1) The document summarizes key concepts about direct current circuits including resistors in series and parallel and calculations for total resistance, current, and voltage in various circuit configurations. 2) Key examples calculate total resistance, current, and voltages for circuits involving resistors connected in series and parallel with batteries and sources of emf. Internal resistances of batteries are also considered. 3) Methods for analyzing more complex circuits involving both series and parallel connections are demonstrated through worked examples. Calculations include determining equivalent resistances, current, voltage, power, and addressing relationships between open circuit potential, terminal potential and internal resistance.

1. Chapter 28. Direct-Current Circuits Physics, 6th Edition
Chapter 28. Direct-Current Circuits
Resistors in Series and Parallel (Ignore internal resistances for batteries in this section.)
28-1. A 5-Ω resistor is connected in series with a 3-Ω resistor and a 16-V battery. What is the
effective resistance and what is the current in the circuit?
3Ω 5Ω
Re = R1 + R2 = 3 Ω +5 Ω; Re = 8.00 Ω
V 16 V
I= = I = 2.00 A
R 8Ω 16 V
28-2. A 15-Ω resistor is connected in parallel with a 30-Ω resistor and a 30-V source of emf.
What is the effective resistance and what total current is delivered?
R1 R2 (15 Ω)(30 Ω)
Re = = ; Re = 10.0 Ω
R1 + R2 15 Ω + 30 Ω 15 Ω 30 Ω
30 V
V 30 V
I= = ; I = 3.00 A
R 10 Ω
28-3. In Problem 28-2, what is the current in 15 and 30-Ω resistors?
30 V
For Parallel: V15 = V30 = 30 V; I15 = ; I15 = 2.00 A
15 Ω
30 V
I 30 = ; I30 = 1.00 A Note: I15 + I30 = IT = 3.00 A
30 Ω
28-4. What is the equivalent resistance of 2, 4, and 6-Ω resistors connected in parallel?
2Ω 4Ω 6Ω
Re = 2 Ω + 4 Ω + 6 Ω; Re = 12.0 Ω
129

2. Chapter 28. Direct-Current Circuits Physics, 6th Edition
28-5. An 18-Ω resistor and a 9-Ω resistor are first connected in parallel and then in series with
a 24-V battery. What is the effective resistance for each connection? Neglecting internal
resistance, what is the total current delivered by the battery in each case?
R1 R2 (18 Ω)(9 Ω) 30 V
Re = = ; Re = 6.00 Ω
R1 + R2 18 Ω + 9 Ω 9Ω
18 Ω
V 24 V
I= = ; I = 4.00 A
R 6.00 Ω
Re = R1 + R2 = 18 Ω +9 Ω; Re = 27.0 Ω 18 Ω 9Ω
V 24 V
I= = I = 0.889 A 24 V
R 27 Ω
28-6. A 12-Ω resistor and an 8-Ω resistor are first connected in parallel and then in series with
a 28-V source of emf. What is the effective resistance and total current in each case?
R1 R2 (12 Ω)(8 Ω)
Re = = ; Re = 4.80 Ω 12 Ω 8Ω
R1 + R2 12 Ω + 8 Ω 28 V
8Ω
12 Ω 28 V
V 28 V
I= = ; I = 5.83 A
R 4.80 Ω
V 28 V
Re = R1 + R2 = 12 Ω +8 Ω; Re = 20.0 Ω I= = I = 1.40 A
R 20 Ω
28-7. An 8-Ω resistor and a 3-Ω resistor are first connected in parallel and then in series with a
12-V source. Find the effective resistance and total current for each connection?
(3 Ω)(8 Ω) V 12 V
Re = ; Re = 2.18 Ω I= = ; I = 5.50 A
3Ω+8Ω R 2.18 Ω
V 12 V
Re = R1 + R2 = 3 Ω +8 Ω; Re = 11.0 Ω I= = I = 1.09 A
R 11 Ω
130

3. Chapter 28. Direct-Current Circuits Physics, 6th Edition
28-8. Given three resistors of 80, 60, and 40 Ω, find their effective resistance when connected
in series and when connected in parallel.
