Physics 104
26.3.
IDENTIFY: The emf of the battery remains constant, but changing the resistance across it changes its power
output.
V2
.
SET UP: The power output across a resistor is P 
R
EXECUTE: With just R1, P1 
R2 added, Rtot  400. P 
26.4.
Assignment 26
V2
and V  P1R1  (360W)(250)  300 V is the battery voltage. With
R1
V 2 (300 V)2

 225 W.
Rtot
400
EVALUATE: The two resistors in series dissipate electrical energy at a smaller rate than R1 alone.
IDENTIFY: For resistors in parallel the voltages are the same and equal to the voltage across the
equivalent resistance.
1
1
1
SET UP: V  IR.
  .
Req R1 R2
1 
 1

EXECUTE: (a) Req  
 32 20 
(b) I 
1
 123 .
V
240 V

 195 A.
Req 123 
V 240 V
V 240 V

 75 A; I 20  
 12 A.
R 32 
R 20 
EVALUATE: More current flows through the resistor that has the smaller R.
IDENTIFY: First do as much series-parallel reduction as possible.
SET UP: The 45.0- and 15.0- resistors are in parallel, so first reduce them to a single equivalent resistance.
Then find the equivalent series resistance of the circuit.
EXECUTE: 1/Rp  1/(45.0 )  1/(15.0 ) and Rp  11.25 . The total equivalent resistance is
(c) I 32 
26.7.
26.8.
18.0   11.25   3.26   32.5 . Ohm’s law gives I  (25.0 V)/(32.5 )  0.769 A.
EVALUATE: The circuit appears complicated until we realize that the 45.0- and 15.0- resistors are in
parallel.
IDENTIFY: Eq. (26.2) gives the equivalent resistance of the three resistors in parallel. For resistors in parallel,
the voltages are the same and the currents add.
(a) SET UP: The circuit is sketched in Figure 26.8a.
EXECUTE: parallel
1
1
1
1
 

Req R1 R2 R3
1
1
1
1



Req 160 240 480
Req  0800
Figure 26.8a
(b) For resistors in parallel the voltage is the same across each and equal to the applied voltage;
V1  V2  V3    280 V
V  IR so I1 
I2 
V1 280 V

 175 A
R1 160 
V2 280 V
V
280 V

 117 A and I 3  3 
 58 A
R2 240 
R3 48 
1
(c) The currents through the resistors add to give the current through the battery:
I  I1  I 2  I3  175 A  117 A  58 A  350 A
EVALUATE: Alternatively, we can use the equivalent resistance Req as shown in Figure 26.8b.
  IReq  0
I

Req

280 V
 350 A, which checks
0800 
Figure 26.8b
(d) As shown in part (b), the voltage across each resistor is 28.0 V.
(e) IDENTIFY and SET UP: We can use any of the three expressions for P:P  VI  I 2R  V 2 /R. They will all
give the same results, if we keep enough significant figures in intermediate calculations.
(280 V) 2
(280 V) 2
 490 W, P2  V22 /R2 
 327 W, and
EXECUTE: Using P  V 2 /R, P1  V12 /R1 
160
240
P3  V32 /R3 
(28.0 V)2
 163W.
4.80 
EVALUATE: The total power dissipated is Pout  P1  P2  P3  980 W. This is the same as the power
Pin   I  (280 V)(350 A)  980 W delivered by the battery.
26.10.
(f) P  V 2 /R. The resistors in parallel each have the same voltage, so the power P is largest for the one with the
least resistance.
(a) IDENTIFY: The current, and hence the power, depends on the potential difference across the resistor.
SET UP: P  V 2 /R
EXECUTE: (a) V  PR  (50 W)(15,000)  274 V
(b) P  V 2 /R  (120 V)2 /(9,000)  16 W
SET UP: (c) If the larger resistor generates 2.00 W, the smaller one will generate less and hence will be safe.
Therefore the maximum power in the larger resistor must be 2.00 W. Use P  I 2 R to find the maximum current
through the series combination and use Ohm’s law to find the potential difference across the combination.
EXECUTE: P  I 2 R gives I  P/R  (200 W)/(150 )  0115 A. The same current flows through both
resistors, and their equivalent resistance is 250 . Ohm’s law gives V  IR  (0.115 A)(250 )  28.8 V.
Therefore P150  2.00 W and P100  I 2R  (0.115 A)2 (100 )  1.32 W.
EVALUATE: If the resistors in a series combination all have the same power rating, it is the largest resistance
that limits the amount of current.
1
1
1
26.13.IDENTIFY: For resistors in parallel, the voltages are the same and the currents add.
so


