23
CIRCUITS
Questions
Q23.14. Reason: All of the current in the circuit goes through bulb D, so it is the brightest. The same current goes
through bulbs A and B, so they are equally bright. Since all the bulbs are identical, the parallel branch with bulbs A and B has
twice the resistance as the parallel branch with bulb C and hence half the current. This allows us to rank the brightness of the
bulb as follows: D  C  A  B.
Assess: The brightness of the bulb depends on the current and the current depends on the resistance.
Q23.18. Reason: (a) The total current splits and some goes through the parallel branch containing the point a (the left
branch) and some goes through the parallel branch containing the point b (the right branch). Notice that the resistance of each
branch is the same (3R), as a result the current splits equally between the two branches. Let’s call the current in the two
branches I. This current in the two branches causes a potential drop of 2IR in the resistor 2R in the left branch and a potential
drop of IR in the resistor R in the right branch. As a result, the electric potential at point b is greater (it is more positive) than
the electric potential at point a and Vab  0.
(b) Since point b is at a higher potential than point a, if they are connected there will be an electric current from point b to
point a.
Assess: The solution of the question requires a good understanding of electric potential difference.
Q23.27. Reason: Increasing the dielectric constant increases the capacitance. Since   RC, increasing the capacitance
increases the charging and discharging time.
Assess: One can therefore determine skin moisture by measuring the charging and discharging time.
Q23.28. Reason: (a) The new resistor has half the length of the old resistor, so the new resistance is half the old.
The capacitance depends on A the area, and halving the distance between Ranvier nodes would halve A and therefore also
halve C.
(b) The new time constant is    RC  ( 12 R)( 12 C)  14  .
(c) Also note that xnode  12 xnode:
 xnode 12 xnode
 1
 2v


4
v 
So the speed of propagation is twice what it was.
Assess: If the distance between the nodes of Ranvier is halved, the propagation speed is doubled. Autoimmune diseases such
as multiple sclerosis and ALS attack and destroy one’s myelin so that the nerve impulses travel much slower.
Q23.34. Reason: First use Ohm’s law to find Req.
Req 
V 8 0 V

 4 0 
I
2 0 A
Now we use the formula for resistors in parallel:
1
1
1
1
 

Req R1 R2 R3
Solve for R3 where Req  40 , R1  10 , and R2  15 .
23-1
23-2
Chapter 23
1
1
 1
1
1   1
1
1 
R3  
   


 12 
 Req R1 R2   40  10  15  


The correct choice is B.
Assess: We know the equivalent resistance is less than any of the individual resistors in parallel, and our answer fits that
criterion.
Problems
P23.6. Prepare: Please refer to Figure P23.6. The batteries and the connecting wires are ideal. Choose the current I to be
in the clockwise direction. If I ends up being a positive number, then the current really does flow in this direction. If I is
negative, the current really flows counterclockwise. There are no junctions, so I is the same for all elements in the circuit.
Solve: (a) With the 9 V battery being labeled 1 and the 6 V battery being labeled 2, Kirchhoff’s loop law is
 V  V
i
bat 1
 VR  Vbat 2  E1  IR  E2  0  I 
E1  E2 9 V  6 V

 0.10 A
R
30 
Note the signs: Potential is gained in battery 1, but potential is lost both in the resistor and in battery 2. Because I is positive,
we can say that I  0.10 A flows from left to right through the resistor.
(b)
Assess: The graph shows 9 V gained in battery 1, VR  IR  3 V lost in the resistor, and another 6 V lost in battery
2. The final potential is the same as the initial potential, as required.
P23.13. Prepare:
For resistors in parallel,
 1

1
Req   
   
R
R
2
 1

1
Adding four of the resistors in parallel gives
1
Solve:
1
 1
1
1
1   4 
Req  



 
  0 25 k
 1 0 k 1 0 k 1 0 k 1 0 k   1 0 k 
We can put four of the resistors in parallel. The total resistance is now 025 k.
Assess: There are other ways to arrive at the same Req using more from our collection of 10 k resistors.
P23.14. Prepare: When resistors are combined in parallel, the combination is always less than the smallest resistor. As a
result, the way to obtain the smallest resistance is to connect the resistors in parallel.
Solve: The equivalent resistance when the six resistors are connected in parallel is obtained by
1
1 1 1 1 1 1 6
      
