Download CHAPTER 28 ELECTRIC CIRCUITS

CHAPTER 28
ELECTRIC CIRCUITS
Problem
10. In Fig. 28-49 all resistors have the same value, R. What will be the resistance measured (a) between A
and B or
(b) between A and C?
FIGURE 28-49
Problems 10 and 11.
Solution
(a) The resistance between A and B is equivalent to two resistors of value R in series with the parallel
combination of resistors of values R and 2R. Thus, RAB  R  R  R(2 R)/( R  2 R)  8R/3. (b) RAC
is equivalent to just one resistor of value R in series with the parallel combination of R and 2R (since the
resistor at point B carries no current, i.e., its branch is an open circuit). Thus
RAC  R  R(2 R)/3R  5R/3.
Problem
16. What possible resistance combinations can you form using three resistors whose values are
10
. , 2.0 , and 3.0 ? (Use all three resistors.)
Solution
There are eight combinations using all three resistors. (a) One in series with two in parallel:
3  2/3  11/3, 2  3/4  11/4, and 1  6/5  11/5 . (b) One in parallel with two in series:
3  3=(3  3)  3/2, 2  4(2  4)  4/3, and 1  5/(1  5)  5/6  . (c) Three in series:
1  2  3  6 . (d) Three in parallel:
(11  21  31 )1  6/11 .
Problem
18. You have a number of 50- resistors, each capable of dissipating 0.50 W without overheating. How
many resistors would you need, and how would you connect them, so as to make a 50- combination
that could be connected safely across a 12-V battery?
Solution
The 50  combination must be capable of dissipating P  E /R  (12 V) /50   2.8 W of power
(when connected across an ideal 12 V battery), so combinations with at least six resistors (capable of
dissipating 0.5 W each) must be considered. In order to get the same total resistance as each individual
2
2
resistor, n parallel branches of n resistors in series are needed (or n branches in series of n resistors in
parallel), making a total of n 2 resistors. The smallest n 2 greater than 6 is for n  3. (Alternatively, one
could argue that the total current is 12 V/50   0.24 A, while the maximum current in each resistor
element is
0.5 W/50   0.1 A, so at least three equal branches in the circuit are needed.)
Problem
22. What is the current through the 3- resistor in the circuit of Fig. 28-51? Hint: This is trivial. Can you
see why?
FIGURE 28-51
Problem 22.
Solution
The current is
I 3  V3 /R3  6 V/3   2 A, from Ohm’s law. The answer is trivial because the
potential difference across the 3  resistor is evident from the circuit diagram. (However, if the 6 V
battery had internal resistance, an argument like that in Example 28-5 must be used.)
Problem
26. In the circuit of Fig. 28-53 find (a) the current supplied by the battery and (b) the current through the
6- resistor.
Solution
(a) With reference to the solution of the next problem, the resistance of the three parallel resistors in
(12/11) , so the current supplied by the battery is I  E/( R1  R|| )  (6 V)/(23/11)   2.87 A.
(b) The voltage drop across the resistors in parallel is V||  E  IR1  IR|| , and the current through the 6 
resistor is I 6   V||/6 . Thus,
FIGURE 28-53
I 6   (2.87 A)(2/11)  522 mA.
Problem 26 Solution
Problem
28. A 50- resistor is connected across a battery, and a 26-mA current flows. When the 50- resistor is
replaced with a 22- resistor, a 43-mA current flows. What are the battery’s voltage and internal
resistance?
Solution
The circuit diagram is like Fig. 28-10, and Kirchhoff’s voltage law is E  IRint  IRL  0. For the two
cases given, this may be written as E  (26 mA) Rint  (26 mA)(50 ), and
E  (43 mA) Rint  (43 mA)(22 ). Solving for E and Rint , we find
26  50  43  22
Rint 
  20.8 , and E  (26 mA)(50  20.8)   184
. V.
43  26