BW5 Ch 23: P. 1-3, 5, 6, 8, 13 -16.
P23.1. Prepare: From the circuit in Figure P23.1, we see that 50  and 100  resistors are connected in
series across the battery. Another resistor of 75  is also connected across the battery.
Solve:
Assess: The series resistors have the same current in them, but the parallel resistors have the same potential
difference across them.
P23.2. Prepare: In Figure P23.2, the positive terminal of the battery is connected to a resistor. The other end
of that resistor is connected to resistor and a capacitor in parallel.
Solve:
Assess: The series elements have the same current in them, but the parallel elements have the same potential
difference across them.
P23.3. Prepare: Circuits are generally drawn with straight lines and 90 ° angles for the wires. We also
usually orient the battery with the positive terminal up so the higher potential part of the circuit is toward the top
of the paper.
Solve:
Assess: The elements that are in series with each other have the same current in them; the elements that are in
parallel with each other have the same potential difference across them.
P23.5. Prepare: Please refer to Figure P23.5. We will use Ohm’s law (DV = IR) for the resistor and the
bulb, use Tactics Box 23.1 for Kirchoff’s loop law, and assume that the connecting wires are ideal.
Solve: Let us assign clock direction to the current in the circuit.
(a): The Kirchoff’s loop law is:
So,
DV12 = V2 - V1 = - I(2.0 W) = -(1.0 A)(2.0 W) = -2.0 V
DV23 = V3 - V2 = -I(1.0 W) = -(1.0 A)(1.0 W) = -1.0 V
DV34 = V4 - V3 = 0 V
(b) When the bulb is removed from the socket, no current flows in the circuit. Thus, DV12 = V2 - V1 = I(2.0 W) =
(0 A)(2.0 W) = 0 V. This means that points 1 and 2 are at the same potential as the positive terminal of the
battery. For the same reason, DV34 = V4 -V3 = 0, implying that points 3 and 4 are at the same potential as the
negative terminal of the battery. Finally, DV23 = V3 -V2 = V4 - V1 = -3.0 V.
Assess: The potential at point 1 is higher than at point 4. This is what you would have expected.
P23.6. Prepare: Please refer to Figure P23.6. The batteries and the connecting wires are ideal. Choose the
current I to be in the clockwise direction. If I ends up being a positive number, then the current really does flow
in this direction. If I is negative, the current really flows counterclockwise. There are no junctions, so I is the
same for all elements in the circuit.
Solve: (a) With the 9 V battery being labeled 1 and the 6 V battery being labeled 2, Kirchhoff’s loop law is
Note the signs: Potential is gained in battery 1, but potential is lost both in the resistor and in battery 2. Because I
is positive, we can say that I  0.10 A flows from left to right through the resistor.
(b)
Assess: The graph shows 9 V gained in battery 1, VR  IR  3 V lost in the
resistor, and another 6 V lost in battery 2. The final potential is the same as the
initial potential, as required.
P23.8. Prepare: Please refer to Figure P23.8. Define the current I as a clockwise flow. There are no junctions,
so conservation of current tells us that the same current flows through each circuit element. As we go
around the circuit in the direction of the current, potential is gained in the battery (Vbat 
 15 V) and
potential is lost in the resistors (Vresistor  IR).
Solve: (a) From Kirchhoff’s loop law,
Vi  Vbat  V10  V20  0
0
– IR1 – IR2 
– I(R1  R2) Þ
Now that we know the current, we can find the potential difference across each resistor:
V10  IR1  (0.50 A)(10 )  5 V
(b)
V20  IR2  (0.50 A)(20 )  10 V
Assess: The graph shows a 15 V gain in the battery, a 5 V loss in the 10  resistor,
and a loss of 10 V in the 20  resistor. The final potential is the same as the initial
potential, as required.
P23.13. Prepare:
For resistors in parallel,
æ 1
ö
1
Req = ç +
+ × × ×÷
è R1 R2
ø
-1
Adding four of the resistors in parallel gives
-1
-1
æ 1
æ 4 ö
1
1
1 ö
Req = ç
+
+
+
=ç
= 0.25 kW
÷
è 1.0 kW 1.0 kW 1.0 kW 1.0 kW ø
è 1.0 kW ÷ø
Solve:
We can put four of the resistors in parallel. The total resistance is now 0.25 kW.
Assess: There are other ways to arrive at the same Req using more from our collection of 1.0 kW resistors.
P23.14. Prepare: When resistors are combined in parallel, the combination is always less than the smallest
resistor. As a result, the way to obtain the smallest resistance is to connect the resistors in parallel.
Solve: The equivalent resistance when the six resistors are connected in parallel is obtained by
1
RTotal
=
1 1 1 1 1 1 6
+ + + + + =
R R R R R R R
RTotal = R/6 = 1000 W/6 = 167 W
or
Assess: Notice that the total resistance is less than the smallest resistor. The smallest resistor is 1000 W and
the total resistance when the six resistors are connected in parallel is 167 .
P23.15. Prepare: The resistance of the three 6.0 Wresistors in parallel is
-1
-1
æ 1
æ 3 ö
1
1 ö
Req = ç
+
+
=ç
= 2.0 W
è 6.0 W 6.0 W 6.0 W ÷ø
è 6.0 W ÷ø
Solve: That parallel combination in series with the 3.0 W resistor adds to an equivalent resistance of 5.0 W.
So that’s the answer: the three 6.0 Wresistors in parallel with each other, and then that combination in series with
the 3.0 Wresistor.
Assess: There are typical resistances that one can buy; generally one can’t (cheaply) buy resistors with every
possible value of resistance. So it is common to combine the resistors you have in combinations of series and
parallel to arrive at a different value of equivalent resistance called for in the circuit.
P23.16. Prepare: When resistors are connected in parallel the combination has less resistance than any of the
individual resistors. Two 1.0 kW resistors in parallel have an equivalent resistance of 0.50kW.
Solve: We can connect three such parallel pairs in series so that the total resistance is
Rtot = 0.50 kW + 0.50 kW + 0.50 kW = 1.5 kW
Assess: It is often necessary to combine standard value resistors in creative ways to arrive at a non-standard
required resistance.