or RTotal  R/6  1000  /6  167 
RTotal R R R R R R R
Assess: Notice that the total resistance is less than the smallest resistor. The smallest resistor is 1000  and the total
resistance when the six resistors are connected in parallel is 167 .
Circuits
23-3
P23.18. Prepare: The connecting wires are ideal with zero resistance. For the first step, the 10  and 30  resistors are in
series; for the second step, the 60  and 40  resistors are in parallel; and for the third step, the
24  and 10  resistors are in series.
Solve:
For the first step, the 10  and 30  resistors are in series and the equivalent resistance is 40 . For the second step, the 60 
and 40  resistors are in parallel and the equivalent resistance is
1
 1
1 


  24 
40

60


For the third step, the 24  and 10  resistors are in series and the equivalent resistance is 34 .
Assess: Have a good understanding of how series and parallel resistances are combined to get an equivalent
resistance.
P23.20. Prepare: Assume ideal batteries and wires. Use Kirchhoff’s loop law (going clockwise from the lower left corner)
and solve for I . Due to the orientation of both batteries, the current must be going clockwise, so the potential differences
shown for the two resistors need to be subtracted.
Solve:
Vi  45 V  30 V  35 V  I (150 )  20V  0V  I 
2V
 13 mA
150 
Assess: This is a small but reasonable current.
P23.38. Prepare: Assume ideal battery, wires, and capacitors.
Solve:
(a) The two on the right are in parallel, so we add them to get 35  F . That in series with 40  F gives
1
 1
1 
Ceq  

  1867  F  19  F
3

5

F
4

0
F 

(b) The amount of charge that flows while charging is given by
Q  CE  (1867 F)(12 V)  22 C
Assess: The values are typical.
23-4
Chapter 23
P23.56. Prepare: The power supply needs a larger emf than the 1.5 V of the battery. 10 min  600 s.
Solve:
(a) The power supply needs an emf that is greater than the battery’s by V  IR for the desired current.
E  15 V  (075 A)(10 )  225 V  23 V
(b) The energy going into the battery is
E  Pt  IVt  (075 A)(225 V)(600 s)  10 kJ
The energy dissipated by the internal resistance of the battery is
Pt  I 2 Rt  (075 A)2 (10 )(600 s)  340 J
Assess: Of the energy put into the battery to recharge it about one third is dissipated as thermal energy.
P23.60. Prepare: The connecting wires are ideal with zero resistance. A visual overview of how to reduce the circuit to
an equivalent resistance is shown.
Solve: In the first step, the resistors 100 , 100 , and 100  in the top branch are in series. Their combined resistance is
300 . In the middle branch, the two resistors, each 100 , are in series. So, their equivalent resistance is 200 . In the second
step, the three resistors are in parallel. Their equivalent resistance is
1
1
1
1
 Req  54.5 



Req 300  200  100 
The equivalent resistance of the circuit is 55 .
ideal. Current and voltage during a capacitor discharge are given by Equations 23.22. Because the charge on a capacitor is Q 
CV, the decay of the capacitor charge is given by Q  Qo e– t/.
Solve: (a) The capacitor discharges through the resistor R as Q  Qo e – t/ For Q  Q0 / 2,
Q0
t
1
 Q0et /15 ms  ln    
 t  (0.015 s)ln(0.5)  10.4 ms  10 ms
2
0.015 s
2
(b) If the initial capacitor energy is U0, we want the time when the capacitor’s energy will be U  U 0 /2. Noting that
U0  Q02 /(2C ), this means Q  Q0 / 2. Applying the equation for the discharging capacitor,
Q0
 1 
t
 1 
 Q0et /15 ms  ln 
 t  (0.015 s)ln 

  5.2 ms
0.015
s
2
2


 2
Assess: A time of 5.2 ms for the energy to become half of its original energy is typical of discharging capacitors.
Notice the answer to part (b) is half the answer to part (a).
P23.74. Prepare: Please refer to Figure P23.74. In an RC circuit, the capacitor voltage discharge is given by Equation
23.22, i.e., V  V0et/.
Solve: From Figure P23.74, we note that V0  30 V and V  10 V at t  2 ms. So,
10 V  (30 V)e2 ms/(R (5010
6
F))
 10 V 
2 103 s
2 103 s
 ln 
R
 36 

6
R(50 10 F)
(50 106 F)ln  13 
 30 V 