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Question

(A) Given: Power delivered $P_{out} = 1 \ kW = 1000 \ W$, Voltage $V = 250 \ V$, Resistance $R = 2 \ \Omega$.The current $I$ flowing through the line

Vedclass · Accepted Answer

$96.9$

Vedclass · Answer

(A) Given: Power delivered $P_{out} = 1 \ kW = 1000 \ W$, Voltage $V = 250 \ V$, Resistance $R = 2 \ \Omega$.The current $I$ flowing through the line is $I = \frac{P_{out}}{V} = \frac{1000}{250} = 4 \ A$.The power loss in the transmission line is $P_{loss} = I^2 R = (4)^2 	imes 2 = 16 	imes 2 = 32 \ W$.The total power generated at the source is $P_{in} = P_{out} + P_{loss} = 1000 + 32 = 1032 \ W$.The efficiency $\eta$ is given by $\eta = \left( \frac{P_{out}}{P_{in}} ight) 	imes 100$.$\eta = \left( \frac{1000}{1032} ight) 	imes 100 \approx 96.898 \% \approx 96.9 \%$.

An electric power line having a total resistance of $2 \ \Omega$ delivers $1 \ kW$ of power at $250 \ V$. The percentage efficiency of the transmission line is: (in $\%$)

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