Req R1 R2
R1R2
, For resistors in series, the currents are the same and the voltages add. Req  R1  R2 .
R1  R2
SET UP: The rules for combining resistors in series and parallel lead to the sequences of equivalent circuits
shown in Figure 26.13.
600 V
EXECUTE: Req  500. In Figure 26.13c, I 
 120 A. This is the current through each of the
500
resistors in Figure 26.13b. V12  IR12  (120 A)(200)  240 V. V34  IR34  (120 A)(300)  360 V.
Req 
Note that V12  V34  600 V. V12 is the voltage across R1 and across R2 , so I1 
2
V12 240 V

 800 A and
R1 300
I2 
V12 240 V
V
360 V

 400 A. V34 is the voltage across R3 and across R4 , so I 3  34 
 300 A and
R2 600
R3 120 
I4 
V34 360 V

 900 A.
R4 400 
EVALUATE: Note that I1  I 2  I3  I 4 .
Figure 26.13
26.17.IDENTIFY: Apply Ohm’s law to each resistor.
SET UP: For resistors in parallel the voltages are the same and the currents add. For resistors in series the
currents are the same and the voltages add.
EXECUTE: The current through the 2.00- resistor is 6.00 A. Current through the 1.00- resistor also is 6.00
A and the voltage is 6.00 V. Voltage across the 6.00- resistor is 12.0 V  6.0 V  18.0 V. Current through the
6.00- resistor is (18.0 V)/(6.00 )  3.00 A. The battery emf is 18.0 V.
EVALUATE: The current through the battery is 6.00 A  3.00 A  9.00 A. The equivalent resistor of the resistor
network is 2.00 , and this equals (18.0 V)/(9.00 A).
26.25.IDENTIFY: Apply Kirchhoff’s point rule at point a to find the current through R. Apply Kirchhoff’s loop rule to
loops (1) and (2) shown in Figure 26.25a to calculate R and  . Travel around each loop in the direction shown.
(a) SET UP:
Figure 26.25a
EXECUTE: Apply Kirchhoff’s point rule to point a:  I  0 so I  400 A  600 A  0
I  2.00 A (in the direction shown in the diagram).
(b) Apply Kirchhoff’s loop rule to loop (1): (600 A)(300)  (200 A) R  280 V  0
180 V  (200) R  280 V  0
280 V  180 V
 500
200 A
(c) Apply Kirchhoff’s loop rule to loop (2): (600 A)(300)  (400 A)(600)    0
  180 V  240 V  420 V
EVALUATE: Can check that the loop rule is satisfied for loop (3), as a check of our work:
280 V    (400 A)(600)  (200 A)R  0
R
3
280 V  420 V  240 V  (200 A)(500)  0
520 V  420 V  100 V
520 V  520 V, so the loop rule is satisfied for this loop.
(d) IDENTIFY: If the circuit is broken at point x there can be no current in the 600- resistor. There is now
only a single current path and we can apply the loop rule to this path.
SET UP: The circuit is sketched in Figure 26.25b.
Figure 26.25b
EXECUTE: 280 V  (300)I  (500)I  0
280 V
 350 A
800 
EVALUATE: Breaking the circuit at x removes the 42.0-V emf from the circuit and the current through the
300- resistor is reduced.
26.27.IDENTIFY: Apply the junction rule at points a, b, c and d to calculate the unknown currents. Then apply the loop
rule to three loops to calculate 1, 2 and R.
(a) SET UP: The circuit is sketched in Figure 26.27.
I
Figure 26.27
EXECUTE: Apply the junction rule to point a: 300 A  500 A  I3  0
I3  800 A
Apply the junction rule to point b: 200 A  I 4  300 A  0
I 4  100 A
Apply the junction rule to point c: I3  I 4  I5  0
I5  I3  I 4  800 A  100 A  700 A
EVALUATE: As a check, apply the junction rule to point d: I5  200 A  500 A  0
I5  700 A
(b) EXECUTE: Apply the loop rule to loop (1): 1  (300 A)(400)  I3 (300)  0
1  120 V  (800 A)(300)  360 V
Apply the loop rule to loop (2):  2  (500 A)(600)  I3 (300)  0
4
 2  300 V  (800 A)(300)  540 V
(c) Apply the loop rule to loop (3): (200 A) R  1   2  0
 2  1 540 V  360 V

