CHAPTER 28
ELECTRIC CIRCUITS
Problem
1.
Sketch a circuit diagram for a circuit that includes a resistor R1 connected to the positive terminal of a battery, a pair of
parallel resistors R2 and R3 connected to the lower-voltage end of R1 , then returned to the battery’s negative terminal,
and a capacitor across R2 .
Solution
A literal reading of the circuit specifications results in connections like those in sketch (a). Because the connecting wires are
assumed to have no resistance (a real wire is represented by a separate resistor), a topologically equivalent circuit diagram is
shown in sketch (b).
Problem 1 Solution (a).
Problem 1 Solution (b).
Problem
2.
A circuit consists of two batteries, a resistor, and a capacitor, all in series. Sketch this circuit. Does the description allow
any flexibility in how you draw the circuit?
Solution
In a series circuit, the same current must flow through all elements. One possibility is shown. The order of elements and the
polarity of the battery connections are not specified.
Problem 2 Solution.
Problem
3.
Resistors R1 and R2 are connected in series, and this series combination is in parallel with R3 . This parallel combination
is connected across a battery whose internal resistance is Rint . Draw a diagram representing this circuit.
660 CHAPTER 28
Solution
The circuit has three parallel branches: one with R1 and R2 in series; one with just R3 ; and one with the battery (an ideal emf
in series with the internal resistance).
Problem 3 Solution.
Section 28-2:
Electromotive Force
Problem
4.
What is the emf of a battery that delivers 27 J of energy as it moves 3.0 C between its terminals?
Solution
From the definition of emf (as work per unit charge), E = W=q = 27 J=3 C = 9 V.
Problem
5.
A 1.5-V battery stores 4.5 kJ of energy. How long can it light a flashlight bulb that draws 0.60 A?
Solution
The average power, supplied by the battery to the bulb, multiplied by the time equals the energy capacity of the battery. For an
ideal battery, P = EI, therefore EIt = 4.5 kJ, or t = 4.5 kJ=(1.5 V )( 0.60 A ) = 5 × 10 3 s = 1.39 h.
Problem
7.
A battery stores 50 W ⋅ h of chemical energy. If it uses up this energy moving 3.0 × 10 4 C through a circuit, what is its
voltage?
Solution
The emf is the energy (work done going through the source from the negative to the positive terminal) per unit charge:
E = (50 W ⋅ h )(3600 s/h )=(3 × 10 4 C) = 6 V. (This is the average emf; the actual emf may vary with time.)
Problem
9.
What resistance should be placed in parallel with a 56-kΩ resistor to make an equivalent resistance of 45 kΩ?
Solution
The solution for R2 in Equation 28-3a is R2 = R1 Rparallel=( R1 − Rparallel ) = (56 kΩ)( 45)=(56 − 45) = 229 kΩ.
Problem
11. In Fig. 28-49, take all resistors to be 1.0 Ω. If a 6.0-V battery is connected between points A and B, what will be the
current in the vertical resistor?
CHAPTER 28 661
Solution
The circuit in Fig. 28-49, with a battery connected across points A and B, is similar to the circuit analysed in Example 28-4. In
this case, R|| = (1 Ω)(2 )=(1 + 2 ) = ( 23 ) Ω, and Rtot = 1 Ω + 1 Ω + 23 Ω = 83 Ω. The total current (that through the battery) is
I tot = E=Rtot = 6 V=( 83 Ω) = ( 94 ) A. The voltage across the parallel combination is I tot R || = ( 94 A)( 23 Ω) = 23 V, which is the
voltage across the vertical 1 Ω resistor. The current through this resistor is then ( 23 V )=(1 Ω) = 1.5 A.
Problem
17. A partially discharged car battery can be modeled as a 9-V emf in series with an internal resistance of 0.08 Ω. Jumper
cables are used to connect this battery to a fully charged battery, modeled as a 12-V emf in series with a 0.02-Ω internal
resistance. How much current flows through the discharged battery?
Solution
Terminals of like polarity are connected with jumpers of negligible resistance. Kirchhoff’s voltage law gives E1 − E2 −
IR1 − IR2 = 0, or I = (E1 − E2 )=( R1 + R2 ) = (12 − 9) V=( 0.02 + 0.08) Ω = 30 A.
Problem 17 Solution.
Problem
19. What is the equivalent resistance between A and B in each of the circuits shown in Fig. 28-50? Hint: In (c), think about
symmetry and the current that would flow through R2 .
