BW6 Ch. 23: P. 17, 19, 20, 27, 29.
P23.17.
Prepare: The connecting wires are ideal with zero resistance. We have to reduce the circuit to a single equivalent resistor by continuing to identify resistors that are in series or parallel combinations.
Solve:
For the first step, the resistors 30  and 45  are in parallel. Their equivalent resistance is
R
1 eq 1
For the second step, resistors 42  and R eq 1
=
30
1
W
+
45
1
W
Þ R eq 1
 18 
 18  are in series. Therefore,
R eq 2
 R eq 1
 42   18   42   60 
For the third step, the resistors 40  and R eq 2
 60  are in parallel. So,
R
1 eq 3
=
60
The equivalent resistance of the circuit is 24  .
1
W
+
40
1
W
Þ
R eq 3
 24 
Assess: Have a good understanding of how series and parallel resistors combine to obtain equivalent resistors.
P23.19
.
Prepare: From Kirchhoff’s junction law we can easily determine that the current through the unknown resistor must be
2
.
5 A
-
1
.
5 A
=
1
.
0 A down. Also, the potential difference across the unknown resistor is the same as across the
300
W resistor. Assume the resistors obey Ohm’s law.
Solve: The potential difference across the unknown resistor must have a resistance of
300
W resistor is D V = IR =
(1
.
5 A)(300
W
)
=
300 V.
The
R = D
I
V
=
300 V
1
.
0 A
=
300
W
Assess: We used a combination of Kirchhoff’s junction law and Ohm’s law to solve this one.
P23.20
.
Prepare: Assume ideal batteries and wires. Use Kirchhoff’s loop law (going clockwise from the lower left corner) and solve for
I .
Due to the orientation of both batteries, the current must be going clockwise, so the potential differences shown for the two resistors need to be subtracted.
Solve:
SD V i
=
4
.
5 V
+
3
.
0 V
-
3
.
5 V
-
Assess: This is a small but reasonable current.
I
(150
W
)
-
2
.
0 V
=
0 V
Þ I =
2 V
150
W
=
13 mA
P23.27.
Prepare: The battery and the connecting wires are ideal. The figure shows how to simplify the circuit in Figure P23.27 using the laws of series and parallel resistances. We have labeled the resistors as R
1
 6.0  , R
2
 15  , R
3
 6.0  , and R
4
 4.0  . Having reduced the circuit to a single equivalent resistance R eq
, we will reverse the procedure and “build up” the circuit using the loop law and the junction law to find the current and potential difference of each resistor.

Solve: R
3
and R
4
are combined to get R
34
 10  , and then R
34
and R
2
are combined to obtain R
234
:
1
R
234
=
1
R
2
+
1
R
34
=
1
15
W
+
1
10
W
Þ
R
234
 6 
Next, R
234
and R
1
are combined to obtain
R eq
 R
234
 R
1
 6.0   6.0   12 
From the final circuit,
Thus, the current through the battery and R
1
is I
R1
 2.0 A and the potential difference across R
1
is I ( R
1
)  (2.0 A)
(6.0  )  12 V.
As we rebuild the circuit, we note that series resistors must have the same current I and that parallel resistors must have the same potential difference  V .
In Step 1 of the previous figure, R eq
 12  is returned to R
1
 6.0  and R
234
 6.0  in series. Both resistors must have the same 2.0 A current as R eq
. We then use Ohm’s law to find
 V
R1
 (2 .0A)(6.0  )  12 V  V
R234
 (2.0 A)(6.0  )  12 V
As a check, 12 V  12 V  24 V, which was  V of the R eq
resistor. In Step 2, the resistance R
234
is returned to R
2 and R
34
in parallel. Both resistors must have the same  V  12 V as the resistor R
234
. Then from Ohm’s law,
I
R2
=
12 V
15
W
=
0.8 A
I
R34
=
12 V
10
W
=
1.2 A
As a check, I
R2
 I
R34
 2.0 A, which was the current I of the R
234 in series. Both resistors must have the same 1.2 A as the R
34
resistor. In Step 3, R
34
is returned to R
3
resistor. We then use Ohm’s law to find
and R
4
(  V )
R3
 (1.2 A)(6.0  )  7.2 V (  V )
R4
 (1.2 A)(4.0  )  4.8 V
As a check, 7.2 V  4.8 V  12 V, which was  V of the resistor R
34
.
Resistor Potential difference (V)
R
R
R
R
1
2
3
4
12
12
7.2
4.8
Current (A)
2.0
0.8
1.2
1.2

Assess:
P23.29.
Prepare: The battery and the connecting wires are ideal. The figure shows how to simplify the circuit in Figure P23.29 using the laws of series and parallel resistances. Having reduced the circuit to a single equivalent resistance, we will reverse the procedure and “build up” the circuit using the loop law and the junction law to find the current and potential difference of each resistor.
Solve: From the last circuit in the diagram,
Thus, the current through the battery is 2.0 A. As we rebuild the circuit, we note that series resistors must have the same current I and that parallel resistors must have the same potential difference  V .
In Step 1, the 6.0  resistor is returned to a 3.0  and 3.0  resistor in series. Both resistors must have the same
2.0 A current as the 6.0  resistance. We then use Ohm’s law to find
 V
3
 (2.0 A)(3.0  )  6.0 V
As a check, 6.0 V  6.0 V  12 V, which was  V of the 6.0  resistor. In Step 2, one of the two 3.0  resistances is returned to the 4.0  , 48  , and 16  resistors in parallel. The three resistors must have the same  V  6.0 V.
From Ohm’s law,
I
4
=
6 V
4
W
=
1.5 A
Resistor
I
48
=
6 V
48
W
=
0.125 A
Potential difference (V)
I
16
=
6 V
16
W
=
0.375 A
Current (A)
3.0 
4.0 
6.0
6.0
2.0
1.5
48  6.0 0.13
16  6.0 0.38
Assess: