Question

A single resistor is connected across the terminals of a battery. Which one or more of the following changes in voltage and current leaves unchanged the electric power dissipated in the resistor? (A) Doubling the voltage and reducing the current by a factor of two. (B) Doubling the voltage and increasing the resistance by a factor of four. (C) Doubling the current and reducing the resistance by a factor of four.

Answer 1

Answer:

All of the choices are correct

Explanation:

The power dissipated in a single resistor connected to a battery is given by:

$P=VI = I^2 R=\frac{V^2}{R}$

where

V is the voltage

I is the current

R is the resistance

Let's analyze each case:

A) Doubling the voltage (V'=2V) and reducing the current by a factor of 2 (I'=I/2). The new power dissipated is:

$P'=V'I'=(2V)(\frac{I}{2})=VI=P$ --> the power is unchanged

B) Doubling the voltage (V'=2V) and increasing the resistance by a factor of 4 (R'=4R). The new power dissipated is:

$P'=\frac{V'^2}{R'}=\frac{(2V)^2}{4R}=\frac{V^2}{R}$ --> the power is unchanged

C) Doubling the current (I'=2I) and reducing the resistance by a factor of four (R'=R/4). The new power dissipated is:

$P'=I'^2 R'=(2I)^2(\frac{R}{4})=I^2 R$ --> the power is unchanged