Question

A "570-W" electric heater is designed to operate from 120-V lines. What current does it draw? 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.)

Answer 1

Answer:

4.75A, 479W

Explanation:

The power P, taken by an appliance is related to the current I, drawn by the appliance and the voltage V, at which it is operating as follows;

P = I x V          -------------------(i)

From the question, the appliance is the electric heater and;

P = 570W

V = 120V

Substitute these values into equation (i) as follows;

570 = I x 120

i = $\frac{570}{120}$

I = 4.75

Therefore, the current that it draws is 4.75A

(b) First, lets find the resistance, R, in the heater as follows;

P = $\frac{V^{2} }{R}$   -----------------------(ii)

Where;

P = Power

V = Voltage

Substitute V=120V and P=570W into equation (ii) as follows;

570 = $\frac{120^{2} }{R}$

Cross multiply and solve for R;

570R = 120²

570R = 14400

R = $\frac{14400}{570}$

R =  25.26Ω

The resistance of the heater is 25.26Ω

Now, if the voltage drops to 110V and resistance is assumed to be constant, then using equation (ii);

P = $\frac{V^{2} }{R}$  

...we can calculate the power the heater takes by substituting V=110V and R = 25.26Ω as follows;

P = $\frac{110^{2} }{25.26}$

P = 479

Therefore, the power the heater takes is 479W

Answer 2

Answer:

Watts=Volt*Amps

So A=570/120=4.75amps

If voltage drops to 110V We get A=570/110=5.(18...)amps