Homework Solutions
Assignment (1-80b)
Do PROBLEMS # (62, 104) Ch. 20 + AP 1991 #4 (handout)
60.0 Ω
62. REASONING
20.0 Ω
The circuit diagram is shown at the right. We can find the
current in the 120.0-Ω resistor by using Ohm’s law, provided
that we can obtain a value for VAB .
A
B
120.0 Ω
To find VAB, we can apply Ohm’s law by multiplying the
current from the battery by the equivalent resistance RAB .
15.0 V
The current from the battery can be obtained by applying Ohm’s
law once again, this time to the entire equivalent circuit.
Once the current in the 120.0-Ω resistor is found, the power dissipated in it can be obtained from one
2
of our power equations – such as: P = I 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
as follows:
1
1 sum of1
=
+
the reciprocals
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, according to the rules
for combining resistors the total equivalent resistance connected across the battery is
REq = 40.0 Ω
+ ???
20.0 Ω = 60.0 Ω
sum of
Applying Ohm’s law to the equivalent circuit, the current from the battery is: I =
15.0 V
= 0.250 A
60.0 Ω
Again applying Ohm’s law, this time to the resistance RAB, we find that
? 0 A )( 40.0 Ω ) = 10.0 V
VAB = ( 0.250 A ) RAB =?(?0.25
Finally, the current in the 120.0-Ω resistor can be found as: I120 =
VAB ? ?10.0
V
–2
= ?
= 8.33 × 10 A
120 Ω 120 Ω
b. The power dissipated in the 120.0-Ω resistor is:
(
2
P = I120
R = 8.33 ×?10?–2? A
)
2
(120.0 Ω ) =
0.833 W
Homework Solutions (cont.)
Assignment (1-80b)
Do PROBLEMS # (62, 104) Ch. 20 + AP 1991 #4 (handout)
104. REASONING
Label the currents with the resistor values.
Take I3 to the right, I2 to the left and I1 to the right.
First, applying the junction rule (KCL) to the left
junction gives
+
-
I1
Current entering node = Current leaving node
-
Then, apply the loop rule (KVL) to the top loop
+
I2
I3
Voltage drops = Voltage rises
+
-
and finally (KVL) to the bottom loop gives
Voltage drops = Voltage rises
Solving Equations (1) , (2) and (3) simultaneously, we find I2 = 0.73 A .
The positive sign shows that the assumed direction is correct. That is, to the left .