Physics 107 HOMEWORK ASSIGNMENT #17
Cutnell & Johnson, 7th edition
Chapter 20: Problems 62, 64, 70, 76, 104
62 A
resistor is connected in parallel with a
resistor. This parallel group is
connected in series with a
resistor. The total combination is connected across a 15.0-V
battery. Find (a) the current and (b) the power delivered to the
resistor.
*64 Concept Simulation 20.5 provides some background pertinent
to this problem. Determine the power supplied to each of the resistors
in the drawing.
70 When a light bulb is connected across the terminals of a battery, the battery delivers 24 W of
power to the bulb. A voltage of 11.8 V exists between the terminals of the battery, which has an
. What is the emf of the battery?
internal resistance of
76 Determine the current (magnitude and direction) in the 8.0- and
resistors in the drawing.
104 For the circuit shown in the drawing, find the current I through the 2.00Ω resistor and the voltage V of the battery to the left of this resistor.
60.0 Ω
62. REASONING The circuit diagram is shown at the
right. We can find the current in the 120.0-Ω resistor by
20.0 Ω
A
B
using Ohm’s law, provided that we can obtain a value
for VAB, the voltage between points A and B in the
120.0 Ω
diagram. To find VAB, we will also apply Ohm’s law,
this time by multiplying the current from the battery
times RAB, the equivalent parallel resistance between A
15.0 V
and B. The current from the battery can be obtained by
applying Ohm’s law again, now to the entire circuit and using the total equivalent resistance
of the series combination of the 20.0-Ω resistor and RAB. Once the current in the 120.0-Ω
resistor is found, the power delivered to it can be obtained from Equation 20.6b, P = I 2 R.
SOLUTION
a. According to Ohm’s law, the current in the 120.0-Ω resistor is I120 = VAB/(120.0 Ω). To
find VAB, we note that the equivalent parallel resistance between points A and B can be
obtained from Equation 20.17 as follows:
1
1
1
=
+
RAB 60.0 Ω 120.0 Ω
or
RAB = 40.0 Ω
This resistance of 40.0 Ω is in series with the 20.0-Ω resistance, so that, according to
Equation 20.16, the total equivalent resistance connected across the battery is 40.0 Ω
+ 20.0 Ω = 60.0 Ω. Applying Ohm’s law to the entire circuit, we can see that the current
from the battery is
I = 15.0V/60.0Ω = 0.250 A
Again applying Ohm’s law, this time to the resistance RAB, we find that
VAB = ( 0.250 A ) RAB = ( 0.250 A )( 40.0 Ω ) = 10.0 V
Finally, we can see that the current in the 120.0-Ω resistor is
I120 =
VAB
10.0 V
=
= 8.33 × 10 –2 A
120 Ω 120 Ω
b. The power delivered to the 120.0-Ω resistor is given by Equation 20.6b as
(
2
P = I120
R = 8.33 × 10 –2 A
2
) (120.0 Ω ) =
0.833 W
64. REASONING The power P dissipated in each resistance R is given by Equation 20.6b as
P = I 2 R , where I is the current. This means that we need to determine the current in each
resistor in order to calculate the power. The current in R1 is the same as the current in the
equivalent resistance for the circuit. Since R2 and R3 are in parallel and equal, the current in
R1 splits into two equal parts at the junction A in the circuit.
SOLUTION To determine the equivalent resistance of the circuit, we note that the parallel
combination of R2 and R3 is in series with R1. The equivalent resistance of the parallel
combination can be obtained from Equation 20.17 as follows:
1
1
1
=
+
RP 576 Ω 576 Ω
or
RP = 288 Ω
This 288-Ω resistance is in series with R1, so that the equivalent resistance of the circuit is
given by Equation 20.16 as
Req = 576 Ω + 288 Ω = 864 Ω
To find the current from the battery we use Ohm’s law:
I=
V
120.0 V
=
= 0.139 A
Req
864 Ω
Since this is the current in R1, Equation 20.6b gives the power dissipated in R1 as
2
P1 = I12 R1 = ( 0.139 A ) ( 576 Ω ) = 11.1 W
R2 and R3 are in parallel and equal, so that the current in R1 splits into two equal parts at the
1
junction A. As a result, there is a current of 2 ( 0.139 A ) in R2 and in R3. Again using
Equation 20.6b, we find that the power dissipated in each of these two resistors is
2
( 576 Ω ) =
2.78 W
2
( 576 Ω ) =
2.78 W
P2 = I 22 R2 =  12 ( 0.139 A ) 
P3 = I32 R3 =  12 ( 0.139 A ) 
70. REASONING The drawing shows the battery (emf = ξ ), its internal resistance r, and the
light bulb (represented as a resistor). The voltage between the terminals of the battery is the
voltage VAB between the points A and B in the drawing. This voltage is not equal to the emf of
the battery, because part of the emf is needed to make the current I go through the internal
resistance. Ohm’s law states that this part of the emf is I r. The current can be determined from
the relation P = I VAB, since the power P delivered to the light bulb and the voltage VAB across it
are known.
