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 73.3 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.

What power will the bulb dissipate if a total of 22.3 A passes through the wires when the motor comes on? Assume negligible change in bulb resistance.

W

What power is consumed by the motor?

kW

Explanation / Answer

P2 = power taken by bulb = V^2/R2

V = voltage across parallel combination and R2 = resistance of bulb

now P2= 120^2/R2 = 73.3 = 196.453 ohms

now given current when motor comes on = I = 22.3 A

so voltage across wire resistance = I*r = 22.3 * 0.4 = 8.92 V = V1

now voltage across parallel combination = V-V1 = 120- 8.92 = 111.08 volts

now power consumed by bulb = V^2/R2 = 62.81 watts

Power consumed = 0.0628 Kilo watt

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