AP Physics Homework Solutions
Unit 9: Electric Circuits
Homework 4 Page 617 problems 4,6,8,9,11,13
18.4
(a) The current through this series combination is
I
 V bc
Rbc

12 V
 2.0 A
6.0 
Therefore, the terminal potential difference of the
power supply is
V  IReq   2.0 A   9.0   6.0    30 V
(b) When connected in parallel, the potential difference across either resistor is the voltage setting of
the power supply. Thus,
V  I9 R9   0.25 A  9.0    2.3 V
18.6
The equivalent resistance of the parallel combination of three resistors is
1
1
1 
 1
Rp  


  3.0 
 18  9.0  6.0  
Hence, the equivalent resistance of the circuit connected to the 30 V source is
Req  R12  R p  12   3.0   15 
18.8
(a) The rules for combining resistors in series and parallel are used to reduce the circuit to an
equivalent resistor in the stages shown below. The result is Req  5.13  .
(b) From P   V  Req , the emf of the power source is
2
V  PReq 
18.9
 4.00 W  5.13    4.53 V
Turn the circuit given in Figure P18.9 90° counterclockwise to observe that it is equivalent to that
shown in Figure 1 below. This reduces, in stages, as shown in the following figures.
From Figure 4,
I
V 25.0 V

 1.93 A
R
12.9 
(b) From Figure 3,
 V ba  IRba
  1.93 A   2.94    5.68 V
(a) From Figures 1 and 2, the current through the 20.0  resistor is
I20 
 V ba 5.68 V
Rbca

25.0 
 0.227 A
18.11
The equivalent resistance is Req  R  R p , where R p is the total resistance of the three parallel branches;
 30   R  5.0  
1
1
1
 1

 1

Rp  



 
 
R  35 
 120  40  R  5.0  
 30  R  5.0  
1
Thus, 75   R 
1
 30   R  5.0   R 2   65   R  150 2
R  35 

R  35 
which reduces to R 2  10  R  2475 2  0 or  R  55  R  45   0 .
Only the positive solution is physically acceptable, so R  55 
18.13
The resistors in the circuit can be combined in the stages shown below to yield an equivalent resistance
of R ad   63 11  .
From Figure 5, I
 V ad
R ad
Then, from Figure 4,

18 V
 3.14 A
 63 11 
 V bd  IRbd   3.14 A   30 11   8.57 V
Now, look at Figure 2 and observe that
I2 
so
 V bd
3.0   2.0 

8.57 V
 1.71 A
5.0 
71 A  3.0   5.14 V
 V be  IR
2 be   1.
Finally, from Figure 1,
I12 
 V be 5.14 V
R12

12 
 0.43 A