Series: Re = 80 Ω + 60 Ω + 40 Ω ; Re = 180 Ω
1 1 1 1 1
Parallel: =∑ = + + ; Re = 18.5 Ω
Re Ri 80 Ω 60 Ω 40 Ω
28-9. Three resistances of 4, 9, and 11 Ω are connected first in series and then in parallel. Find
the effective resistance for each connection.
Series: Re = 4 Ω + 9 Ω + 11 Ω ; Re = 24.0 Ω
1 1 1 1 1
Parallel: =∑ = + + ; Re = 2.21 Ω
Re Ri 4 Ω 9 Ω 11 Ω
*28-10. A 9-Ω resistor is connected in series with two parallel resistors of 6 and 12 Ω. What is
the terminal potential difference if the total current from the battery is 4 A?
(6 Ω)(12 Ω) 9Ω
Re = = 4 Ω; Re = 4 Ω + 9 Ω = 13 Ω 6Ω
6 Ω + 12 Ω
12 Ω
VT = IR = (4 A)(13 Ω); VT = 52.0 V VT
4A
*28-11. For the circuit described in Problem 28-10, what is the voltage across the 9-Ω resistor
and what is the current through the 6-Ω resistor?
V9 = (4 A)(9 Ω) = 36 V; V9 = 36.0 V
The rest of the 52 V drops across each of the parallel resistors:
V6 = V7 = 52 V – 36 V; V6 = 16 V
V6 16 V
I6 = = ; I6 = 2.67 A
R6 6 Ω
131

4. Chapter 28. Direct-Current Circuits Physics, 6th Edition
*28-12. Find the equivalent resistance of the circuit drawn in Fig. 28-19.
Start at far right and reduce circuit in steps: R’ = 1 Ω + 3 Ω + 2 Ω = 6 Ω;
(6 Ω)(3 Ω)
R '' = = 2 Ω ; Re = 2 Ω + 4 Ω + 2 Ω ; Re = 8 Ω
6Ω+3Ω
4Ω 1Ω 4Ω 4Ω
3Ω 3Ω 3Ω 6Ω 2Ω 8Ω
2Ω 2Ω 2Ω 2Ω
*28-13. Find the equivalent resistance of the circuit shown in Fig. 28-20.
Start at far right and reduce circuit in steps: R = 1 Ω + 2 Ω = 3 Ω;
(6 Ω)(3 Ω)
R' = = 2 Ω ; R’’ = 2 Ω + 3 Ω = 5 Ω
6Ω+3Ω
(5 Ω)(4 Ω)
Re = = 2.22 Ω ; Re = 2.22 Ω
5Ω+4Ω
1Ω
4Ω
4Ω 4Ω 6Ω 3Ω 4Ω 5Ω Re
6Ω 2Ω
3Ω 2Ω 3Ω 3Ω
*28-14. If a potential difference of 24 V is applied to the circuit drawn in Fig. 28-19, what is the
4Ω 1Ω 4Ω
current and voltage across the 1-Ω resistor?
24 V 24 V
24 V 3Ω 3Ω 3Ω 6Ω
Re = 8.00 Ω; I= = 3.00 A;
8Ω
2Ω 2Ω 2Ω
The voltage across the 3 and 6-Ω parallel 4Ω
connection is found from It and the 2-Ω combination resistance: 24 V
2Ω
6V
V3 = V6 = (2 Ω)(3.00 A); V6 = 6.00 V; I6 = = 1A
6Ω 2Ω
Thus, I1 = I6 = 1.00 A, and V1 = (1 A)(1 Ω) = 1 V; V1 = 1 V; I1 = 1 A
132

5. Chapter 28. Direct-Current Circuits Physics, 6th Edition
*28-15. If a potential difference of 12-V is applied to the free ends in Fig. 28-20, what is the
current and voltage across the 2-Ω resistor? 1Ω
12 V 12 V 4Ω
Re = 2.22 Ω; I= = 5.40 A; 6Ω
2.22 Ω
3Ω 2Ω
12 V
Note that V5 = 12 V; I5 = = 2.40 A
5Ω 12 V 4Ω
4.8 V 3Ω
6Ω
V3,6 = (2.4 A)(2 Ω) = 4.80 V; I3 = = 1.6 A
3Ω
3Ω
4 Ω 5Ω
I2 = I1 = 1.60 A; V2 = (1.6 A)(2 Ω) = 3.20 V
12 V 4Ω 12 V
I2 = 1.60 A; V2 = 3.20 V 2Ω
3Ω
EMF and Terminal Potential Difference
28-16. A load resistance of 8 Ω is connected in series with a 18-V battery whose internal
resistance is 1.0 Ω. What current is delivered and what is the terminal voltage?
