junction point, the sum of all the currents

entering the junction must equal the sum of all currents leaving the junction.

Kirchhoff’s Loop Law (Conservation of energy)The sum of the changes in potential around any

closed path of a circuit must be zero.

How to Use the Kirchhoff’s Laws• Draw the circuit and draw currents with an arrow in every separate branch of the

circuit. (A branch is a section where the current does not change.)• Apply the junction law to enough junctions so that every current is used at least

once.• Apply the loop law to enough closed loops so that each current appears at least

once. Remember the sign convention for the potential changes:

Sign Conventions

Batteries + if moving around the loop you pass from the − terminal to the + terminal.− if moving around the loop you pass from the + terminal to the − terminal.

Resistors − if moving around the loop you are moving in the direction of the defined current (your arrows).+ if you are moving against the defines current.Caution: If your circuit has multiple branches your loop path may go with the defined current in one branch and against it in another.

Example: Analyzing circuits with more that one loop

++

330 Ω 150 Ω 270 Ω

19 V 12 V

Find the current and its direction through the 150 Ω resistor and the potential difference across it.


++

330 Ω 150 Ω 270 Ω

19 V 12 V I2

I1

I3

1 2 3

in outI I

I I I

junction eq.

• Draw the circuit and draw currents with an arrow in every separate branch of the circuit. (A branch is a section where the current does not change.)

• Apply the junction law to enough junctions so that every current is used at least once.

We could also have used this junction. We do not need to use both, because the information is redundant.

2 3 1

in outI I

I I I


++

330 Ω 150 Ω 270 Ω

19 V 12 V I2

I1

I3

1 3

loop A: 0

19 V 330 150 12 V

ii

V

I I

• Apply the loop law to enough closed loops so that each current appears at least once. Remember the sign convention for the potential changes:

Sign Conventions

Batteries + if moving around the loop you pass from the − terminal to the + terminal.− if moving around the loop you pass from the + terminal to the − terminal.

Resistors − if moving around the loop you are moving in the direction of the

defined current (your arrows).+ if you are moving against the defines current.Caution: If your circuit has multiple branches your loop path may go with the defined current in one branch and against it in another.

units check: V A A V

V AV IR A Ω is the same as a V so we can add them.

A


++

330 Ω 150 Ω 270 Ω

19 V 12 V I2

I1

I3

1 3

loop A: 0

19 V 330 150 12 V

ii

V

I I

loop equations

++

330 Ω 150 Ω 270 Ω

19 V 12 V I2

I1

I3

3 2

loop B: 0

12 V 150 270

ii

V

I I

++

330 Ω 150 Ω 270 Ω

19 V 12 V I2

I1

I3

1 2

loop C: 0

19 V 330 270

ii

V

I I

The third loop is redundant and provides no new information. There are n−1 unique loops. Here, we can chose any two loop.

A

B

C


++

330 Ω 150 Ω 270 Ω

19 V 12 V I2

I1

I31 2 3I I I

1 30 19 V 330 150 12 VI I

3 20 12 V 150 270 I I

1 2 3I I I

2 3 3

2 3

2 3

2 3

0 19 V 330 150 12 V

19 V 12 V 330 330 150

7 V 330 480

7 V 330 480

I I I

I I

I I

I I

3 2

2 3

0 12 V 150 270

12 V 270 150

I I

I I

We have e equations and 3 unknowns.

Eliminate I1 first, let’s keep I3 because that is what we are asked to find.


++

330 Ω 150 Ω 270 Ω

19 V 12 V I2

I1

I3

1 2 3I I I

1 30 19 V 330 150 12 VI I

3 20 12 V 150 270 I I

2 37 V 330 480 I I

2 3

33012 V 270 150

270I I

37.667 V 0 663.3 I

3

7.667 V11.56 mA

663.3 I

2 37 V 330 480 I I

2 3

48012 V 270 150

150I I

245.40 V 1194 0I

2

45.40 V38.02 mA

1194 I

1 2 3 38.02 mA 11.56 mA 26.46 mAI I I

1

2

3

26 mA

38 mA

12 mA

I

I

I


Find the current and its direction through the 150 Ω resistor and the potential difference across it.

1

2

3

26 mA

38 mA

12 mA

I

I

I

++

330 Ω 150 Ω 270 Ω

19 V 12 V I2

I1

I3

150 11.56 mA 150 1.734 VV IR

150 1.7 VV

Clicker Question

Which is not a loop equation for the following circuit?

A)

C)

B)

E)

D)

1 A 1 A 2 B 3I R I R I R E

2 A 1 A 2 1I R I R E E

+

+

R1

E1

E2

R2

R3 IC

IA IB

B 3 2I R E

1 2 A 1 A 2I R I R E E

1 A 1 A 2 B 3 0I R I R I R E

Ideal and Real Batteries

The ideal battery is a voltage source. • It maintains a constant voltage across its terminals. • It can supply any amount of current required to maintain its terminal voltage. • The current the battery supplies is determined by the rest of the circuit.

