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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?

Urb
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Swapnil MZS
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2 Answers2

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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.

Philip Wood
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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.

Dale
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