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

The power dissipated is reduced by a factor of 2

Step-by-step 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=(V)/(R)`, we can rewrite the previous equation in terms of the voltage applied across the resistor (V):

`P=((V)/(R))^2R=(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'=(V^2)/(2R)=(1)/(2)(V^2)/(R)=(P)/(2)`

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