A real batteries have internal resistance.• Real batteries are simply an ideal voltage source (battery) with a series

resistor representing its internal resistance. • The internal resistance limits the maximum current the battery can supply.

+E + rE

ideal battery real battery

AmmetersAmmeter. •An ammeter measures current. • It must be placed in series with the circuit in which the current is to be measured.• The ideal ammeter has zero internal resistance.• The ideal ammeter has zero voltage drop (the same as zero series resistance).•A real ammeter has internal resistance and is modeled as an ideal ammeter in series

with a resistor, hence does contribute a voltage drop in the circuit it is measuring.

comments• The ammeter symbol is ALWAYS an ideal ammeter.• In problems always assume an ideal ammeter unless you are explicitly told otherwise.

Aideal ammeter real ammeter

Ar

VoltmetersVoltmeter. •A voltmeter measures voltage (potential difference). • It must be placed in parallel with the circuit in which the voltage is to be measured.• The ideal voltmeter has infinite internal resistance. • The ideal voltmeter has draws no current (the same as infinite resistance). •A real voltmeter has internal resistance and is modeled as an ideal voltmeter in

parallel with a resistor, hence does draw current from the circuit it is measuring.

comments• The voltmeter symbol is ALWAYS an ideal voltmeter.• In problems always assume an ideal voltmeter unless you are explicitly told

otherwise.

real voltmeter

Videal voltmeter

Vr

Example: A real battery with EMF E is connected to an external resistor R. A voltmeter measures the potential difference across the resistor ΔVR. What is the internal resistance of the battery?What would the short circuit current of this battery be?E = 9.0 V; R = 17 Ω; ΔVR = 8.5 V

0 Ri i

V Ir IR Ir V E E

9.0 V

1 17 1 1.0 8.5 V

R

R RR

R R R

V Ir

V V Rr V R

VI V VR

E

E E EE

RVIR

1.0 r

short

9.0 V9.0 A

1.0 I

r

E

short 9.0 AI

Clicker QuestionWhat four identical resistor values R will have an equivalent resistance of 100 Ω when connected as shown?

A)

C)

B)

D)

E)

R

R

R

R

Req = 100 Ω

100 Ω

50 Ω

25 Ω

200 Ω

none of the above

Light Bulbs and BrightnessLight Bulbs. • Electrical devices that emit light. • Incandescent lamps are resistors, that get hot enough to emit black body radiation.• The brightness of a lamp is proportional to the power dissipated in the lamp.

comments• The relative brightness of a light bulb in a circuit may be determined by finding the

power dissipated in each bulb.•Well will ignore the changes in resistance with temperature of incandescent lamps. The

filament in an incandescent lamp is typically made from tungsten because of its very high melting point. The resistivity of metals increases with increasing temperature. Because the filament operates at thousands of degrees above room temperature, the change in resistance from on to off can be very significant.

Light bulb

Light Bulb BrightnessThe following circuit has 3 identical bulbs, a battery, and a switch.

Switch open:When the switch is open, obviously bulb C is off.We can simplify the circuit. What can we say about the brightness of bulbs A and B?

Assumptions: The voltage across the battery will be constant. The resistances R of all three bulbs are the same.

I

Bulbs A & B are in series, thus the same current flows through each. Because the two bulbs also have the same resistance, the power (and therefore their brightness) will be the same.

What is the actual power in each bulb?2 2

2;2 2 4

I P I R RR R R R R

E E E E 2

A B

1

4P P

R

E

Light Bulb BrightnessSwitch closed:We can simplify the circuit. What can we say about the brightness of bulbs A, B, and C?

• Bulbs B & C are in parallel, thus the same voltage across each. Because the two bulbs also have the same resistance, the power (and therefore their brightness) will be the same.

IB IC

IA

• Bulb A is in series with the B & C. All the current flowing in the circuit passes through A. From Kirchhoff’s junction law we know that IA = IB + IC. Because the bulbs all have the same resistance, the current through A is twice that though B and C. Thus the power (and therefore their brightness) will be four times that of bulbs B and C.

Light Bulb BrightnessSwitch closed:We can simplify the circuit. What can we say about the brightness of bulbs A, B, and C?

What is the actual power in each bulb?

2 221 1

2 2 32

;3 3 9B C A B CI I I P P I R R

R R R R

E E E E

IB IC

IA

1

eq31

2 21 1

AB C

R R R R RR R

22

2

eq

493 3

2 2

;A AI P I R RR R R R

E E E E

24

9AP R

E

21

9B CP PR

E

Equivalent resistance of the circuit:

Power in bulb A.

Power in bulbs B and C.

Light Bulb BrightnessThe following circuit has 3 identical bulbs, a battery, and a switch.

What can we say about the change in brightness of bulbs A and B when the switch is closed?

Qualitatively we can say: • The brightness of A increases because the current in the circuit increases as C

is brought in parallel with B. • The brightness of B decreases because more voltage is dropped across A as

the current in the circuit increases.

