P21.25
(a)
P=
(ΔV )2
R
(11.6 V )2
becomes
20.0 W =
so
R = 6.73 Ω .
R
FIG. P21.25
(b)
ΔV = IR
so
11.6 V = I (6.73 Ω )
and
I = 1.72 A
e = IR + Ir
15.0 V = 11.6 V + (1.72 A ) r
so
r = 1.97 Ω .
P21.27
(a)
Rp =
1
= 4.12 Ω
1
7.00
Ω
(
) + (1 10.0 Ω )
R s = R1 + R 2 + R 3 = 4.00 + 4.12 + 9.00 = 17.1 Ω
(b)
ΔV = IR
FIG. P21.27
34.0 V = I (17.1 Ω )
I = 1.99 A
for 4.00 Ω , 9.00 Ω resistors.
Applying ΔV = IR ,
(1.99 A )(4.12 Ω ) = 8.18 V
8.18 V = I (7.00 Ω )
so
I = 1.17 A for 7.00 Ω resistor
8.18 V = I (10.0 Ω )
so
I = 0.818 A for 10.0 Ω resistor.
P21.29 If we turn the given diagram on its side, we find that it is the same
as figure (a). The 20.0 Ω and 5.00 Ω resistors are in series, so the
first reduction is shown in (b). In addition, since the 10.0 Ω , 5.00 Ω ,
and 25.0 Ω resistors are then in parallel, we can solve for their
equivalent resistance as:
Req =
(
1
1
10.0 Ω
+
1
5.00 Ω
+
1
25.0 Ω
)
= 2.94 Ω .
This is shown in figure (c), which in turn reduces to the circuit
shown in figure (d).
Next, we work backwards through the diagrams applying I =
ΔV
R
and ΔV = IR alternately to every resistor, real and equivalent. The
12.94 Ω resistor is connected across 25.0 V, so the current through
the battery in every diagram is
I=
25.0 V
ΔV
=
= 1.93 A
12.94 Ω
R
.
In figure (c), this 1.93 A goes through the 2.94 Ω equivalent
resistor to give a potential difference of:
ΔV = IR = (1.93 A )(2.94 Ω ) = 5.68 V .
From figure (b), we see that this potential difference is the same
across ΔV ab , the 10 Ω resistor, and the 5.00 Ω resistor.
(b)
Therefore, ΔV ab = 5.68 V .
(a)
Since the current through the 20.0 Ω resistor is also the current
through the 25.0 Ω line ab,
I=
ΔV ab 5.68 V
=
= 0.227 A = 227 mA .
25.0 Ω
R ab
FIG. P21.29
P21.30
(a)
Since all the current in the circuit must pass through the series
100 Ω resistor, P = I 2 R
P m ax = RI m2 ax
P
=
R
I max =
so
FIG. P21.30
25.0 W
= 0.500 A
100 Ω
1 ⎞
⎛ 1
+
R eq = 100 Ω + ⎜
⎝ 100 100 ⎟⎠
−1
Ω = 150 Ω
ΔV max = R eq I max = 75.0 V
P = IΔV = (0.500 A )(75.0 V ) = 37.5 W
(b)
total power
P1 = 25.0 W
P 2 = P 3 = RI 2 = (100 Ω )(0.250 A ) = 6.25 W
2
P21.31
1 ⎞
⎛ 1
Rp = ⎜
+
⎝ 3.00 1.00 ⎟⎠
−1
= 0.750 Ω
R s = (2.00 + 0.750 + 4.00 ) Ω = 6.75 Ω
ΔV 18.0 V
=
= 2.67 A
Rs
6.75 Ω
I battery =
P = I 2R
P 2 = (2.67 A ) (2.00 Ω )
2
:
P 2 = 14.2 W in 2.00 Ω
P 4 = (2.67 A ) (4.00 A ) = 28.4 W in 4.00 Ω
2
ΔV 2 = (2.67 A )(2.00 Ω ) = 5.33 V,
ΔV 4 = (2.67 A )(4.00 Ω ) = 10.67 V
ΔV p = 18.0 V − ΔV 2 − ΔV 4 = 2.00 V (= ΔV 3 = ΔV1 )
P3 =
(ΔV 3 )2 = (2.00 V )2
P1 =
(ΔV1 ) = (2.00 V )2
R3
R1
3.00 Ω
1.00 Ω
= 1.33 W in 3.00 Ω
= 4.00 W in 1.00 Ω
FIG. P21.31
P21.35 We name currents I1 , I 2 , and I 3 as shown.
From Kirchhoff’s current rule, I3 − I1 − I 2 = 0 .
Applying Kirchhoff’s voltage rule to the loop containing I 2 and
I3 ,
12.0 V − (4.00 )I 3 − (6.00 )I 2 − 4.00 V = 0
8.00 = (4.00 )I 3 + (6.00 )I 2
FIG. P21.35
Applying Kirchhoff’s voltage rule to the loop containing I1 and
I2 ,
− (6.00 ) I 2 − 4.00 V + (8.00 ) I1 = 0
(8.00 )I1 = 4.00 + (6.00 )I 2 .
Solving the above linear system, we proceed to the pair of
simultaneous equations:
⎧8 = 4I1 + 4I 2 + 6I 2
⎨
⎩8I1 = 4 + 6I 2
or
⎧8 = 4I1 + 10I 2
⎨
⎩I 2 = 1.33I1 − 0.667
and to the single equation 8 = 4I1 + 13.3I1 − 6.67
I1 =
I 2 = 1.33 (0.846 A ) − 0.667
14.7 V
= 0.846 A .
17.3 Ω
I 3 = I1 + I 2
and
Then
give
I1 = 846 mA, I 2 = 462 mA, I3 = 1.31 A .
All currents are in the directions indicated by the arrows in the circuit diagram.
P21.40 Using Kirchhoff’s rules,
12.0 − (0.010 0 )I1 − (0.060 0 )I 3 = 0
10.0 + (1.00 )I 2 − (0.060 0 )I 3 = 0
and I1 − I 2 − I3 = 0 or
I1 = I 2 + I 3
then 12.0 − (0.010 0)I 2 − (0.070 0)I3 = 0
and I 2 = 0.06I3 − 10
Solving simultaneously,
12 − (0.01)(0.06I3 − 10 ) − 0.07I3 = 0
I 2 = 0.283 A downward
in the dead battery
FIG. P21.40
and I3 = 171 A downward in the starter.
The currents are forward in the live battery and in the starter, relative to normal
starting operation. The current is backward in the dead battery, tending to charge it up.