chapter 19 dc circuits. units of chapter 19 emf and terminal voltage resistors in series and in...

Chapter 19

DC Circuits

Units of Chapter 19

• EMF and Terminal Voltage

• Resistors in Series and in Parallel

• Kirchhoff’s Rules

• EMFs in Series and in Parallel; Charging a Battery

• Circuits Containing Capacitors in Series and in Parallel

Units of Chapter 19

• RC Circuits – Resistor and Capacitor in Series

• Electric Hazards

• Ammeters and Voltmeters

19.1 EMF and Terminal Voltage

Electric circuit needs battery or generator to produce current – these are called sources of emf.

Battery is a nearly constant voltage source, but does have a small internal resistance, which reduces the actual voltage from the ideal emf:

(19-1)

19.1 EMF and Terminal Voltage

This resistance behaves as though it were in series with the emf.

Example 19-1A 65.0 Ω resistor is connected to the terminals of a battery whose emf is 12.0 V and whose internal resistance is 0.5 Ω. Calculate (a) the current in the circuit, (b) the terminal voltage of the battery, Vab, and (c) the power dissipated in the battery’s internal resistance r.

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(a) Vab = ε - Ir and Vab = IR so IR = ε - Ir

I =ε

R +r=

12.0 V

65.0 Ω +0.5 Ω= 0.183 A

(b) Vab = ε - Ir =12.0 V - (0.183 A)(0.5 Ω) =11.9 V

(c) PR = I2R = (0.183 A) 2(65.0 Ω) = 2.18 W

Pr = I2r = (0.183 A) 2(0.5 Ω) = 0.02 W

19.2 Resistors in Series and in Parallel

A series connection has a single path from the battery, through each circuit element in turn, then back to the battery.


The current through each resistor is the same; the voltage depends on the resistance. The sum of the voltage drops across the resistors equals the battery voltage.

(19-2)


From this we get the equivalent resistance (that single resistance that gives the same current in the circuit).

(19-3)


A parallel connection splits the current; the voltage across each resistor is the same:


The total current is the sum of the currents across each resistor:


This gives the reciprocal of the equivalent resistance:

(19-4)


An analogy using water may be helpful in visualizing parallel circuits:

Example 19-3Two 100 Ω resistors are connected (a) in parallel, and (b) in series, to a 24.0 V battery. What is the current through each resistor and what is the equivalent resistance of each circuit?

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(a) I = I1 +I2 =V

R1

+V

R2

=24.0 V

100 Ω+

24.0 V

100 Ω= 0.48 A

R eq =V

I=

24.0 V

0.48 A= 50 Ω

(b) V = V1 +V2 = IR1 +IR2

I =V

R1 +R2

=24.0 V

100 Ω +100 Ω= 0.120 A

R =V

I=

24.0 V

0.120 A= 200 Ω

Example 19-4How much current is drawn from the battery shown in the figure?

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1

RP

=1

500 Ω+

1

700 Ω= 0.0034 Ω−1

RP =1

0.0034 Ω−1= 290 Ω

R eq = 400 Ω +290 Ω = 690 Ω

I =V

R eq

=12.0 V

690 Ω= 0.017 A =17 mA

Example 19-5What is the current through the 500 Ω resistor in the figure?

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Vab = IR = (0.017 A)(400 Ω) = 7.0 V

Vbc =12.0 V - 7.0 V = 5.0 V

I1 =5.0 V

500 Ω=1.0x10-2 A =10 mA

I2 =5.0 V

700 Ω= 7.0x10-3 A = 7 mA

Example 19-7A 9.0 V battery whose internal resistance r is 0.50 Ω is connected in the circuit shown. (a) How much current is drawn from the battery? (b) What is the terminal voltage of the battery? (c) What is the current in 6.0 Ω resistor?

