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### College Physics (Urone)

Book edition 1st Edition
Author(s) Paul Peter Urone
Pages 1272 pages
ISBN 9781938168000

# Refer to Figure 21.7 and the discussion of lights dimming when a heavy appliance comes on. (a) Given the voltage source is 120 V , the wire resistance is${\mathbf{0}}{\mathbf{.}}{\mathbf{400}}{\mathbf{\Omega }}$, and the bulb is nominally 75.0W , what power will the bulb dissipate if a total of 15.0 A passes through the wires when the motor comes on? Assume negligible change in bulb resistance. (b) What power is consumed by the motor?

(a)The power will the bulb dissipate if a total of 15.0 A passes through the wires when the motor comes on is $66.8\mathrm{W}$.

(b) The power consumed by motor is 1644 W.

See the step by step solution

## Step 1: Definition of Power

Electric power is the percentage of time it takes to work or generate energy and is expressed as the work W or the transmitted energy divided by the time interval t.

## Step 2: Calculating P2

(a)

Calculating resistance in the bulb,

${\mathrm{R}}_{\mathrm{Bulb}}=\frac{{\mathrm{E}}^{2}}{{\mathrm{P}}_{1}}\phantom{\rule{0ex}{0ex}}=\frac{{\left(120\mathrm{V}\right)}^{2}}{75\mathrm{W}}\phantom{\rule{0ex}{0ex}}=192\mathrm{\Omega }$

Next step is calculating current flowing through bulb when motor comes on:

${\mathrm{I}}_{2}=\frac{\mathrm{E}-{\mathrm{IR}}_{\mathrm{wire}}}{{\mathrm{R}}_{\mathrm{Bulb}}}\phantom{\rule{0ex}{0ex}}=\frac{120\mathrm{V}-\left(15.0\mathrm{A}×0.400\mathrm{\Omega }\right)}{192\mathrm{\Omega }}\phantom{\rule{0ex}{0ex}}=0.59\mathrm{A}$

Finally, power in the bulb with motor on is:

${\mathrm{P}}_{2}={\mathrm{I}}_{2}^{2}{\mathrm{R}}_{\mathrm{b}}\phantom{\rule{0ex}{0ex}}={\left(0.59\mathrm{A}\right)}^{2}×192\mathrm{\Omega }\phantom{\rule{0ex}{0ex}}=66.8\mathrm{W}$

Therefore, the power will the bulb dissipate if a total of 15.0 A passes through the wires when the motor comes on is 66.8 W.

## Step 3: Power consumed by motor

(b)

To calculate power consumed by the motor we need to calculate current going through the motor:

${\mathrm{I}}_{3}=\mathrm{I}-{\mathrm{I}}_{2}\phantom{\rule{0ex}{0ex}}=15.0\mathrm{A}-0.59\mathrm{A}\phantom{\rule{0ex}{0ex}}=14.41\mathrm{A}$

Next, resistance of the motor can be calculated as:

${\mathrm{R}}_{\mathrm{m}}=\frac{\mathrm{E}-{\mathrm{IR}}_{1}}{{\mathrm{I}}_{3}}\phantom{\rule{0ex}{0ex}}=\frac{120\mathrm{V}-\left(15.0\mathrm{A}×0.400\mathrm{\Omega }\right)}{14.41\mathrm{A}}\phantom{\rule{0ex}{0ex}}=7.92\mathrm{\Omega }$

Finally, power of the motor is:

${\mathrm{P}}_{\mathrm{m}}={\mathrm{I}}_{3}^{2}\mathrm{Rm}\phantom{\rule{0ex}{0ex}}={\left(14.41\mathrm{A}\right)}^{2}×\left(7.92\mathrm{\Omega }\right)\phantom{\rule{0ex}{0ex}}=1644\mathrm{W}$

Hence, the power consumed by motor is 1644 W.