Question

A 540 w electric heater is designed to operate from 120 v lines. (a) what is its resistance? (b) what current does it draw? (c) if the line voltage drops to 110 v, what power does the heater take? (assume

Answer 1

P=540w,V=120v
1)R=?
P=VI
P=V(V/R) ..by ohms law
P=V^2/R
R=V^2/P
R=120x120 / 540
R=14400 / 540 =26.66
R=27ohm(approx)

2)I=?
V=IR
I=V/R=120/27=4.44
I=4.5A(approx)

3)if V=110,P=?
P=VI---(1)
540=110xI
I=4.90
from (1)..,
P=110x4.9
P=539W

Answer 2

Concept:

If all physical conditions and temperatures stay constant, Ohm's law asserts that the voltage across a conductor is directly proportional to the current flowing through it.

Given:

$P=540 \mathrm{~W} and \\ V=120 \mathrm{~V} .$

To Find:

(a) electric heater's resistance.

(b) the current does it draws.

(c) if the line voltage drops to 110 v, power of the heater.

Answer:

(a). The heater's power consumption is determined by its resistance. This resistance can be determined using:

                                      $R=\frac{V^{2}}{P}$

Substituting the values, we get

$R=\frac{V^{2}}{P}=\frac{(120 \mathrm{~V})^{2}}{540 \mathrm{~W}}$

$R= 26.7 ohm$

(b). Current through the heater can be calculated by

$P=I^{2} R \\I=\sqrt{\frac{P}{R}}$

Substituting the values, we get

$I=\sqrt{\frac{P}{R}}=\sqrt{\frac{540 \mathrm{~W}}{26.7 \Omega}}$

$I= 4.50 A$

(c). The line voltage drops to $110 V$, assuming the voltage is constant, power will be

$P=\frac{V^{2}}{R}$

Substituting the values, we get

$P=\frac{V^{2}}{R}=\frac{(110 \mathrm{~V})^{2}}{26.7 \Omega}$

$P = 453 W$

The resistance of electric heater is $26.7 ohm$ . the current drawn by the electric heater is $4.50A$ . The power when voltage drops to 110V and constant resistance is $453W$.

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