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(a) 1

R eq1

=1

8.0 Ω+

1

4.0 Ω=

3

8.0 Ω so R = 2.7 Ω

R eq2 = 6.0 Ω +2.7 Ω = 8.7 Ω

1

R eq3

=1

10.0 Ω+

1

8.7 Ω= 0.21 Ω−1 so R = 4.8 Ω

R eq = 4.8 Ω +5.0 Ω +0.50 Ω =10.3 Ω

I =ε

R eq

=9.0 V

10.3 Ω= 0.87 A

(b) Vab = ε - Ir = 9.0 V - (0.87 A)(0.50 Ω) = 8.6 V

(c) Current across 6.0 Ω resistor is same as that

across 8.7 Ω resistor.

V8.7 = 9.0 V - (0.87 A)(0.50 Ω +5.0 Ω)

I'=9.0 V - (0.87 A)(0.50 Ω +5.0 Ω)

8.7 Ω= 0.48 A

19.3 Kirchhoff’s Rules

Some circuits cannot be broken down into series and parallel connections.


For these circuits we use Kirchhoff’s rules.

Junction rule: The sum of currents entering a junction equals the sum of the currents leaving it.


Loop rule: The sum of the changes in potential around a closed loop is zero.


Problem Solving: Kirchhoff’s Rules

1. Label each current.

2. Identify unknowns.

3. Apply junction and loop rules; you will need as many independent equations as there are unknowns.

4. Solve the equations, being careful with signs.

Example 19-8Calculate the currents I1, I2, and I3 in the three branches of the circuit shown .

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Junction rule : I3 = I1 +I2

Loop rule : - 30I1 + 45 - (40 +1)I3 = 0

Loop rule : - 30I1 +(20 +1)I2 - 80 = 0

From the 3rd equation, we have

I2 =80 + 30I1

21= 3.8 +1.4I1

From the 2nd equation, we have

I3 =45 - 30I1

41=1.1- 0.73I1

Using these in the first equation,

I1 = I3 - I2 =1.1- 0.73I1 - 3.8 -1.4I1

I1 = -0.87 A

I2 = 3.8 +1.4I1 = 3.8 +1.4(-0.87) = 2.6 A

I3 =1.1- 0.73I1 =1.1- 0.73(-0.87) =1.7 A

19.4 EMFs in Series and in Parallel; Charging a Battery

EMFs in series in the same direction: total voltage is the sum of the separate voltages


EMFs in series, opposite direction: total voltage is the difference, but the lower-voltage battery is charged.


EMFs in parallel only make sense if the voltages are the same; this arrangement can produce more current than a single emf.

Example 19-9A good car battery is being used to jump start a car with a weak battery. The good battery has an emf of 12.5 V and internal resistance 0.020 Ω. Suppose the weak battery has an emf of 10.1 V and internal resistance 0.10 Ω. Each copper jumper cable is 3.0 m long and 0.50 cm in diameter, and can be attached as shown. Assume the starter motor can be represented as a resistor Rs=0.15 Ω. Determine the current through the starter motor (a) if only the weak battery is connected to it, and (b) if the good battery is also connected.

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(a) R eq = 0.10 Ω +0.15 Ω = 0.25 Ω

I = V/R = (10.1 V)/(0.25 Ω) = 40 A

(b) R J = ρL

A= (1.68x10-8 Ωm)

(3.0 m)

(0.25x10-2 m)2= 0.0026 Ω

Loop rule : 12.5 V - I1(2R J +r1) - I3R s = 0

12.5 V - I1(0.025 Ω) - I3(0.15 Ω) = 0

Loop rule : 10.1 V - I3(0.15 Ω) - I2(0.10 Ω) = 0

Junction rule : I1 +I2 = I3

I3 = 71 A, I2 = -5A, I1 = 76 A

19.5 Circuits Containing Capacitors in Series and in Parallel

Capacitors in parallel have the same voltage across each one:


In this case, the total capacitance is the sum:

(19-5)


Capacitors in series have the same charge:


In this case, the reciprocals of the capacitances add to give the reciprocal of the equivalent capacitance:

(19-6)

Example 19-10Determine the capacitance of a single capacitor that will have the same effect as the combination shown in the figure. Take C1=C2=C3=C.