 900
200 A
200 A
EVALUATE: Apply the loop rule to loop (4) as a check of our calculations:
(200 A)R  (300 A)(400)  (500 A)(600)  0
(200 A)(900)  120 V  300 V  0
180 V  180 V  0
IDENTIFY: Use Kirchhoff’s rules to find the currents.
SET UP: Since the 10.0-V battery has the larger voltage, assume I1 is to the left through the 10-V battery, I 2
R
26.28.
is to the right through the 5 V battery, and I 3 is to the right through the 10- resistor. Go around each loop in
the counterclockwise direction.
EXECUTE: Upper loop: 100 V  (200   300 ) I1  (100   400 ) I 2  500 V  0. This gives
50 V  (500 ) I1  (500 ) I 2  0, and  I1  I 2  100 A.
Lower loop: 500 V  (100   400 ) I 2  (100 ) I3  0. This gives 500 V  (500 ) I 2  (100 ) I3  0,
and I 2  2I3  100 A.
Along with I1  I 2  I3 , we can solve for the three currents and find:
I1  0800 A, I 2  0200 A, I3  0600 A.
(b) Vab  (0200 A)(400 )  (0800 A)(300 )  320 V.
EVALUATE: Traveling from b to a through the 400- and 300- resistors you pass through the resistors in
the direction of the current and the potential decreases; point b is at higher potential than point a.
26.40.IDENTIFY: An uncharged capacitor is placed into a circuit. Apply the loop rule at each time.
SET UP: The voltage across a capacitor is VC  q /C.
EXECUTE: (a) At the instant the circuit is completed, there is no voltage across the capacitor, since it has no
charge stored.
(b) Since VC  0, the full battery voltage appears across the resistor VR    245 V.
(c) There is no charge on the capacitor.

245 V
(d) The current through the resistor is i 

 00327 A.
Rtotal 7500
(e) After a long time has passed the full battery voltage is across the capacitor and i  0. The voltage across the
capacitor balances the emf: VC  245 V. The voltage across the resistor is zero. The capacitor’s charge is
26.41.
q  CVC  (460  106 F) (245 V)  113  103 C. The current in the circuit is zero.
EVALUATE: The current in the circuit starts at 0.0327 A and decays to zero. The charge on the capacitor starts
at zero and rises to q  113  103 C.
IDENTIFY: The capacitor discharges exponentially through the voltmeter. Since the potential difference across
the capacitor is directly proportional to the charge on the plates, the voltage across the plates decreases
exponentially with the same time constant as the charge.
SET UP: The reading of the voltmeter obeys the equation V  V0e –t /RC , where RC is the time constant.
EXECUTE: (a) Solving for C and evaluating the result when   4.00s gives
t
400 s
C

 8.49  107 F
R ln(V /V0 )
 120 V 
6
(340  10 )ln 
 300 V 
(b)   RC  (3.40  106 )(8.49  107 F)  2.89 s
EVALUATE: In most laboratory circuits, time constants are much shorter than this one.
26.46.IDENTIFY: The charge is increasing while the current is decreasing. Both obey exponential equations, but they are
not the same equation.
5
SET UP: The charge obeys the equation Q  Qmax (1  et /RC ), but the equation for the current is
I  I maxe–t /RC .
EXECUTE: When the charge has reached
1
4
of its maximum value, we have Qmax /4  Qmax (1– e–t /RC ), which
says that the exponential term has the value e – t /RC  34 . The current at this time is
I  I maxe–t /RC  Imax (3/4)  (3/4)[(10.0 V)/(12.0 )]  0.625 A.
EVALUATE: Notice that the current will be
3
,
4
not
1
,
4
of its maximum value when the charge is
1
4
of its
maximum. Although current and charge both obey exponential equations, the equations have different forms for
a charging capacitor.
6