AP Physics Homework Solutions
Unit 9: Electric Circuits
Homework 5 Page 617 problems 5, 10, 12,14,16,18
18.5
(a) The equivalent resistance of the two parallel
resistors is
1
1 
 1
Rp  

  4.12 
 7.00  10.0  
Thus,
Rab  R 4  R p  R9   4.00  4.12  9.00   17.1 
(b) Iab 
18.10
 V ab 34.0 V

R ab
17.1 
 1.99 A , so I4  I9  1.99 A
Also,
 V p  IabR p  1.99 A  4.12   8.18 V
Then,
I7 
and
I10 
 V p
R7
 V p
R10

8.18 V
 1.17 A
7.00 

8.18 V
 0.818 A
10.0 
First, consider the parallel case. The resistance of resistor B is
RB 
 V B
IB

6.0 V
 3.0 
2.0 A
In the series combination, the potential difference across B is given by
 V B   V battery   V A  6.0 V-4.0 V  2.0 V
The current through the series combination is then
Is 
 V B
RB

2.0 V 2
 A
3.0  3
and the resistance of resistor A is R A 
 V A
Is

4.0 V
 6.0 
23A
18.12
(a) The total current from a to b is equal to the maximum
current allowed in the 100  series resistor adjacent
to point a. This current has a value of
Im ax 
Pm ax
25.0 W

 0.500 A
R
100 
The total resistance is
1
1 
 1
R ab  100   

  100   50.0   150 
 100  100  
Thus,  V m ax  Im axR ab   0.500 A 150    75.0 V
(b) The power dissipated in the series resistor is P1  Im2 axR  25.0 W , and the power dissipated in
each of the identical parallel resistors is
P2  P3   Im ax 2 R   0.250 A  100    6.25 W
2
2
The total power delivered is
P  P1  P2  P3   25.0  6.25  6.25 W  37.5 W
18.14
The resistance of the parallel combination of the
3.00  and 1.00  resistors is
1
1 
 1
Rp  

  0.750 
 3.00  1.00  
The equivalent resistance of the circuit connected to
the battery is
Req  2.00   R p  4.00   6.75 
and the current supplied by the battery is
I
V 18.0 V

 2.67 A
R eq 6.75 
The power dissipated in the 2.00- resistor is
P2  I2 R2   2.67 A   2.00    14.2 W
2
and that dissipated in the 4.00- resistor is
P4  I2 R 4   2.67 A   4.00    28.4 W
2
The potential difference across the parallel combination of the 3.00  and 1.00  resistors is
 V p  IR p   2.67 A   0.750   2.00 V
Thus, the power dissipation in these resistors is given by
 V p  2.00 V 2
2
P3 

R3
3.00 
 V p  2.00 V 2
 1.33 W
2
P1 
and
18.16
R1

1.00 
 4.00 W
Going counterclockwise around the upper loop,
applying Kirchhoff’s loop rule, gives
15.0 V   7.00 I1   5.00  2.00 A   0
I1 
or
15.0 V  10.0 V
 0.714 A
7.00 
From Kirchhoff’s junction rule, I1  I2  2.00 A  0
I2  2.00 A  I1  2.00 A  0.714 A  1.29 A
so
Going around the lower loop in a clockwise direction gives
 E  2.00 I2   5.00  2.00 A   0
E   2.00   1.29 A    5.00    2.00 A   12.6 V
or
18.18
Observe that the center branch of this circuit, that is the branch containing points a and b, is not a
continuous conducting path, so no current can flow in this branch. The only current in the circuit flows
counterclockwise around the perimeter of this circuit. Going counterclockwise around the this outer
loop and applying Kirchhoff’s loop rule gives
8.0 V   2.0  I  3.0  I+12 V  10  I  5.0  I 0
or
I
12 V  8.0 V
 0.20 A
20 
Now, we start at point b and go around the upper panel of the circuit to point a, keeping track of
changes in potential as they occur. This gives
Vab  Va  Vb  4.0 V+  6.0  0   3.0  0.20 A   12 V  10  0.20 A   5.4 V
Since V ab  0 , pointa is 5.4 V higherin potentialthan pointb