Question

A certain resistor dissipates 0.5 W when connected to a 3 V potential difference. When connected to a 1 V potential difference, this resistor will dissipate:

Answer 1

Answer:

0.056 W

Explanation:

$Power = IV$

From ohms law we know that

$V= IR\\\\I= \frac{V}{R} \\\\Power= \frac{V}{R}*V\\\\Power= \frac{V^2}{R}$

Given data

P1 = 0.5 Watt

P2 = ?

V1= 3 Volts

V2= 1 Volt

Thus we can solve for the power dissipated as follows

$P1= \frac{V1^2}{R1}\\\\P2= \frac{V2^2}{R2}$

$\frac{P1}{P2} = \frac{V1^2}{V2^2}\\\\ P2=\frac{ V2^2}{ V1^2} *P1\\\\ P2=\frac{ 1^2}{ 3^2} *0.5= 0.055= 0.056 W$

The  resistor will dissipate 0.056 Watt