CHAPTER 28 ELECTRIC CIRCUITS

CHAPTER 28 ELECTRIC CIRCUITS

Problem

1. Sketch a circuit diagram for a circuit that includes a resistor R ₁ connected to the positive terminal of a battery, a pair of parallel resistors R ₂ and R ₃ connected to the lower-voltage end of R ₁ , then returned to the battery’s negative terminal, and a capacitor across R ₂ .

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 R ₁ and R ₂ are connected in series, and this series combination is in parallel with R ₃ . This parallel combination

is connected across a battery whose internal resistance is R _int . Draw a diagram representing this circuit.

Solution

The circuit has three parallel branches: one with R ₁ and R ₂ in series; one with just R ₃ ; 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 C = 3 = 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 = E I, 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 R 2 in Equation 28-3a is R ₂ = R R _{1 parallel} = ( R ₁ − R _parallel ) = ( 56 k Ω )( 45 ) ( = 56 − 45 ) = 229 k . Ω

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?

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 ) = ( ) ² ₃ Ω , and R _tot = 1 Ω + 1 Ω + ² ₃ Ω = ⁸ ₃ Ω. The total current (that through the battery) is

I _tot = E= R _tot = 6 V = ( ⁸ ₃ Ω ) = ( ) ⁹ ₄ A . The voltage across the parallel combination is I _tot R _|| = ( ⁹ ₄ A )( ² ₃ Ω ) = ³ ₂ V , which is the voltage across the vertical 1 Ω resistor. The current through this resistor is then ( ³ ₂ 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 E ₁ − E ₂ − IR ₁ − IR ₂ = 0 , or I = ( E ₁ − E = ₂ ) ( R ₁ + R ₂ ) = ( 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 R ₂ .

Solution

(a) There are two parallel pairs ( ¹ ₂ R in series, so R ₁ ) _AB = ¹ ₂ R 1 + 1 R = R

2 1 1 . (b) Here, there are two series pairs ( 2 R in ₁ ) parallel, so R _AB = ( 2 R ₁ )( 2 R ₁ ) ( = 2 R ₁ + 2 R ₁ ) = R ₁ . (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 R ₂ , 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.

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 ⁺ _– 9.0 V

FIGURE 28-51 Problem 22.

Solution

The current is I _{3 Ω} = V _{3 Ω} = R _{3 Ω} = 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 R ₁ = 270 Ω in the voltage divider of Fig. 28-5. (a) What should be the value of R ₂ in order that 4.5 V appear across R ₂ ? (b) What will be the power dissipation in R ₂ ?

Solution

(a) For this voltage divider, Equation 28-2b gives V ₂ = R ₂ E = ( R ₁ + R ₂ ), or R ₂ = R V _{1 2} = ( E − V ₂ ) = ( 270 Ω )( . ) ( 4 5 12 = − 4 5 . ) = 162 Ω . (b) The power dissipated (Equation 27-9b) is P ₂ = V ₂ ² = R ₂ = ( . 4 5 V ) ² = 162 Ω = 125 mW.

Problem

25. In the circuit of Fig. 28-52, R ₁ is a variable resistor, and the other two resistors have equal resistances R. (a) Find an expression for the voltage across R ₁ , and (b) sketch a graph of this quantity as a function of R ₁ as R ₁ varies from 0 to 10R. (c) What is the limiting value as R ₁ → ∞ ?

FIGURE 28-52 Problem 25.

Solution

(a) The resistors in parallel have an equivalent resistance of R _|| = RR 1 = ( R + R 1 ). The other R, and R _|| , is a voltage divider in

series with E , so Equation 28-2 gives V _|| = E R _|| = ( R + R _|| ) = E R ₁ = ( R + 2 R ₁ ). (b) and (c) If R ₁ = 0 (the second resistor shorted

out), V _|| = 0 while if , R ₁ = ∞ (open circuit), V _|| = ¹ ₂ E (the value when R ₁ is removed). If R ₁ = 10 R V , _|| = ( 10 21 = ) E (as in

Problem 24).

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= ( R ₁ + R _|| ) = ( 6 V ) ( = 23 11 = ) Ω = 2 87 . A . (b) The voltage drop across the resistors in parallel is V _|| = − E IR ₁ = IR _|| , and the current through the 6 Ω resistor is I _{6 Ω} = V _|| = 6 Ω . Thus, I _{6 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 E ₃ 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 R ₁ and R ₂ . Kirchhoff’s voltage law applied to the outer loop, and to the left- hand loop, gives E ₁ − IR ₁ − IR ₂ + E ₂ = 0 , and E ₁ − IR ₁ + E ₃ = 0 , respectively. Therefore,

E E E E

E E E

3 1 1

1 2

1 2

1 1

2 1 1 2

1 2

= − = +

+

F HG I

KJ ^{− =} + ⁻

IR R R R R R

R R .

FIGURE 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, ¹ ₂ I must go through each of the 2 Ω -resistors connected at node A (because

V _A − V _C = ¹ ₂ I ( 2 Ω ) = V _A − V _D ). At node B, the 2 Ω -resistor inputs twice the current of the 4 Ω -resistor, or ² ₃ I and ¹ ₃ I

respectively (because V _C − V _B = ² ₃ I ( 2 Ω ) = ¹ ₃ I ( 4 Ω ) = V _D − V _B ). Therefore ¹ ₆ 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

( V _C = V _D ) as is the lower pair. The effective resistance between A and B is R _eff = × 2 2 Ω = ( 2 + 2 ) + × 2 4 Ω ÷

( 2 + 4 ) = 1 Ω + ( ) ⁴ ₃ Ω = ( ) ⁷ ₃ Ω . Thus I = V R = _eff , and the ammeter current is ¹ ₆ I = ¹ ₆ ( 6 V ) ( ) = ⁷ ₃ Ω = ( ) ³ ₇ A = 0 .429 A .

