# Seemingly contradictory situation in electrical system loss

Physics Asked by Swapnil MZS on December 18, 2020

In a power supply system, we know that we decrease the current and increase the potential difference. If we decrease the current by a factor of 10 and increase potential difference by a factor of 10, the system loss (emitted heat) decreases following the formula $$P=I^2R$$. But according to $$P=V^2/R$$, the system loss is being increased by a factor of 100. It seems contradictory. Now what is the conclusion?

There is more than one relevant potential difference. You must distinguish between the potential difference, $$V_L$$, across the load (i.e. the 'user' that we are aiming to supply) and the potential difference, $$V_W$$, across just the transmitting wires (of resistance $$R_W$$ taken together). The load and the wires are in series across the supply so $$V_text{supply}=V_L+V_W$$.

The power received by the load is $$IV_L$$.

The power dissipated in the wires ('system loss') is $$IV_W$$. We can also write this as $$I^2R_W$$ or as $$frac {V_W^2}{R_W}.$$

If we increase $$V_L$$ by a factor of 10, we can get the same power, $$IV_L$$, to the load using only a tenth of the current. [The load has to be of higher resistance now, and is, in practice, the primary of a loaded step-down transformer, but that doesn't affect our argument.] So using $$I^2R_W$$, the power dissipated in the wires drops to $$frac{1}{100}$$ of its previous value. But suppose we use $$frac {V_W^2}{R_W}$$ ... That's fine too, because if $$I$$ is $$frac{1}{10}$$ its previous value, so is $$V_W$$ (since $$V_W=IR_W$$), so again we find that the power dissipation in the wires drops to $$frac{1}{100}$$ of its previous value.

Correct answer by Philip Wood on December 18, 2020

If we decrease the current by a factor of 10 and increase potential difference by a factor of 10, the system loss (emitted heat) decreases

This is incorrect. The power is given by $$P=IV$$, so if $$I$$ decreases and $$V$$ increases by the same factor then power remains constant.

the system loss (emitted heat) decreases following the formula ?=?2?. But according to ?=?2/?, the system loss is being increased

The formulas $$P=I^2 R$$ and $$P=V^2/R$$ both require you to know the resistance. You have incorrectly assumed that the resistance is constant. In fact, since $$R=V/I$$ in this case $$R$$ increases by a factor of 100, which gives the correct answer (no change) with both formulas.

Answered by Dale on December 18, 2020

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