- #1

- 238

- 26

- Homework Statement
- An electric resistance space heater rated at 1500 W for a voltage source of v(t)= 120 * (2)^(0.5) sin (2*pi*60t) V has a thermostatically controlled switch. The heater periodically switches on for 5 min and off for 7 min. Determine (a) the maximum instantaneous power, (b) the average power over the 12-min cycle, and (c) the electric energy converted to heat in each 12-min cycle

- Relevant Equations
- p(t) = v(t)*i(t)

The v(t) = 120##\sqrt(2)\sin(2\pi 60t)##. Let the resistance of the heater be "R", then

the rated power = 1500W. The rated power is the average power

1500 = ##\frac{120^2 *5 * 60} {12*60 * R} ## ->eq1

R = 4 Ohm.

a. Instantaneous power is

##P = \frac{v^2} R ## -> eq2

##P = \frac{{(120\sqrt2})^2} {R} ##

##P = 7200 W ##

b. The average power should be the rated power which is 1500W.

c. The energy is E = P*t Joules

##E = \frac{5*60 *(120\sqrt2)^2} {4} J = 2.16*10^6J##

Are the solutions correct? My main doubt is the rated power the same as average power?

the rated power = 1500W. The rated power is the average power

1500 = ##\frac{120^2 *5 * 60} {12*60 * R} ## ->eq1

R = 4 Ohm.

a. Instantaneous power is

##P = \frac{v^2} R ## -> eq2

##P = \frac{{(120\sqrt2})^2} {R} ##

##P = 7200 W ##

b. The average power should be the rated power which is 1500W.

c. The energy is E = P*t Joules

##E = \frac{5*60 *(120\sqrt2)^2} {4} J = 2.16*10^6J##

Are the solutions correct? My main doubt is the rated power the same as average power?