Solution
(a) There are two parallel pairs ( 12 R1 ) in series, so RAB = 12 R1 + 12 R1 = R1 . (b) Here, there are two series pairs (2 R1 ) in
parallel, so RAB = (2 R1 )(2 R1 )=(2 R1 + 2 R1 ) = R1 . (c) Symmetry requires that the current divides equally on the right and left
sides, so points C and D are at the same potential. Thus, no current flows through R2 , and the circuit is equivalent to (b).
(Note that the reasoning in parts (a) and (b) is easily generalized to resistances of different values; the generalization in
part (c) requires the equality of ratios of resistances which are mirror images in the plane of symmetry.)
FIGURE 28-50
Problem 19 Solution.
662 CHAPTER 28
Problem
22. What is the current through the 3-Ω resistor in the circuit of Fig. 28-51? Hint: This is trivial. Can you see why?
5.0 Ω
+
6.0 V –
3.0 Ω
FIGURE 28-51
+
9.0 V
–
Problem 22.
Solution
The current is I3Ω = V3Ω =R3Ω = 6 V=3 Ω = 2 A, from Ohm’s law. The answer is trivial because the potential difference
across the 3 Ω resistor is evident from the circuit diagram. (However, if the 6 V battery had internal resistance, an argument
like that in Example 28-5 must be used.)
Problem
23. Take E = 12 V and R1 = 270 Ω in the voltage divider of Fig. 28-5. (a) What should be the value of R2 in order that
4.5 V appear across R2 ? (b) What will be the power dissipation in R2 ?
Solution
(a) For this voltage divider, Equation 28-2b gives V2 = R2 E=( R1 + R2 ), or R2 = R1V2 =(E − V2 ) = (270 Ω)( 4.5)=(12 − 4.5) =
162 Ω. (b) The power dissipated (Equation 27-9b) is P2 = V22 =R2 = ( 4.5 V) 2=162 Ω = 125 mW.
Problem
25. In the circuit of Fig. 28-52, R1 is a variable resistor, and the other two resistors have equal resistances R. (a) Find an
expression for the voltage across R1 , and (b) sketch a graph of this quantity as a function of R1 as R1 varies from 0 to
10R. (c) What is the limiting value as R1 → ∞ ?
FIGURE
28-52 Problem 25.
Solution
(a) The resistors in parallel have an equivalent resistance of R|| = RR1=( R + R1 ). The other R, and R|| , is a voltage divider in
series with E, so Equation 28-2 gives V|| = ER||=( R + R|| ) = ER1=( R + 2 R1 ). (b) and (c) If R1 = 0 (the second resistor shorted
out), V|| = 0, while if R1 = ∞ (open circuit), V|| = 12 E (the value when R1 is removed). If R1 = 10 R, V|| = (10=21)E (as in
Problem 24).
CHAPTER 28 663
Problem 25 Solution.
Problem
26. In the circuit of Fig. 28-53 find (a) the current supplied by the battery and (b) the current through the 6-Ω resistor.
Solution
(a) With reference to the solution of the next problem, the resistance of the three parallel resistors in (12=11) Ω, so the current
supplied by the battery is I = E=( R1 + R|| ) = (6 V )=(23=11) Ω = 2.87 A. (b) The voltage drop across the resistors in parallel is
V|| = E − IR1 = IR|| , and the current through the 6 Ω resistor is I6 Ω = V||=6 Ω. Thus, I6 Ω = (2.87 A )(2=11) = 522 mA.
FIGURE 28-53
Problem 26 Solution, and Problem 27.
Problem
29. In the circuit of Fig. 28-54 it makes no difference whether the switch is open or closed. What is E3 in terms of the other
quantities shown?
Solution
If the switch is irrelevant, then there is no current through its branch of the circuit. Thus, points A and B must be at the same
potential, and the same current flows through R1 and R2 . Kirchhoff’s voltage law applied to the outer loop, and to the lefthand loop, gives E1 − IR1 − IR2 + E2 = 0, and E1 − IR1 + E3 = 0, respectively. Therefore,
E3 = IR1 − E1 =
FG E + E IJ R
HR + R K
1
1
1
FIGURE
2
2
− E1 =
E2 R1 − E1 R2
.
R1 + R2
28-54 Problem 29 Solution.
