In the figure below, the voltage source is 120 V, the wire resistance is 0.400 ,

Question

In the figure below, the voltage source is 120 V, the wire resistance is 0.400 , and the bulb is nominally 61.3 W. When the motor (which has a low resistance) comes on, a large current flows causing a significant voltage drop in the wire. This reduces the voltage received by the bulb and causes it to dim noticeably.

(a) What power will the bulb dissipate if a total of 19.8 A passes through the wires when the motor comes on? Assume negligible change in bulb resistance.
W==???

(b) What power is consumed by the motor?
kW==????

Explanation / Answer

a)

Resistance of the light bulb

RBulb=V2/PBulB =1202/61.3=234.91 ohms

Voltage across the wire resistance R2 is

V1=IR2 =19.8*0.4 =7.92 Volts

Now Voltage across parallel combination motor and bulb

V2=120-7.92 =112.08 Volts

Power dissipated in the bulb is

P=V22/RBulb =112.082/234.91

P=53.5 Watts

b)

Input power

Pinput =120*19.8=2376 W

Power dissipated in wire

PWire=I2R2 =19.82*0.4 =156.8 W

Power Consumed by the motor

Pmotor=Pinput-Pbulb-Pwire = 2376-156.8-53.5

Pmotor=2165.7 W

Related Questions

Navigate

• Browse (All)
• Browse I
• Subjects