Question

An electric heater converts 1 kJ of electrical energy to heat energy every second with 100% efficiency. The power produced is:

A. 1 W

B. 1kW

C. 1MW

D. 1 kJ

Answer 1

An electric heater is an electrical device which converts heat into an electrical current. The heating element is an electrical resistor inside each electric heater, and works on the Joule heating principle: an electrical current that passes through a resistor will convert that electrical energy into heat energy]. Calculate ${{\text{P}}_{\text{out }}}$, ${{P}_{\text{in }}}$ and Take out the efficiency by taking out the ratios of two powers.

Formula used:
Efficiency $=\dfrac{P_{\text {out }}}{P_{\text {in }}}$

Complete solution Step-by-Step:
The ratio of total output power to input power, expressed in percent, is efficiency. At full load and nominal input voltage, this is usually specified. The efficiency of the power supply is the amount of the actual power supplied to the components divided by the electrical power drawn from the socket of the mains supply.
As we know that Efficiency $=\dfrac{P_{\text {out }}}{P_{\text {in }}}$
$\eta=100 \%$
So, P $_{\text {out }}=P_{\text {in }} \times \eta$
${{\text{P}}_{\text{out }}}=\dfrac{1000\times 100}{100}$
$\therefore {{\text{P}}_{\text{out }}}=1\text{kW}$
$\therefore$ The power produced is: $\mathrm{P}_{\text {out }}=\dfrac{1000 \times 100}{100}=1 \mathrm{kW}$

Hence, the correct option is (B).

Note:
There is no constant efficiency of the power supplies; it varies with different factors such as environmental conditions and load conditions. When operated at 50 percent of their load, the supplies achieve their maximum efficiency. Indeed, manufacturers only guarantee maximum efficiency when the supply runs at a load of 50 per cent. The efficiency of the power supply is the amount of the actual power supplied to the components divided by the electrical power drawn from the socket of the power supply. This is usually specified at full load and nominal input voltage.