0
$\begingroup$

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?

$\endgroup$
3

2 Answers 2

2
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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 𝑃=𝐼2𝑅. But according to 𝑃=𝑉2/𝑅, 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.

$\endgroup$
2
  • $\begingroup$ Why I can't assume resistance is constant? $\endgroup$ Commented Oct 14, 2020 at 21:17
  • $\begingroup$ @Swapnil MZS because you already specified V and I and so R=V/I is also specified and cannot be assumed to be anything other than what it is specified to be $\endgroup$
    – Dale
    Commented Oct 14, 2020 at 21:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.