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C23 = C2 +C3 = C +C = 2C

1

Ceq

=1

C1

+1

C23

=1

C+

1

2C=

3

2C

Ceq =2

3C

Example 19-11Determine the charge on each capacitor in the figure, and the voltage across each, assuming C=3.0 µF and the battery voltage is V=4.0 V.

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Q = CV = (2.0 μF)(4.0 V) = 8.0 μC

This Q arrives at the negative plate of C1 so C1 = 8.0 μC.

The charge that leaves the positive plate is split between

C2 and C3, so they each equal 0.5Q1 = 4.0 μC.

V1 = Q1/C1 = (8.0 μC)/(3.0 μF) = 2.7 V

V2 = Q2/C2 = (4.0 μC)/(3.0 μF) =1.3 V

V3 = Q3/C3 = (4.0 μC)/(3.0 μF) =1.3 V

19.6 RC Circuits – Resistor and Capacitor in Series

When the switch is closed, the capacitor will begin to charge.


The voltage across the capacitor increases with time:

This is a type of exponential.


This curve has a characteristic time constant:

The charge follows a similar curve:

(19-7)


If an isolated charged capacitor is connected across a resistor, it discharges:

Example 19-12If a charged capacitor, C= 35 µF, is connected to a resistance R=120 Ω as in the figure, how much time will elapse until the voltage falls to 10% of its original (maximum) value?

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τ = RC = (120 Ω)(35x10-6 F) = 4.2x10-3 s

VC = V0(e -t/RC) so 0.10V0 = V0e-t/RC

e-t/RC = 0.10 and ln(e-t/RC) =-t

RC= ln0.10 = -2.3

t = 2.3(RC) = (2.3)(4.2x10-3 s) = 9.7x103 s = 9.7 ms

19.7 Electric Hazards

Even very small currents – 10 to 100 mA can be dangerous, disrupting the nervous system. Larger currents may also cause burns.

Household voltage can be lethal if you are wet and in good contact with the ground. Be careful!


A person receiving a shock has become part of a complete circuit.


Faulty wiring and improper grounding can be hazardous. Make sure electrical work is done by a professional.


The safest plugs are those with three prongs; they have a separate ground line.

Here is an example of household wiring – colors can vary, though! Be sure you know which is the hot wire before you do anything.

19.8 Ammeters and Voltmeters

An ammeter measures current; a voltmeter measures voltage. Both are based on galvanometers, unless they are digital.

The current in a circuit passes through the ammeter; the ammeter should have low resistance so as not to affect the current.


A voltmeter should not affect the voltage across the circuit element it is measuring; therefore its resistance should be very large.


An ohmmeter measures resistance; it requires a battery to provide a current


If the meter has too much or (in this case) too little resistance, it can affect the measurement.

Summary of Chapter 19

• A source of emf transforms energy from some other form to electrical energy

• A battery is a source of emf in parallel with an internal resistance

• Resistors in series:


• Resistors in parallel:

• Kirchhoff’s rules:

1. sum of currents entering a junction equals sum of currents leaving it

2. total potential difference around closed loop is zero


• Capacitors in parallel:

• Capacitors in series:


• RC circuit has a characteristic time constant:

• To avoid shocks, don’t allow your body to become part of a complete circuit

• Ammeter: measures current

• Voltmeter: measures voltage

Homework - Ch. 19

• Questions #’s 1, 2, 4, 5, 15, 16

• Problems #’s 3, 9, 11, 13, 17, 19, 21, 23, 25, 27, 29, 31, 35, 37, 41, 43, 51

chapter 19 dc circuits. units of chapter 19 emf and terminal voltage resistors in series and in...

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