232 Lecture supplement 12
67.
Three 60.0-W, 120-V lightbulbs are connected across a 120-V power
source, as shown in Figure P28.67. Find
(a) the total power delivered to the three bulbs and
(b) the voltage across each. Assume that the resistance of each bulb is constant
(even though in reality the resistance might increase markedly with current).
(c) Now the switch S is opened. Compare the brightness of bulbs R1, R2 and R3
before and after the switch is opened.
S
P28.67
(a)
First determine the resistance of each light bulb: P =
R=
( ∆V ) 2
( ∆V ) 2 ( 120 V ) 2
R
=
= 240 Ω .
P
60.0 W
We obtain the equivalent resistance R eq of the network of light
bulbs by identifying series and parallel equivalent resistances:
1
R eq = R1 +
= 240 Ω + 120 Ω = 360 Ω .
R
1
+
( 2 ) (1 R 3 )
The total power dissipated in the 360 Ω is
P=
(b)
( ∆V ) 2 ( 120 V ) 2
R eq
=
360 Ω
= 40.0 W
.
The current through the network is given by P = I2R eq :
I=
P
=
Req
40.0 W
1
= A.
360 Ω
3
The potential difference across R1 is
1 
∆V1 = IR1 =  A  ( 240 Ω) = 80.0 V .
3 
FIG. P28.67
The potential difference ∆V 23 across the parallel combination of R 2 and R 3 is

1
1 
∆V 23 = IR 23 =  A  
= 40.0 V .
 3   (1 240 Ω ) + (1 240 Ω ) 
You can also determine ∆V23 by using:
120V = ∆V23 + ∆V1
⇒ ∆V23 = 120 − 80 = 40V
(c) The power dissipated in each bulb before S is opened:
P1 =
(∆V1 ) 2
80 2
=
W = 26.67 W
240
R1
(∆V23 ) 2
40 2
=
W = 6.67 W
R2
240
When the switch is opened R3 is off. Now R1 and R2 are in series and the equivalent
resistance is R1+R2 = 480 Ω. And the current through both bulbs:
∆V 120
I=
=
= 0.25 A
Req 480
So now the new potential differences and power dissipated for R1 and R2:
P2 = P3 =
∆V1 = IR1 = 0.25.240 = 60 V
∆V2 = IR2 = 0.25.240 = 60 V
P1 = P2 =
(∆V1 ) 2
(∆V2 ) 2
60 2
=
=
W = 15 W
240
R1
R2
So when S is opened R1 becomes less bright while R2 becomes brighter.