E 18 V
I= = ; I = 2.00 A
r + RL 1.0 Ω + 8 Ω
28-17. A resistance of 6 Ω is placed across a 12-V battery whose internal resistance is 0.3 Ω.
What is the current delivered to the circuit? What is the terminal potential difference?
E 12 V
I= = ; I = 1.90 A
r + RL 0.3 Ω + 6 Ω
VT = E – Ir = 12 V – (1.90 A)(0.3 Ω); VT = 11.4 V
133

6. Chapter 28. Direct-Current Circuits Physics, 6th Edition
28-18. Two resistors of 7 and 14 Ω are connected in parallel with a 16-V battery whose internal
resistance is 0.25 Ω. What is the terminal potential difference and the current in
delivered to the circuit? 16 V
7Ω 14 Ω
(7 Ω)(14 Ω) 0.25 Ω
R' = = 4.67 Ω ; Re = 0.25 Ω + 4.67 Ω
7 Ω + 14 Ω
E 16 V
I= = ; I = 3.25 A VT = E – Ir = 16 V – (3.25 A)(0.25 Ω);
r + R ' 4.917 Ω
VT = 15.2 V; I = 3.25 A
28-19. The open-circuit potential difference of a battery is 6 V. The current delivered to a 4-Ω
resistor is 1.40 A. What is the internal resistance?
E = IRL + Ir; Ir = E - IRL
E − IRL 6 V - (1.40 A)(4 Ω)
r= = ; r = 0.286 Ω
I 1.40 A
28-20. A dc motor draws 20 A from a 120-V dc line. If the internal resistance is 0.2 Ω, what is
the terminal voltage of the motor?
VT = E – Ir = 120 V – (20A)(0.2 Ω); VT = 116 V
28-21. For the motor in Problem 28-21, what is the electric power drawn from the line? What
portion of this power is dissipated because of heat losses? What power is delivered by
the motor?
Pi = EI = (120 V)(20 A); Pi = 2400 W
PL = I2r = (20 A)2(0.2); PL = 80 W
Po = VTI = (116 V)(20 A); Po = 2320 W;
Note: Pi = PL + Po; 2400 W = 80 W + 2320 W
134

7. Chapter 28. Direct-Current Circuits Physics, 6th Edition
28-22. A 2-Ω and a 6-Ω resistor are connected in series with a 24-V battery of internal
resistance 0.5 Ω. What is the terminal voltage and the power lost to internal resistance?
E 24 V
Re = 2 Ω + 6 Ω + 0.5 Ω = 8.50 Ω; I= = = 2.82 A
Re 8.5 Ω
VT = E – Ir = 24 V – (2.82 A)(0.5 Ω); VT = 22.6 V
PL = I2 r = (2.82 A)2 (0.5 Ω); PL = 3.99 W
*28-23. Determine the total current and the current through each resistor for Fig. 28-21 when
E = 24 V, R1 = 6 Ω, R2 = 3 Ω, R3 = 1 Ω, R4 = 2 Ω, and r = 0.4 Ω.
R3 = 1 Ω
24 V
(3 Ω)(6 Ω)
R1,2 = = 2 Ω ; R1,2,3 = 2 Ω + 1 Ω = 3 Ω R1 R2 2Ω
3Ω + 6 Ω 6Ω R4
3Ω 0.4 Ω
(3 Ω)(2 Ω) R3 = 1 Ω
Re = = 1.20 Ω; 24 V 24 V
3Ω + 2 Ω 3Ω
2Ω 2Ω
2Ω R4 R4 0.4 Ω
0.4 Ω
Re = 1.20 Ω + 0.4 Ω = 1.60 Ω
24 V 24 V 24 V
IT = ; IT = 15.0 A
1.60 Ω 1.2 Ω 1.6 Ω
0.4 Ω
18 V
V4 = V1.2= (1.2 Ω)(15 A) = 18 V I4 =
2Ω
I4 = 9.0 A; I3 = 15 A – 9 A = 6 A; V3 = (6 A)(1 Ω) = 6 V; V1 = V2 = 18 V – 6 V;
12 V 12 V
V1 = V2 = 12 V; I2 = = 4 A; I1 = =2A;
3Ω 6Ω
IT = 15 A, I1 = 2 A, I2 = 4 Ω, I3 = 6 Α, I4 = 9 A.
