Question

A "590-W" electric heater is designed to operate from 120-V lines.

A)What is its operating resistance?
b)What current does it draw?
c)If the line voltage drops to 110 V, what power does the heater take? (Assume that the resistance is constant. Actually, it will change because of the change in temperature.)
d)The heater coils are metallic, so that the resistance of the heater decreases with decreasing temperature. If the change of resistance with temperature is taken into account, will the electrical power consumed by the heater be larger or smaller than what you calculated in the previous part?
a. It will be smaller. The resistance will be smaller so the current drawn will increase, decreasing the power.
b. It will be smaller. The resistance will be smaller so the current drawn will decrease, decreasing the power.
c. It will be larger. The resistance will be smaller so the current drawn will increase, increasing the power.
d. It will be larger. The resistance will be smaller so the current drawn will decrease, increasing the power.

Answer 1

a) 24.4 Ω

b) 4.92 A

c) 495.9 W

d)

c. It will be larger. The resistance will be smaller so the current drawn will increase, increasing the power.

Step-by-step explanation:

b)

The formula for power is:

P = IV

where,

P = Power of heater = 590 W

V = Voltage it takes = 120 V

I = Current Drawn = ?

Therefore,

590 W = (I)(120 V)

I = 590 W/120 V

I = 4.92 A

a)

From Ohm's Law:

V = IR

R = V/I

Therefore,

R = 120 V/4.92 A

R = 24.4 Ω

c)

For constant resistance and 110 V the power becomes:

P = V²/R

Therefore,

P = (110 V)²/24.4 Ω

P = 495.9 W

d)

If the resistance decreases, it will increase the current according to Ohm's Law. As a result of increase in current the power shall increase according to formula (P = VI). Therefore, correct option is:

c. It will be larger. The resistance will be smaller so the current drawn will increase, increasing the power.