Light bulb
I
−
A
−
+
+
r = 0.10 Ω
ξ
B
SOLUTION The terminal voltage VAB is equal to the emf ξ of the battery minus the
voltage across the internal resistance r, which is I r (Equation 20.2): VAB = ξ − I r. Solving
this equation for the emf gives
ξ = VAB + I r
(1)
The current also goes through the light bulb, and it is related to the power P delivered to the
bulb and the voltage VAB according to I = P/VAB (Equation 20.6a). Substituting this
expression for the current into Equation (1) gives
 P 
r
V
 AB 
ξ = VAB + I r = VAB + 
 24 W  (
= 11.8 V + 
 0.10 Ω ) = 12.0 V
 11.8 V 
76. REASONING This problem can be solved by using Kirchhoff’s loop rule. We begin by
drawing a current through each resistor. The drawing shows the directions chosen for the
currents. The directions are arbitrary, and if any one of them is incorrect, then the analysis
will show that the corresponding value for the current is negative.
V1 = 4.0 V
A − +
F
E
R1 = 8.0 Ω
−
+
R2 = 2.0 Ω
−
+
I2
V2 = 12 V
+ −
B
I1
C
D
We mark the two ends of each resistor with plus and minus signs that serve as an aid in
identifying the potential drops and rises for the loop rule, recalling that conventional current
is always directed from a higher potential (+) toward a lower potential (–). Thus, given the
directions chosen for I1 and I 2 , the plus and minus signs must be those shown in the
drawing. We then apply Kirchhoff's loop rule to the top loop (ABCF) and to the bottom loop
(FCDE) to determine values for the currents I1 and I2.
SOLUTION Applying Kirchhoff’s loop rule to the top loop (ABCF) gives
V1 + I 2 R2 = I1R1
Potential
drop
Potential
rises
(1)
Similarly, for the bottom loop (FCDE),
V2 = I 2 R2
Potential
rise
Potential
drop
(2)
Solving Equation (2) for I2 gives
I2 =
V2
R2
=
12 V
= 6.0 A
2.0 Ω
Since I2 is a positive number, the current in the resistor R2 goes from left to right , as
shown in the drawing. Solving Equation (1) for I1 and substituting I2 = V2/R2 into the
resulting expression yields
V 
V1 +  2  R2
V +I R
V +V
4.0 V + 12V
 R2 
I1 = 1 2 2 =
= 1 2 =
= 2.0 A
R1
R1
R1
8.0 Ω
Since I1 is a positive number, the current in the resistor R1 goes from left to right , as
shown in the drawing.
104. REASONING First, we draw a current I1 (directed to the right) in the 6.00-Ω resistor. We
can express I1 in terms of the other currents in the circuit, I and 3.00 A, by applying the
junction rule to the junction on the left; the sum of the currents into the junction must equal
the sum of the currents out of the junction.
I
Current into
junction on
left
= I1 + 3.00 A
or
I1 = I − 3.00 A
Current out of
junction on
left
In order to obtain values for I and V we apply the loop rule to the top and bottom loops of
the circuit.
SOLUTION Applying the loop rule to the top loop (going clockwise around the loop), we
have
24.0 V + ( I − 3.00 A )( 6.00 Ω )
( 3.00 A )( 4.00 Ω ) + ( 3.00 A )(8.00 Ω) = Potential drops
Potential rises
This equation can be solved directly for the current; I = 5.00 A .
Applying the loop rule to the bottom loop (going counterclockwise around the loop), we
have
+ 24.0 V + I ( 2.00 Ω ) =
V
( I − 3.00 A )( 6.00 Ω)
Potential drops
Potential rises
Substituting I = 5.00 A into this equation and solving for V gives V = 46.0 V .