The solution is easier using Kirchhoff’s laws, developed later in this chapter.
135

8. Chapter 28. Direct-Current Circuits Physics, 6th Edition
*28-24. Find the total current and the current through each resistor for Fig. 28-21 when E = 50 V,
R1 = 12 Ω, R2 = 6 Ω, R3 = 6 Ω, R4 = 8 Ω, and r = 0.4 Ω. R3 = 6 Ω
50 V
(12 Ω)(6 Ω) R1 R2 8Ω
R1,2 = = 4 Ω ; R1,2,3 = 4 Ω + 6 Ω = 10 Ω 12 Ω R4
12 Ω + 6 Ω 6Ω 0.4 Ω
R3 = 6 Ω
(10 Ω)(8 Ω) 50 V 10 Ω
50 V
R= = 4.44 Ω
10 Ω + 8 Ω 4Ω 8Ω
4Ω R4 R4 0.4 Ω
0.4 Ω
Re = 4.44 Ω + 0.4 Ω = 4.84 Ω
24 V 50 V
50 V 1.6 Ω
IT = ; IT = 10.3 A 2.86 Ω
4.84 Ω 0.4 Ω
45.9 V
V4 = Vp= (4.44 Ω)(10.3 A) = 45.9 V I4 =
8Ω
I4 = 5.73 A; I3 = 10.3 A – 5.73 A = 4.59 A; V3 = (4.59 A)(6 Ω) = 27.5 V;
18.4 V 18.4 V
V1 = V2 = 45.9 V – 27.5 V = 18.4 V; I2 = = 3.06 A; I1 = = 1.53 A ;
6Ω 12 Ω
IT = 10.3 A, I1 = 1.53 A, I2 = 3.06 Ω, I3 = 4.59 Α, I4 = 5.73 A.
Kirchhoff’s Laws
28-25. Apply Kirchhoff’s second rule to the current loop in Fig. 28-22. What is the net voltage
around the loop? What is the net IR drop? What is the current in the loop?
Indicate output directions of emf’s, assume direction of 20 V
+
2Ω
current, and trace in a clockwise direction for loop rule:
I
ΣE = ΣIR; 20 V – 4 V = I(6 Ω) + I(2 Ω);
6Ω
4V
16 V
(8 Ω)I = 16 V; I= ; I = 2.00 A
8Ω
Net voltage drop = Σ E = 16 V; ΣIR = (8 Ω)(2 A) = 16 V
136

9. Chapter 28. Direct-Current Circuits Physics, 6th Edition
28-26. Answer the same questions for Problem 28-25 where the polarity of the 20-V battery is
changed, that is, its output direction is now to the left? ( Refer to Fig. in Prob. 28-25. )
Σ E = -20 V – 4 V = -24 V; ΣIR = I(2 Ω) + I(6 Ω = (8Ω)I
ΣE = ΣIR; -24 V = (8 Ω)I; I = -4.00 A; ΣIR = (8 Ω)(-4 A) = -24 V
The minus sign means the current is counterclockwise (against the assume direction)
*28-27. Use Kirchhoff’s laws to solve for the currents through the circuit shown as Fig. 28-23.