FIGURE 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: R _AB = 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., V _A − V _C = V _D − V _B . Equivalently, V _A − V _D = V _C − V _B . Since V _A − V _C = I R V 1 , _A − V _D =

I ₂ ( 2 R ), etc., the symmetry argument requires that both R-resistors on the perimeter carry the same current, I ₁ , and both 2R- resistors carry current I ₂ . Then Kirchhoff’s current law implies that the current through E is I ₁ + I ₂ , and the current through the central resistor is I ₁ − I ₂ (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 − I R ₁ − I ₂ ( 2 R ) = 0 , and ACDA,

− I R ₁ − ( I ₁ − I ₂ ) R + I ₂ ( 2 R ) = 0 . These equations may be rewritten as I ₁ + 2 I ₂ = E= R and − 2 I ₁ + 3 I ₂ = 0 , with solution I ₁ = 3 7 E= R and I ₂ = 2 E= 7 R . Therefore, I = I ₁ + I ₂ = 5 E= 7 R , and R _AB = 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 E ₁ = 6 0 . V , E ₂ = 1 5 . V , E ₃ = 4 5 . V , R ₁ = 270 Ω , R ₂ = 150 Ω , R ₃ = 560 Ω , and R ₄ = 820 Ω . Find the current in R ₃ , 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

I _b = − − + −

+ = −

( . )( ) ( . . )( )

( )( ) ( )( ) . .

6 1 5 820 4 5 1 5 420

560 820 420 1380 4 77

V V V V

Ω Ω mA

Ω Ω Ω Ω

A negative current is downward through E ₂ in Fig. 28-57.

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?

+

– 10 kΩ

150 V

5.0 kΩ

FIGURE 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 k Ω resistor, its effective resistance is changed to R _|| = ( 10 k Ω )( 200 k Ω ) ÷ ( 210 k Ω ) = 9 52 . k Ω , 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 R _m connected as indicated, the circuit reduces to two pairs of parallel resistors in series. The total resistance is R _tot = ( 30 k Ω ) R _m = ( 30 k Ω + R _m ) + 40 k Ω = 2 . The voltage reading is V _m = R I _{m m} = R _m ( 30 k Ω ) I _tot ÷

( 30 kΩ + R _m ), where I _tot = ( 100 V ) = R _tot (the expression for V _m follows from Equation 28-2, with R ₁ and R ₂ 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 V _m = 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

points A and B?

+ 30 V –

10 kΩ

20 kΩ

A

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 k Ω 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 V _C =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.

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 . F µ .

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 of the battery voltage? (b) If the capacitor is charging through a 22-kΩ ³ 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 k} Ω = 0 227 . F µ .

Problem

51. A 1 0 . - F µ 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, U _C ( ) t = ¹ ₂ CV t ( ) ² = ¹ ₂ CV e ₀ ^{2 − 2 t RC =} = U _C ( ) 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 k Ω )( 1 F µ )( . 0 693 ) = 347 ms . (b) U t U _c ( ) = _c ( ) 0 = 1 2 = implies t = ¹ ₂ 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 R 2 (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 R 3 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 R ₁ and C ₁ , and none goes through R ₂ and R ₃ . Thus, I ₁ ( ) 0 = E= R , and I ₂ ( ) 0 = I ₃ ( ) 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 R ₁ and R ₂ in series, and none goes through R ₃ . Thus I ₁ ( ) ∞ = I ₂ ( ) ∞ = E= 2 R , and

I ₃ ( ) ∞ = 0 . (c) One can easily guess that I ₁ and I ₂ respectively decrease and increase monotonically from their initial to their

final values, and that I ₃ 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.

R +

–

2R

2R R

C I

FIGURE 28-65 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 = 3 E= 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 ( ) ¹ ₂ × ( R + 2 R ) = 3 R = 2 . The battery current is I ( ) ∞ = 2 E= 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 R R C R _{1 2} = ( ₁ + R ₂ ).

Solution

Consider the loops and node added to Fig. 28-71. Kirchhoff’s laws are E = I R _{1 1} + I R _{2 2} , V _C = I R _{2 2} , and I _C = I 1 − I 2 . Since V _C = q C = and I _C = dq dt = , the equations can be combined to yield

E − − = − E + − = − E − F E

HG I

KJ ^{+ = − − + =}

I R I R I I R I R I R V

R R R I R q

CR R R

C C

C

C

1 1 2 2 2 1 2 2 1

2

1 2 1

2 1 2

0

( ) ( )

( ) .

=

This is exactly in the same form as the first equation, solved in the text, in the section “The RC Circuit: Charging” (with I → I _C , R → R ₁ and C → CR ₂ = ( R ₁ + R ₂ )), so the time constant for the circuit is τ = CR R _{1 2} = ( R ₁ + R ₂ ) (the ratio of the coefficients of I _C 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

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 R _eq , therefore R _eq = R + RR _eq = ( R + R _eq ).

Solving for R _eq , one finds R _eq ² − RR _eq − R ² = 0 , or R _eq = ¹ ₂ ( 1 + 5 ) (only the positive root is physically meaningful for a R

resistance).