Problem
30. What is the current through the ammeter in Fig. 28-55?
Solution
If the ammeter has zero resistance, the potential difference across it is zero, or nodes C and D are at equal potentials. If I is the
current through the battery, 12 I must go through each of the 2 Ω -resistors connected at node A (because
VA − VC = 12 I (2 Ω) = VA − VD ). At node B, the 2 Ω -resistor inputs twice the current of the 4 Ω -resistor, or 23 I and 13 I
respectively (because VC − VB = 23 I (2 Ω) = 13 I ( 4 Ω) = VD − VB ). Therefore 16 I must go through the ammeter from D to C,
as required by Kirchhoff’s current law. To find the value of I, note that the upper pair of resistors are effectively in parallel
(VC = VD ) as is the lower pair. The effective resistance between A and B is Reff = 2 × 2 Ω=(2 + 2 ) + 2 × 4 Ω ÷
664 CHAPTER 28
(2 + 4 ) = 1 Ω + ( 43 ) Ω = ( 73 ) Ω. Thus I = V=Reff , and the ammeter current is
FIGURE
1
6
I =
1
6
(6 V)=( 73 ) Ω = ( 37 ) A = 0.429 A.
28-55 Problem 30 Solution.
Problem
31. In Fig. 28-56, what is the equivalent resistance measured between points A and B?
Solution
The effective resistance is determined by the current which would flow through a pure emf if it were connected between
A and B: RAB = E=I . Since I is but one of six branch currents, the direct solution of Kirchhoff’s circuit laws is tedious (6 × 6
determinants). (The method of loop currents, not mentioned in the text, involves more tractable 3 × 3 determinants.)
However, because of the special values of the resistors in Fig. 28-56, a symmetry argument greatly simplifies the calculation.
The equality of the resistors on opposite sides of the square implies that the potential difference between A and C equals
that between D and B, i.e., VA − VC = VD − VB . Equivalently, VA − VD = VC − VB . Since VA − VC = I1 R, VA − VD =
I2 (2 R), etc., the symmetry argument requires that both R-resistors on the perimeter carry the same current, I1 , and both 2Rresistors carry current I2 . Then Kirchhoff’s current law implies that the current through E is I1 + I2 , and the current through
the central resistor is I1 − I2 (as added to Fig. 28-56). Now there are only two independent branch currents, which can be
found from Kirchhoff’s voltage law, applied, for example, to loops ACBA, E − I1 R − I2 (2 R) = 0, and ACDA,
− I1 R − ( I1 − I 2 ) R + I 2 (2 R) = 0. These equations may be rewritten as I1 + 2 I 2 = E=R and −2 I1 + 3 I2 = 0, with solution
I1 = 3E=7 R and I 2 = 2 E=7 R. Therefore, I = I1 + I 2 = 5E=7 R, and RAB = E=I = 7 R=5. (The configuration of resistors in
Fig. 28-56 is called a Wheatstone bridge.)
FIGURE
28-56 Problem 31 Solution.
Problem
34. In Fig. 28-57, take E1 = 6.0 V, E2 = 1.5 V, E3 = 4.5 V, R1 = 270 Ω, R2 = 150 Ω, R3 = 560 Ω, and R4 = 820 Ω. Find
the current in R3 , and give its direction.
Solution
The general expressions for the branch currents can be found from the solution to the next problem. Here, we only need
Ib = −
(6 V − 1.5 V )(820 Ω) + ( 4.5 V − 1.5 V )( 420 Ω)
= −4.77 mA.
(560 Ω)(820 Ω) + ( 420 Ω)(1380 Ω)
A negative current is downward through E2 in Fig. 28-57.
CHAPTER 28 665
FIGURE
28-57 Problem 34 Solution, and Problems 35 and 36.
Problem
39. A voltmeter with 200-kΩ resistance is used to measure the voltage across the 10-kΩ resistor in Fig. 28-59. By what
percentage is the measurement in error because of the finite meter resistance?
5.0 kΩ
+
150 V –
FIGURE
10 kΩ
28-59 Problems 39 and 40.
Solution
The voltage across the 10 kΩ resistor in Fig. 28-59 is (150 V)(10)=(10 + 5) = 100 V (the circuit is just a voltage divider as
described by Equations 28-2a and b), as would be measured by an ideal voltmeter with infinite resistance. With the real
voltmeter connected in parallel across the 10 kΩ resistor, its effective resistance is changed to R|| = (10 kΩ)(200 kΩ) ÷
(210 kΩ) = 9.52 kΩ, and the voltage reading is only (150 V )(9.52 )=(9.52 + 5) = 98.4 V, or about 1.64% lower.