4Ω
First law at point P: I1 + I2 = I3 Current rule
I1
nd
Loop A (2 law): ΣE = ΣIR Loop rule 2 ΩP A 5V
Ω
6I
5 V – 4 V = (4 Ω)I1 + (2 Ω)I1 – (6 Ω)I2 2
4V I2
Simplifying we obtain: (1) 6I1 – 6I2 = 1 A
3Ω B I3
1Ω
Loop B: 4 V – 3 V = (6 Ω)I2 + (3 Ω)I3 + (1 Ω)I3
Simplifying: (2) 6I2 + 4I3 = 1 A, but I3 = I1 + I2 3V
Substituting we have: 6I2 + 4(I1 + I2) = 1 A or (3) 4I1 + 10I2 = 1 A
From which, I1 = 0.25 A – 2.5 I2 ; Substituting into (1): 6(0.25 A – 2.5I2 ) – 6 I2 = 1 A
1.5 A – 15I2 – 6I2 = 1 A; -21I2 = -0.5 A; I2 = 0.00238 A; I2 = 23.8 mA
Putting this into (1), we have: 6I1 – 6(0.0238 A) = 1 A, and I1 = 190 mA
Now, I1 + I2 = I3 so that I3 = 23.8 mA + 190 mA or I3 = 214 mA
3Ω
*28-28. Use Kirchhoff’s laws to solve for the currents in Fig. 28-24. I1
Current rule: I1 + I3 = I2 or (1) I3 = I2 – I1 20 V A 5V
4Ω
I2
Loop A: 20 V = (3 Ω)I1 + (4 Ω)I2 ; (2) 3I1 + 4I2 = 20 A P
4V I2
Loop B: 8 V = (6 Ω)I3 + (4 Ω)I2; (3) 3I3 + 2I2 = 4 A 2Ω B 4Ω
I3
Outside Loop: 20 V – 8 V = (3 Ω)Ι1 – (6 Ω)Ι3 or I1 – 2 I3 = 4 A
8V
137

10. Chapter 28. Direct-Current Circuits Physics, 6th Edition
*28-28. (Cont.) (3) 3I3 + 2I2 = 4 A and I3 = I2 – I1
3Ω
3(I2 – I1) + 2I2 = 4 A; 3I1 = 5I2 – 4 A I1
(2) 3I1 + 4I2 = 20 A; (5I2 – 4 A) + 4I2 = 20; 20 V A 5V
4Ω
I2
P
I2 = 2.67 A; 3I1 = 5(2.67 A) – 4 A; I1 = 3.11 A
4V I2
I3 = I2 – I1 = 2.67 A – 3.11 A = -0.444 A 2Ω B 4Ω
I3
Note: I3 goes in opposite direction to that assumed.
8V
I1 = 3.11 A, I2 = 2.67 A, I3 = 0.444 A
*28-29. Apply Kirchhoff’s laws to the circuit of Fig. 28-25. Find the currents in each branch.
Current rule: (1) I1 + I4 = I2 + I3 1.5 Ω 3V
Ω
Applying loop rule gives six possible equations: 3Ω I1
(2) 1.5I1 + 3I2 = 3 A; (3) 3I2 – 5I3 = 0 I2
5Ω I3
(4) 5I3 + 6I4 = 6A; (5) 1.5I1 – 6I4 = -3A
I4
6V 6Ω
(6) 6I4 + 3I2 = 6 A (7) 1.5I1 + 5I3 = 3A
Put I4 = I2 + I3 – I1; into (4): 5I3 + 6(I2 + I3 – I1) = 6 A  -6I1 + 6I2 + 11I3 = 6 A
Now, solving (2) for I1 gives: I1 = 2 A – 2I2, which can be used in the above equation.
-6(2 A – 2I2) + 6I2 + 11I3 = 6 A, which gives: 18I2 + 11I3 = 18 A
But, from (3), we put I2 = 5
3 I 3 into above equation to find that: I3 = 0.439 A
From (2): 1.5I1 + 3(0.439 A) = 3 A; and I1 = 0.536 A
From (3): 3I2 – 5(0.439 A) = 0; and I2 = 0.736 A
From (4): 5(0.439 A) + 6I4 = 6 A; and I4 = 0.634 A
Currents in each branch are: I1 = 536 mA, I2 = 732 mA, I3 = 439 mA, I4 = 634 mA
Note: Not all of the equations are independent. Elimination of two may yield another.
It is best to start with the current rule, and use it to eliminate one of the currents quickly.
