2
$\begingroup$

Let's say I have equipment of resistance $R$ (maybe a bulb or heater). If I apply a voltage $V$ to it, the power used by the equipment will be $V^2/R$ assuming Ohm's law holds for the equipment.
If I change the voltage, the power used by the equipment changes.

But now, I have equipment with a power rating that says $P$. Now; if I apply voltage $V$ to the equipment, I get a current $I$ given by $I = P/V$. But now, if I change the voltage, the power won't change; instead, the current will change.

Why is this so? So how is the power across the equipment constant in the second case? Is there something else that I am missing here?

Improve this question
$\endgroup$

1 Answer 1

Reset to default
3
$\begingroup$

The power rating only applies at a specific voltage, e.g. 110V or 240V. If you change the voltage, the equipment will consume more or less power, as determined by its resistance.

If the equipment uses $P$ power at $V$ voltage, then it uses a current of $I=P/V$. Its resistance is thus $R=V/I$.

If we now change the voltage to $V_1$, the resistance R (which does not change) means a new current $I_1=V_1/R$ flows, and the new power is $P=V_1I_1$.

Improve this answer
$\endgroup$
3
  • $\begingroup$ Ahh, Got it! So, the power rating is given for a specific voltage, and changing voltage changes the power consumed. Thanks. $\endgroup$
    – Sumit Gupta
    Commented Jul 27, 2023 at 7:48
  • $\begingroup$ Just remember that, with most equipment, changing the voltage is not a good idea. Increasing the voltage (say, plugging 110V equipment into a 220V socket) may be destructive. Lowering the voltage may cause erratic or faulty behaviour. That said, a lot of low-power equipment like plug-packs will happily work from 100 to 240V. $\endgroup$
    – hdhondt
    Commented Jul 27, 2023 at 9:37
  • $\begingroup$ > Increasing the voltage (say, plugging 110V equipment into a 220V socket) may be destructive I'd like to add that decreasing the voltage may also be destructive, as in putting a 6 volt battery in a 12 volt car and trying to engage the starter motor. Since the 6 volts will spin the motor very slowly (if at all) the motor (which, while spinning, acts as a generator) doesn't produce enough back-voltage to limit the current, essentially a dead short. $\endgroup$
    – Arthur Kalliokoski
    Commented Jul 27, 2023 at 15:15

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.