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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