138

11. Chapter 28. Direct-Current Circuits Physics, 6th Edition
The Wheatstone Bridge
28-30. A Wheatstone bridge is used to measure the resistance Rx of a coil of wire. The
resistance box is adjusted for 6 Ω, and the contact key is positioned at the 45 cm mark
when measured from point A of Fig. 28-13. Find Rx. ( Note: l1 + l2 = 100 cm )
R3l2 (6 Ω)(55 cm)
Rx = = ; Rx = 7.33 Ω
l1 (45 cm)
28-31. Commercially available Wheatstone bridges are portable and have a self-contained
galvanometer. The ratio R2/R1 can be set at any integral power of ten between 0.001 and
1000 by a single dual switch. When this ratio is set to 100 and the known resistance R is
adjusted to 46.7 Ω, the galvanometer current is zero. What is the unknown resistance?
R2
Rx = R3 = (46.7 Ω)(100) ; Rx = 4670 Ω
R1
28-32. In a commercial Wheatstone bridge, R1 and R2 have the resistances of 20 and 40 Ω,
respectively. If the resistance Rx is 14 Ω, what must be the known resistance R3 for zero
galvanometer deflection?
R2 20 Ω
Rx = R3 = (14 Ω) ; Rx = 7.00 Ω
R1 40 Ω
Challenge Problems:
28-33. Resistances of 3, 6, and 9 Ω are first connected in series and then in parallel with an 36-V
source of potential difference. Neglecting internal resistance, what is the current leaving
the positive terminal of the battery?
36 V
Re = ΣRi = 3 Ω + 6 Ω + 9 Ω = 18 Ω; I= ; I = 2.00 A
18 Ω
139

12. Chapter 28. Direct-Current Circuits Physics, 6th Edition
1 1 1 1 1 36 V
28-33. (Cont.) =∑ = + + ; Re = 1.64 Ω; I = ; I = 22.0 A
Re Ri 3 Ω 6 Ω 9 Ω 1.64 Ω
28-34. Three 3-Ω resistors are connected in parallel. This combination is then placed in series
with another 3-Ω resistor. What is the equivalent resistance?
3Ω
1 1 1 1
= + + = 1 Ω; Re = R’ + 3 Ω
R' 3 Ω 3 Ω 3 Ω V
3Ω 3Ω 3Ω
Re = 1 Ω + 3 Ω; Re = 4 Ω
*28-35. Three resistors of 4, 8, and 12 Ω are connected in series with a battery. A switch allows
the battery to be connected or disconnected from the circuit? When the switch is open, a
voltmeter across the terminals of the battery reads 50 V. When the switch is closed, the
voltmeter reads 48 V. What is the internal resistance in the battery?
RL = 4 Ω + 8 Ω + 12 Ω = 24 Ω; E = 50 V; VT = 48 V = IRL
48 V
I= = 2.00 A ; E – VT = Ir; 50 V – 48 V = Ir
24 V
50 V - 48 V
r= ; r = 1.00 Ω
2.00 A
*28-36. The generator in Fig. 28-26 develops an emf of E1 = 24 V and has an internal resistance
of 0.2 Ω. The generator is used to charge a battery E2 = 12 V whose internal resistance
is 0.3 Ω. Assume that R1 = 4 Ω and R2 = 6 Ω. What is the terminal voltage across the
generator? What is the terminal voltage across the battery?
24 V - 12 V
I= = 1.14 A
6 Ω + 4 Ω + 0.2 Ω + 0.3 Ω 12 V
4Ω
V1= E1 – Ir = 24 V – (1.14 A)(0.2 Ω) = 23.8 V; 24 V r
r 6Ω
V2 = 12 V + (1.14 A)(0.3 Ω) = 12.3 V
140

13. Chapter 28. Direct-Current Circuits Physics, 6th Edition
*28-37. What is the power consumed in charging the battery for Problem 28-36. Show that the
power delivered by the generator is equal to the power losses due to resistance and the
power consumed in charging the battery.
P = E I = (24 V)(1.143 A) Pe = 27.43 W 12 V
4Ω
PR= I2 Re = (1.143 A)2 (10.5 Ω); PR = 13.69 W 24 V r
r 6Ω
PV = (12 V)(1.143 A) = 13.72 W; Pe = PR + PV;
27.4 W = 13.7 W + 13.7 W
*28-38. Assume the following values for the parameters of the circuit illustrated in Fig. 28-8:
E1 = 100 V, E2 = 20 V, r1 = 0.3 Ω. r2 = 0.4 Ω, and R = 4 Ω. What are the terminal
voltages V1 and V2? What is the power lost through the 4-Ω resistor?
