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(B) The power delivered is $P_{out} = 1 \, kW = 1000 \, W$ at a voltage $V = 220 \, V$.The current $I$ flowing through the line is given by $I = \frac

Vedclass · Accepted Answer

$96$

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(B) The power delivered is $P_{out} = 1 \, kW = 1000 \, W$ at a voltage $V = 220 \, V$.The current $I$ flowing through the line is given by $I = \frac{P_{out}}{V} = \frac{1000}{220} = \frac{50}{11} \, A$.The power loss in the transmission line due to its resistance $R = 2 \, \Omega$ is $P_{loss} = I^2 R$.$P_{loss} = \left( \frac{50}{11} ight)^2 	imes 2 = \frac{2500}{121} 	imes 2 = \frac{5000}{121} \approx 41.32 \, W$.The total power generated at the source is $P_{in} = P_{out} + P_{loss} = 1000 + 41.32 = 1041.32 \, W$.The efficiency $\eta$ is given by $\eta = \left( \frac{P_{out}}{P_{in}} ight) 	imes 100$.$\eta = \left( \frac{1000}{1041.32} ight) 	imes 100 \approx 96.03 \%$.Therefore, the efficiency is approximately $96 \%$.

An electrical power line, having a total resistance of $2 \, \Omega$, delivers $1 \, kW$ at $220 \, V$. The efficiency of the transmission line is approximately $....... \%$.

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