A line having a total resistance of 0.20 Ω delivers 10.00 kW at 250 V to a small factory. What

Step 1: Identify the delivered power and voltage. The factory receives 10.00 kW (which is 10,000

Question

A line having a total resistance of $0.20 \Omega$ delivers $10.00 \mathrm{~kW}$ at 250 $\mathrm{V}$ to a small factory. What is the efficiency of the transmission? The line dissipates power due to its resistance. Consequently we'll need to find the current in the line. Use $\mathrm{P}=V I$ to find $I=\mathrm{P} / \mathrm{V}$. Then $$ \begin{array}{l} \text { Power lost in line }=I^{2} R=\left(\frac{P}{V}\right)^{2} R=\left(\frac{10000 \mathrm{~W}}{250 \mathrm{~V}}\right)^{2}(0.20 \Omega)=0.32 \mathrm{~kW} \\ \text { Efficiency }=\frac{\text { Power delivereed by linc }}{\text { Power supplied to line }}=\frac{10.00 \mathrm{~kW}}{(10.00+0.32) \mathrm{kW}}=0.970=97.0 \% \end{array} $$

          A line having a total resistance of $0.20 \Omega$ delivers $10.00 \mathrm{~kW}$ at 250 $\mathrm{V}$ to a small factory. What is the efficiency of the transmission?
The line dissipates power due to its resistance. Consequently we'll need to find the current in the line. Use $\mathrm{P}=V I$ to find $I=\mathrm{P} / \mathrm{V}$. Then
$$
\begin{array}{l}
\text { Power lost in line }=I^{2} R=\left(\frac{P}{V}\right)^{2} R=\left(\frac{10000 \mathrm{~W}}{250 \mathrm{~V}}\right)^{2}(0.20 \Omega)=0.32 \mathrm{~kW} \\
\text { Efficiency }=\frac{\text { Power delivereed by linc }}{\text { Power supplied to line }}=\frac{10.00 \mathrm{~kW}}{(10.00+0.32) \mathrm{kW}}=0.970=97.0 \%
\end{array}
$$

Added by Andrea B.

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