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

All of the choices are correct

Step-by-step explanation:

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

`P=VI = I^2 R=(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)((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'=(V'^2)/(R')=((2V)^2)/(4R)=(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((R)/(4))=I^2 R` --> the power is unchanged