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(D) The power delivered to the factory is $P_{out} = 10 	ext{ kW} = 10000 	ext{ W}$ at a voltage $V = 200 	ext{ V}$.The current $I$ flowing through

Vedclass · Accepted Answer

$95$

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(D) The power delivered to the factory is $P_{out} = 10 	ext{ kW} = 10000 	ext{ W}$ at a voltage $V = 200 	ext{ V}$.The current $I$ flowing through the cable is given by $I = \frac{P_{out}}{V} = \frac{10000}{200} = 50 	ext{ A}$.The power loss in the cable due to its resistance $R = 0.2 \Omega$ is $P_{loss} = I^2 R = (50)^2 	imes 0.2 = 2500 	imes 0.2 = 500 	ext{ W}$.The total power generated at the source is $P_{in} = P_{out} + P_{loss} = 10000 + 500 = 10500 	ext{ W}$.The efficiency of transmission is $\eta = \frac{P_{out}}{P_{in}} 	imes 100 = \frac{10000}{10500} 	imes 100 \approx 95.24 \%$.Rounding to the nearest given option, the efficiency is $95 \%$.

An electrical cable having a resistance of $0.2 \Omega$ delivers $10 \text{ kW}$ at $200 \text{ V}$ $D$.$C$. to a factory. What is the efficiency of transmission (in $\%$)?

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