Quantitatively, we simply ratio the powers2

closedB

2openB

1 149 9

11 944

P RP

R

E

E

2

closedA

2openA

4 4169 9

11 944

P RP

R

E

E

closedB

openB

4

9

P

P

closedA

openA

16

9

P

P

Clicker Question

In the circuit to the right. The switch open.Compare the brightness of bulbs A and B.

A)

C)

B)

D)

A = B

A > B

A < B

There is no way to tell.

Clicker Question

In the circuit to the right. The switch open.Compare the brightness of bulbs B and C.

A)

C)

B)

D)

B = C

B > C

B < C


Clicker Question

In the circuit to the right. When the switch is closed,the brightness of bulb B ______ .

A)

C)

B)

D)

increases

decreases

does not change


Clicker Question

In the circuit to the right. The switch is now closed.Compare the brightness of bulbs B and C.

A)

C)

B)

D)

B = C

B > C

B < C


GroundingGrounding provides a convenient reference point for potential differences. All potentials we measure are potential differences. However when we say a point in a circuit has a certain potential, we mean it has that potential with respect to a common reference point. Typically a ground connection defines that point.

4 VV IR

6 VV IR

Example: Adding a ground wire does not change the function of the circuit. It only changes the potentials with respect to ground. This is true when there is only ONE connection to ground so that no current can flow between the circuit and ground.In this example, only the point of grounding is changed. Note that the currents and the potential differences with in the circuit are unchanged.

RC Circuits and TransientsSo far we have considered capacitors that were charged and discharged. Now we will described how circuits with capacitors evolve from the charged state to the discharged state and vice versa.The capacitor initially charged to Q0 = VC.While the switch is open no current flows in the circuit.

The switch is closed at t = 0current begins to flow and the capacitor begins to discharge.I0 = V/RWe can use Kirchhoff’s Loop Law

; 0 C R

dQ Q Q dQI V V IR R

dt C C dt

C RI = 0

before switch is closed

ΔVR = 0ΔVC = Q0/C

000

0

1ln exp

Q t

Q

dQ Q t tdt Q Q

Q RC Q RC RC

0; expt

RC Q Q

ΔVC = Q/C

C R

ΔVR = IR

I

after switch is closed

τ is the time constant, the rate the capacitor charges and discharges

units:s = FRC

RC Circuits and Transients

00

0

exp exp

exp exp expC C

QdQ d t tI Q

dt dt

V C Vt t tI

RC R

0; expt

RC Q Q

The time constant τ is determined by R and C only. It is the rate the circuit evolves from the initial state (charged) to the final state (discharged).

The exponential decay of the charge decreases by a factor e−1 over each time interval τ.

The current also decays exponentially with time.

0 expt

I I

RC Circuits and Transients

The capacitor initially discharged with Q0 = 0.While the switch is open no current flows in the circuit. The switch is closed at t = 0.I0 = V/RThe capacitor charges to Qmax following a complimentary exponential.

max max max; exp 1 expt t

RC Q Q Q Q

The charging process also follows exponential behavior with time constant τ.

The exponential decay of the charge difference (Qmax − Q) decreases by a factor e−1 for each time interval τ.

max; 1 expt

RC Q Q

Example: The switch has been in position a for a long time. It is changed to position b at t = 0. What is the time constant of the circuit? What is the charge on the capacitor and the current through the circuit after time t = 4.7 μs? How long does it take for the capacitor to loose half its charge?E = 9.0 V; R = 17 Ω; C = 1.5 μF

6 517 1.5 10 F 2.55 10 sRC E C R

26 μs

6

bat0 6

4.7 10 s9.0 Vexp exp exp 0.4403 A

17 17 1.5 10 F

Vt tI I

R RC

0 bat

6

6 5

6

exp exp

4.7 10 s9.0 V 1.5 10 F exp 1.1228 10 C

17 1.5 10 F

t tQ Q V C

RC

11 μCQ

0.44 AI

Example: The switch has been in position a for a long time. It is changed to position b at t = 0. What is the time constant of the circuit? What is the charge on the capacitor and the current through the circuit after time t = 4.7 μs? How long does it take for the capacitor to loose half its charge?E = 9.0 V; R = 17 Ω; C = 1.5 μF

E C R

0

0

6 5

exp

1exp

2

1ln ln 2

2

ln 2 17 1.5 10 F ln 2 1.7675 10 s

tQ Q

Q t

Q RC

t

RC

t RC

18 μst

Clicker Question

Which of the following are not units of RC (resistance × capacitance)?

A)

C)

B)

D)

E)

Ω F

1C A

s

1 1kg m s N

1C V

Stop here

Ammeters and VoltmetersAmmeter. •An ammeter measures current. • It must be placed in series with the circuit in which the current is to be measured.• The ideal ammeter has zero internal resistance.

Voltmeter. •A voltmeter measures voltage (potential difference). • It must be placed in parallel with the circuit in which the voltage is to be measured.• The ideal voltmeter has infinite internal resistance.

A Videal ammeter ideal voltmeter

Documents

Chapter 32 Fundamentals of Circuits L.A. Bumm. Schematic Symbols ideal wire with zero resistance