If the potential difference applied to a fixed resistance is doubled, the power dissipated by that resistance

1 answer:

5 0

The power dissipated by a resistor is
$P=I^2 R = \frac{V^2}{R}$
where I is the current, V the voltage and R the resistance.

We can see that if the voltage V is doubled and R is kept fixed, the new dissipated power is
$P' = \frac{(2V)^2}{R}= \frac{4V^2}{R}=4 P$
so, it quadruples.

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