Chapter 18
Direct-Current Circuits
Problem Solutions
18.1
From ∆V = I ( R + r ) , the internal resistance is
r=
18.3
∆V
9.00 V
−R=
− 72.0 Ω = 4.92 Ω
I
0.117 A
For the bulb in use as intended,
Rbulb =
( ∆V ) 2
=
( 120 V ) 2
75.0 W
= 192 Ω
Now, presuming the bulb resistance is unchanged,
the current in the circuit shown is
I=
∆V
120 V
=
= 0.620 A
Req 0.800 Ω + 192 Ω + 0.800 Ω
and the actual power dissipated in the bulb is
= I 2 Rbulb = ( 0.620 A )
2
( 192 Ω ) = 73.8 W
99
100 CHAPTER 18
18.5
(a) The equivalent resistance of the two parallel
resistors is
1 
 1
+
Rp = 

 7.00 Ω 10.0 Ω 
−1
= 4.12 Ω
Thus,
Rab = R4 + Rp + R9 = ( 4.00 + 4.12 + 9.00 ) Ω = 17.1 Ω
(b) I ab =
( ∆V ) ab
Also,
Rab
18.7
34.0 V
= 1.99 A , so I 4 = I 9 = 1.99 A
17.1 Ω
( ∆V ) p = I ab Rp = ( 1.99 A ) ( 4.12 Ω ) = 8.18 V
Then, I7 =
and
=
I10 =
( ∆V ) p
R7
( ∆V ) p
R10
=
8.18 V
= 1.17 A
7.00 Ω
=
8.18 V
= 0.818 A
10.0 Ω
Ω
Ω
If a potential difference is applied between points a and b, the vertical resistor with a
free end is not part of any closed current path. Hence, it has no effect on the circuit and
can be ignored. The remaining four resistors between a and b reduce to a ∆single
equivalent resistor, Req = 2.5R , as shown below:
Direct-Current Circuits 101
18.13
The resistors in the circuit can be combined in the stages shown below to yield an
equivalent resistance of Rad = ( 63 11) Ω .
( ∆V ) ad
I=
Then, from Figure 4,
( ∆V ) bd = I Rbd = ( 3.14 A ) ( 30 11 Ω ) = 8.57 V
Rad
=
18 V
= 3.14 A
( 63 11 ) Ω
From Figure 5,
Now, look at Figure 2 and observe that
I2 =
so
( ∆V ) bd
3.0 Ω + 2.0 Ω
=
8.57 V
= 1.71 A
5.0 Ω
( ∆V ) be = I 2 Rbe = ( 1.71 A ) ( 3.0 Ω ) = 5.14 V
Finally, from Figure 1,
I12 =
( ∆V ) be
R12
=
5.14 V
= 0.43 A
12 Ω
102 CHAPTER 18
18.17
We name the currents I1 , I 2 , and I 3 as shown.
Using Kirchhoff’s loop rule on the rightmost
loop gives
+12.0 V- ( 1.00+3.00) I 3
− ( 5.00+ 1.00) I 2 − 4.00 V= 0
or
( 2.00 ) I 3 + ( 3.00 ) I2 = 4.00 V
(1)
Applying the loop rule to the leftmost loop yields
+4.00 V+ ( 1.00+5.00 ) I 2 − ( 8.00 ) I1 = 0
or
( 4.00 ) I1 − ( 3.00 ) I 2 = 2.00 V
(2)
From Kirchhoff’s junction rule, I1 +I 2 =I 3
(3)
Solving equations (1), (2) and (3) simultaneously gives
I 1 =0.846 A, I 2 =0.462 A, and I 3 = 1.31 A
All currents are in the directions indicated by the arrows in the circuit diagram.
18.21
First simplify the circuit
by combining the series
resistors. Then, apply
Kirchhoff’s junction rule
at point a to find
I1 + I 2 = 2.00 A
Next, we apply Kirchhoff’s loop rule to the rightmost loop to obtain
−8.00 V + ( 6.00 ) I1 − ( 12.0 ) I 2 = 0
or
−8.00 V + ( 6.00 ) I1 − ( 12.0 ) ( 2.00 A − I1 ) = 0
This yields
Finally, apply Kirchhoff’s loop rule to the leftmost loop to obtain
+ ε 1− ( 4.00 ) ( 2.00 A ) − ( 6.00 ) I 1 + 8.00 V = 0
or
ε 1 = ( 4.00 ) ( 2.00 A ) + ( 6.00 ) ( 1.78 A ) − 8.00 V = 10.7 V
I1 = 1.78 A
Direct-Current Circuits 103
18.26
Using Kirchhoff’s loop rule on the outer
perimeter of the circuit gives
+ 12 V − ( 0.01) I1 − ( 0.06 ) I 3 = 0
or
I1 + 6 I 3 = 1.2 × 103 A
(1)
For the rightmost loop, the loop rule gives
+10 V + ( 1.00 ) I 2 − ( 0.06 ) I 3 = 0
or
I 2 − 0.06 I 3 = −10 A
(2)
Applying Kirchhoff’s junction rule at either junction gives
I1 = I 2 + I 3
(3)
Solving equations (1), (2), and (3) simultaneously yields
I 2 = 0.28 A ( in dead battery ) and I 3 = 1.7 × 10 2 A (in starter)
18.31
6
−6
(a) τ = RC = ( 2.0 × 10 Ω ) ( 6.0 × 10 F ) = 12 s
(b) Qmax = C ε = ( 6.0 × 10 −6 F ) ( 20 V ) = 1.2 × 10− 4 C
18.33
Qmax = C ε = ( 5.0 × 10 −6 F ) ( 30 V ) = 1.5 × 10 −4 C , and
τ = RC = ( 1.0 × 106 Ω ) ( 5.0 × 10 −6 F ) = 5.0 s
Thus, at t = 10 s = 2τ
Q = Qmax ( 1 − e −t τ ) = ( 1.5 × 10−4 C ) ( 1 − e −2 ) = 1.3 × 10 −4 C
104 CHAPTER 18
18.35
−t τ
From Q = Qmax ( 1 − e ) , we have at t = 0.900 s ,
Q
− 0.900 s τ
=1−e
= 0.600
Qmax
Thus, e −0.900 s τ = 0.400 , or −
0.900 s
= ln ( 0.400 )
τ
giving the time constant as τ = −
R (Ω)
1.00
5.00
10.0
15.0
20.0
25.0
30.0
The curve peaks at
L
0.900 s
= 0.982 s
ln ( 0.400)
L (W)
1.19
3.20
3.60
3.46
3.20
2.94
2.70
= 3.60 W at a load resistance of R = 10.0 Ω .