My Notes Ask Your Toae In the gure below the voltage source is 120 V the w the b

Question

My Notes Ask Your Toae In the gure below the voltage source is 120 V the w the bulb and causes it to dm noticeaby e esistance is 0.400 and the bulb is non ally 71.3 w when the motor which has a lo resistance comes on, a large cu ent no s causing a significant oitage drop the ire. This educes the voit ge eceived y Refrigerator Low Bulb dims Large IR drop in wines resistance Voltage source Ry draws large a) What power will the bulb dissipate if a total of 26.8 A passes hrcugh the wires when the motor comes on? Assume negligible change in bulb resistance ) What power is consumed by the motor?

Explanation / Answer

A)

R bulb = 120^2/71.3 = 201.96 ohms

now

Req = V/ I = 120 /26.8 = 4.48 ohms

Req - 0.4 = Rp = 4.08 ohms

so V across bulb = 120 - 26.8*0.4 = 109.28 V

so Power = V^2/R = 109.28^2/201.96 = 59.13 W

B)

Input power, Pinput = 120*26.8 = 3216 W

Power dissipated in wire, PWire= I2R2 = 26.82*0.4 = 287.30 W

Power Consumed by the motor, Pmotor= Pinput - Pbulb - Pwire = 3216 - 287.30 - 59.13

Pmotor = 2869.57 W = 2.869 kW

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