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It is known that when a heavy appliance comes on, lights that are connected to the...

It is known that when a heavy appliance comes on, lights that are connected to the same circuit will dim. Consider the case of a motor in a refrigerator that is in parallel to a bulb.

a)         Given the voltage source is 120 V, the wire resistance is 0.400 Ω, and the bulb is nominally 75.0 W, what power will the bulb dissipate if a total of 30.0 A passes through the wires when the motor comes on? Assume negligible change in bulb resistance.

b)         What is the power consumed by the motor?

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Answer #1

a)RBulb = V2 / PBulB =1202 / 75 = 192 ohms

Voltage across the wire resistance R2 , V1= I R2 = 30 * 0.4 = 12 V

Voltage across parallel combination motor and bulb, V2 = 120 - 12 = 108 V

Power dissipated in the bulb , P = V22 / RBulb = 1082 / 192

= 60.75 W

b) Pinput = 120 * 30 = 3600 W

PWire = I2 R2 = 302 * 0.4 = 360 W

Pmotor = Pinput - Pbulb - Pwire

= 3600 - 360 - 60.75

= 3179 W = 3.17 kW

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