100 V - 20 V 20 V
I= = 17.0 A 100 V
4 Ω + 0.3 Ω + 0.4 Ω
r 4Ω r
V1= E1 – Ir = 100 V – (17.0 A)(0.3 Ω) = 94.9 V;
V2 = 20 V + (17.0 A)(0.4 Ω) = 26.8 V; P = I2 R = (17 A)2(4 Ω) = 1160 W
*28-39. Solve for the currents in each branch for Fig. 28-27. 5Ω 12 V
Current rule: I1 = I2 + I3 ; Loops: Σ E = Σ IR’s I1
A
(1) 5I1 + 10I2 = 12 A; (2) -10I2 + 20I3 = 6 A 10 Ω
I2
P
(2) -5I2 + 10I3 = 3 A; (3) 5I1 + 20I3 = 18 A I2
4V
From (1): 5(I2 + I3) + 10 I2 = 12 A  15I2 + 5I3 = 12 A
20 Ω B I3
Multiplying this equation by –2: -30I2 – 10I3 = -24 A
6V
Now add this to (2): -35I2 + 0 = -21 A and I2 = 0.600 A
Now, from (1) and from (2): I1 = 1.20 A and I3 = 0.600 A
141

14. Chapter 28. Direct-Current Circuits Physics, 6th Edition
*28-40. If the current in the 6-Ω resistor of Fig. 28-28 is 2 A, what is the emf of the battery?
Neglect internal resistance. What is the power loss through the 1-Ω resistor?
2A 1Ω
V6 = (2 A)(6 Ω) = 12 V; V3 = 12 V = I3(3 Ω)
IT E
12 V 6Ω 3Ω
I3 = = 4 A ; IT = 2 A + 4 A = 6 A; I3
3Ω
V1 = (1 Ω)(6 A) = 6 V; E = 6 V + 12 V = 18 V; P = IT2 R = (6 A)2(1 Ω) = 36 W
Critical Thinking Problems
*28-41. A three-way light bulb uses two resistors, a 50-W filament and a 100-W filament. A
three-way switch allows each to be connected in series and provides a third possibility by
connecting the two filaments in parallel? Draw a possible arrangement of switches than
will accomplish these tasks. Assume that the household voltage is 120 V. What are the
resistances of each filament? What is the power of the parallel combination?
I1
Switch can be set at A, B, or C to give three possibilities:
IT
V2 (120 V) 2 (120 V) 2 R1 R2
P= ; R1 = = 288 Ω; R2 = = 144 Ω I2
R 50 W 100 W B 120 V
(288 Ω)(144 Ω) A C
For Parallel: Re = = 96 Ω
288 Ω + 144 Ω
V 2 (120 V) 2
P= = ; P = 150 W
R 96 Ω
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15. Chapter 28. Direct-Current Circuits Physics, 6th Edition
*28-42. The circuit illustrated in Fig. 28-7 consists of a 12-V battery, a 4-Ω resistor, and a
switch. When new, the internal resistance of the battery is 0.4 Ω, and a voltmeter is
placed across the terminals of the battery. What will be the reading of the voltmeter
when the switch is open and when it is closed? After a long period of time, the
experiment is repeated and it is noted that open circuit reading is unchanged, but the
terminal voltage has reduced by 10 percent. How do you explain the lower terminal
voltage? What is the internal resistance of the old battery?
E 12 V
When new: I = = ; I = 2.73 A 4Ω
R + r 4 Ω + 0.4 Ω
VT = E – Ir = 12 V – (2.73 A)(0.4 Ω); VT = 10.9 V
VT reduced by 10% due to increase of rint. V’ = 10.9 V – 0.1(10.9 V); V’ = 9.81 V
9.81 V E −V '
V’ = IRL = 9.81 V; I= = 2.45 A; V ' = E − Ir ; r=
4Ω I
E − V ' 12 V - 9.81 V
r= = ; r = 0.893 Ω
I 2.45 A
*28-43. Given three resistors of 3, 9, and 18 Ω, list all the possible equivalent resistances that
can be obtained through various connections?
