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I am solving a question about power transmission efficiency and encountered two different results using two methods.

The problem states:

An electric power line, having a total resistance of $2 \, \Omega$, delivers $1 \, \mathrm{kW}$ at $220 \, \mathrm{V}$. The efficiency of the transmission line is approximately __________:


If I use the standard method:

  1. Calculate current using $I = \dfrac{P}{V}$: $$ I = \dfrac{1000}{220} = \dfrac{50}{11} \, \mathrm{A} $$
  2. Calculate power loss as $I^2 R$: $$ \text{Power loss} = \left( \dfrac{50}{11} \right)^2 \times 2 = \dfrac{5000}{121} \, \mathrm{W} $$
  3. Efficiency is then given by: $$ \text{Efficiency} = \dfrac{\text{Delivered Power}}{\text{Delivered Power} + \text{Power Loss}} \times 100\% = \dfrac{1000}{1000 + \dfrac{5000}{121}} \times 100\% $$ Simplifying this gives: $$ \text{Efficiency} \approx 96\% $$

Using this method, I get the correct answer matching the textbook: 96%.

However, I tried another approach using $P = \frac{V^2}{R}$. Here’s what I did:

  1. Input power must be $\mathrm{P}=\dfrac{\mathrm{V^2}}{\mathrm{R}} = \dfrac{220 \times 220}{2} = 24200 \mathrm{W}$.

  2. Efficiency must therefore be $\eta = \mathrm{\dfrac{P_o}{P_i}}=\dfrac{1000}{24200}$.

  3. Percentage efficiency is $\dfrac{1000}{24200} \times 100 = 4.13 \%$

    Why does using $P = \dfrac{V^2}{R}$ give a drastically different efficiency value compared to the textbook solution? Am I applying the formula incorrectly, or is there something I’m missing conceptually?

Any clarification would be appreciated!

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1 Answers1

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This was a doubt that I had for a long time. To answer it, we must fall back to the definition of power.

Power is defined as the rate of change of work, in its simplest terms.

$$P = \frac{dW}{dt}$$

Work done on an electric charge is $qV$, where $q$ is charge, and $V$ is potential difference. Substituting this, we get

$$P = \dot{q} V = iV$$

This would be the basic definition. From here, we could use Ohm's law to substitute $V=iR$ in the equation, giving us its other forms.

Note that Ohm's law is valid only for certain materials, aptly named Ohmic conductors. Other substances are Non-Ohmic conductors such as a semiconductor diode. These don't have a straight $V-i$ graph, and hence $V=iR$ is NOT valid.

However, $P = i V$ is still valid, even for these materials.

So, the best way to get the correct answer would be to fall back on the definition. Sure, you could use Ohm's law here, but note when it is valid.

$P = {V^2 \over R}$ is only valid when the potential difference is constant. $P = i^2 R$ is only valid when the current is constant. Can you guess why? Hint: We need to integrate $P = \frac{dW}{dt}$ to get $P = iV$. Along these lines...

If any other doubts, feel free to comment

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