Question

A constant voltage is applied across a circuit element. If the resistance of the element is doubled, what is the effect on the power dissipated by this element? The power dissipated is reduced by a factor of 2. The power dissipated remains constant. The power dissipated is doubled. The power dissipated is quadrupled. The power dissipated is reduced by a factor of 4.

Answer 1

Answer:

The power dissipated is reduced by a factor of 2

Explanation:

The power dissipated by a resistor is given by:

$P=I^2 R$

where

I is the current

R is the resistance

by using Ohm's law, $I=\frac{V}{R}$, we can rewrite the previous equation in terms of the voltage applied across the resistor (V):

$P=(\frac{V}{R})^2R=\frac{V^2}{R}$

In this problem, the resistance of the element is doubled, while the voltage is kept constant. So we have $R'=2R$ while V remains the same; substituting into the formula, we have:

$P'=\frac{V^2}{2R}=\frac{1}{2}\frac{V^2}{R}=\frac{P}{2}$

so, the power dissipated is reduced by a factor 2.