A 500, W heating unit is designed to operate | vedclass

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(C) The power consumed by a heating element is given by $P = \frac{V^2}{R}$.Since the resistance $R$ of the heating unit remains constant,the power is

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$8.6$

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(C) The power consumed by a heating element is given by $P = \frac{V^2}{R}$.Since the resistance $R$ of the heating unit remains constant,the power is proportional to the square of the voltage: $P \propto V^2$.Therefore,the ratio of the new power $P_{new}$ to the rated power $P_{rated}$ is given by $\frac{P_{new}}{P_{rated}} = \left( \frac{V_{new}}{V_{rated}} ight)^2$.Given $P_{rated} = 500\, W$,$V_{rated} = 115\, V$,and $V_{new} = 110\, V$,we have:$P_{new} = \left( \frac{110}{115} ight)^2 	imes 500 = (0.9565)^2 	imes 500 \approx 0.9149 	imes 500 = 457.46\, W$.The drop in power is $\Delta P = P_{rated} - P_{new} = 500 - 457.46 = 42.54\, W$.The percentage drop in heat output is $\frac{\Delta P}{P_{rated}} 	imes 100 = \frac{42.54}{500} 	imes 100 = 8.508\% \approx 8.6\% $.

$A$ $500\, W$ heating unit is designed to operate from a $115\, V$ line. If the line voltage drops to $110\, V$,the percentage drop in heat output will be ............... $\%$

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