Problem
42. The voltage across the 30-kΩ resistor in Fig. 28-60 is measured with (a) a 50-kΩ voltmeter, (b) a 250-kΩ voltmeter,
and (c) a digital meter with 10-MΩ resistance. To two significant figures, what does each read?
FIGURE
28-60 Problem 42 Solution.
Solution
With a meter of resistance Rm connected as indicated, the circuit reduces to two pairs of parallel resistors in series. The total
resistance is Rtot = (30 kΩ) Rm =(30 kΩ + Rm ) + 40 kΩ=2. The voltage reading is Vm = Rm I m = Rm (30 kΩ) I tot ÷
(30 kΩ + Rm ), where I tot = (100 V )=Rtot (the expression for Vm follows from Equation 28-2, with R1 and R2 as the
above pairs, or from I m as a fraction of I tot , as in the solution to Problem 65). For the three voltmeters specified,
I tot = 2.58 mA, 2.14 mA, and 2.00 mA, while Vm = 48.4 V, 57.3 V, and 59.9 V, respectively. (After checking the
calculations, round off to two figures. Of course, 60 V is the ideal voltmeter reading.)
Problem
43. In Fig. 28-61 what are the meter readings when (a) an ideal voltmeter or (b) an ideal ammeter is connected between
666 CHAPTER 28
points A and B?
10 kΩ
A
+
30 V –
20 kΩ
B
FIGURE 28-61
Problem 43.
Solution
(a) An ideal voltmeter has infinite resistance, so AB is still an open circuit (as shown on Fig. 28-61) when such a voltmeter is
connected. The meter reads the voltage across the 20 kΩ resistor (part of a voltage divider), or (30 V )20=(20 + 10) = 20 V
(see Equation 28-2a or b). (b) An ideal ammeter has zero resistance, and thus measures the current through the
points A and B when short-circuited (i.e., no current flows through the 20 kΩ resistor). In Fig. 28-61, this would be I AB =
30 V=10 Ω = 3 mA. (Such a connection does not measure the current in the original circuit, since an ammeter should be
connected in series with the current to be measured.)
Problem
45. Show that the quantity RC has the units of time (seconds).
Solution
The SI units for the time constant, RC, are (Ω)( F ) = (V=A)(C=V) = (s=C)(C) = s, as stated.
Problem
47. Show that a capacitor is charged to approximately 99% of the applied voltage in five time constants.
Solution
After five time constants, Equation 28-6 gives a voltage of VC =E = 1 − e −5 = 1 − 6.74 × 10 −3 ' 99.3% of the applied
voltage.
Problem
49. Figure 28-62 shows the voltage across a capacitor that is charging through a 4700-Ω resistor in the circuit of
Fig. 28-29. Use the graph to determine (a) the battery voltage, (b) the time constant, and (c) the capacitance.
CHAPTER 28 667
FIGURE
28-62 Problem 49 Solution.
Solution
(a) For the circuit considered, the voltage across the capacitor asymptotically approaches the battery voltage after a long time
(compared to the time constant). In Fig. 28-62, this is about 9 V. (b) The time constant is the time it takes the capacitor
voltage to reach 1 − e −1 = 63.2% of its asymptotic value, or 5.69 V in this case. From the graph, τ ' 1.5 ms. (c) The time
constant is RC, so C = 1.5 ms=4700 Ω = 0.319 µ F.
Problem
50. The voltage across a charging capacitor in an RC circuit rises to 1 − 1=e of the battery voltage in 5.0 ms. (a) How long
will it take to reach 1 − 1=e 3 of the battery voltage? (b) If the capacitor is charging through a 22-kΩ resistor, what is its
capacitance?
Solution
(a) Equation 28-6 and the given circuit characteristics imply that the time constant is τ = RC = 5.0 ms. Therefore, in three
time constants, or 15 ms, the capacitor is charged to 1 − e −3 of the battery voltage. (b) Evidently, C = τ=R = 5 ms=22 kΩ =
0.227 µ F.
Problem
51. A 1.0 - µ F capacitor is charged to 10.0 V. It is then connected across a 500-kΩ resistor. How long does it take (a) for the
capacitor voltage to reach 5.0 V and (b) for the energy stored in the capacitor to decrease to half its initial value?