1 1 1 1
All in parallel: = + + ; Re = 2 Ω
Re 3 Ω 9 Ω 18 Ω
All in series: Re = R1 + R2 + R3 = 3 Ω + 9 Ω + 18 Ω; Re = 30 Ω
(3 Ω)(9 Ω)
Parallel (3,9) in series with (18): Re = + 18 Ω; Re = 20.2 Ω
3Ω+9Ω
(3 Ω)(18 Ω)
Parallel (3,18) in series with (9): Re = + 9 Ω; Re = 11.6 Ω
3 Ω + 18 Ω
(9 Ω)(18 Ω)
Parallel (9,18) in series with (3): Re = + 3 Ω; Re = 9.00 Ω
9 Ω + 18 Ω
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16. Chapter 28. Direct-Current Circuits Physics, 6th Edition
(12 Ω)(18 Ω)
*28-43 (Cont.)Series (3 + 9) in parallel with (18): Re = ; Re = 20.2 Ω
12 Ω + 18 Ω
(9 Ω)(21 Ω)
Series (3 + 18) in parallel with (9): Re = ; Re = 6.30 Ω
9 Ω + 21 Ω
(3 Ω)(27 Ω)
Series (9 + 18) in parallel with (3): Re = ; Re = 2.70 Ω
3 Ω + 27 Ω
*28-44. Refer to Fig. 28-21, assume that E = 24 V, R1 = 8 Ω, R2 = 3 Ω, R3 = 2 Ω, R4 = 4 Ω, and
r = 0.5 Ω. What current is delivered to the circuit by the 24-V battery? What are the
voltage and current for the 8-Ω resistor? R3 = 2 Ω
24 V
(3 Ω)(8 Ω) 4Ω
R1,2 = = 2.18 Ω ; R1,2,3 = 2.18 Ω + 2 Ω = 4.18 Ω 8Ω R2
3 Ω+8 Ω R4 0.5 Ω
R1 3Ω
R3 = 2 Ω
(4.18 Ω)(4 Ω)
Re = = 2.04 Ω; 24 V 4.18 Ω
24 V
4.18 Ω + 4 Ω R4 4Ω 4Ω
2.18 Ω R4 0.5 Ω
0.5 Ω
Re = 2.04 Ω + 0.5 Ω = 2.54 Ω
24 V 24 V
24 V
IT = ; IT = 9.43 A 2.54 Ω
2.54 Ω 2.04 Ω
0.5 Ω
V4 = V1,2,3 = (9.43 A) (2.04 Ω) = 19.3 V
19.3 V
I4 = = 4.82 A ; I3 = 9.43 A – 4.82 A = 4.61 A; V3 = (4.61 A)(2 Ω) = 9.23 V
4Ω
10.1 V
V1 = V2 = 19.3 V – 9.23 V; V1 = V2 = 10.1 V; I1 = = 1.26 A ;
8Ω
Finally, for the 8-Ω resistor: V1 = 10.1 V and I1 = 1.26 A
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17. Chapter 28. Direct-Current Circuits Physics, 6th Edition
*28-45. What is the effective resistance of the external circuit for Fig. 28-29 if internal resistance
is neglected. What is the current through the 1-Ω resistor?
4Ω 1Ω 4Ω 1Ω 4Ω 1Ω
R1 R3 R5
24 V 24 V 24 V
6Ω 3Ω 6Ω 8Ω 3Ω 6Ω 2.18 Ω
8Ω R2 R4
(3 Ω)(8 Ω)
R4,5 = = 2.18 Ω ; R3,4,5 = 2.18 Ω + 1 Ω = 3.18 Ω 4Ω
3 Ω+8 Ω
24 V
6Ω 3.18 Ω
(3.18 Ω)(6 Ω)
R1,2,3,4 = ; R1,2,3,4 = 2.08 Ω
3.18 Ω + 6 Ω
Re = 2.08 Ω + 4 Ω; Re = 6.08 Ω 4Ω
24 V 24 V
I3 in the 1-Ω resistor is same as I in R3,4,5
24 V
IT = = 3.95 A
6.08 Ω
V2,3,4,5 = (3.95 A)(2.08 Ω) = 8.21 V; Also V3,4,5 = 8.21 V
8.21 V
I 3,4,5 = = 2.58 A; Therefore I3 = 2.58 A in 1-Ω resistor
3.18 Ω
Re = 6.08 Ω; I3 = 2.58 A
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