Solution
A capacitor discharging through a resistor is described by exponential decay, with time constant RC (see Equation 28-8), and,
of course, UC (t ) = 12 CV (t ) 2 = 12 CV 02 e −2 t=RC = UC ( 0)e −2 t=RC is the energy stored (see Equation 26-8b).
(a) V (t )=V ( 0) = 1=2 implies t = RC ln 2 = (500 kΩ)(1 µ F )(0.693) = 347 ms. (b) U c (t )=U c (0 ) = 1=2 implies
t = 12 RC ln 2 = 173 ms.
Problem
56. In the circuit of Fig. 28-64 the switch is initially open and both capacitors initially uncharged. All resistors have the same
value R. Find expressions for the current in R2 (a) just after the switch is closed and (b) a long time after the switch is
closed. (c) Describe qualitatively how you except the current in R3 to behave after the switch is closed.
FIGURE 28-64
Problem 56 Solution.
Solution
(a) An uncharged capacitor acts instantaneously like a short circuit (see Example 28-9), so initially (t = 0 ) all of the current
from the battery goes through R1 and C1, and none goes through R2 and R3. Thus, I1 (0 ) = E=R, and I 2 (0 ) = I3 (0) = 0.
(b) A fully charged capacitor acts like an open circuit (when responding to a constant applied emf ), so after a long time
(t = ∞) , all of the current goes through R1 and R2 in series, and none goes through R3. Thus I1 (∞) = I 2 (∞) = E=2 R, and
I3 (∞) = 0. (c) One can easily guess that I1 and I2 respectively decrease and increase monotonically from their initial to their
668 CHAPTER 28
final values, and that I3 first increases from, and then decreases to zero. (One can use the loop and node equations to solve for
the currents. They turn out to be linear combinations of two decaying exponentials with different time constants.)
Problem
57. In the circuit for Fig. 28-65 the switch is initially open and the capacitor is uncharged. Find expressions for the current I
supplied by the battery (a) just after the switch is closed and (b) a long time after the switch is closed.
I
R
+
–
2R
C
2R
FIGURE 28-65
R
Problem 57.
Solution
(a) Just after the switch is closed, the uncharged capacitor acts instantaneously like a short circuit and the resistors act like two
parallel pairs in series. The effective resistance of the combination is 2 × ( R)(2 R)=( R + 2 R) = 4 R=3, and the current supplied
by the battery is I (0) = 3E=4 R. (b) A long time after the switch is closed, the capacitor is fully charged and acts like an open
circuit. Then the resistors act like two series pairs in parallel, with an effective resistance of ( 12 ) × ( R + 2 R) = 3R=2. The
battery current is I (∞ ) = 2E=3 R.
Problem
75. Write the loop and node laws for the circuit of Fig. 28-71, and show that the time constant for this circuit is
R1 R2 C=( R1 + R2 ).
Solution
Consider the loops and node added to Fig. 28-71. Kirchhoff’s laws are E = I1 R1 + I 2 R2 , VC = I2 R2 , and IC = I1 − I2 . Since
VC = q=C and IC = dq=dt, the equations can be combined to yield
E − I1 R1 − I2 R2 = E − ( I C + I 2 ) R1 − I2 R2 = E − IC R1 −
FG V IJ ( R
HR K
C
1
+ R2 ) = E − I C R1 −
2
q
= 0.
CR2 =( R1 + R2 )
This is exactly in the same form as the first equation, solved in the text, in the section “The RC Circuit: Charging” (with
I → IC , R → R1 and C → CR2 =( R1 + R2 )), so the time constant for the circuit is τ = CR1 R2 =( R1 + R2 ) (the ratio of the
coefficients of IC and q).
FIGURE 28-71
Problem 75 Solution.
Problem
76. The circuit in Fig. 28-72 extends forever to the right, and all the resistors have the same value R. Show that the equivalent
CHAPTER 28 669
resistance measured across the two terminals at left is
1
2
R(1 + 5 ). Hint: You don’t need to sum an infinite series.
...
...
FIGURE 28-72
Problem 76.
Solution
Since the circuit line is infinite, the addition or deletion of one more element leaves the equivalent resistance unchanged.
Diagrammatically:
Problem 78 Solution.
The right-hand picture represents R in series with the parallel combination R and Req, therefore Req = R + RReq =( R + Req ).
2
Solving for Req , one finds Req
− RReq − R 2 = 0, or Req =
resistance).
1
2
(1 + 5 ) R (only the positive root is physically meaningful for a