Question

A circuit is set up such that it has a current of 8 A. What would be the new current if the resistance was increased by a factor of 2?

Answer 1

Answer:

4 A

Explanation:

The relationship between current, voltage and resistance in a circuit is given by Ohm's law:

$V=RI$

where

V is the voltage

R is the resistance

I is the current

The equation can also be rewritten as

$I=\frac{V}{R}$

from which we see that the current is inversely proportional to the resistance, R.

In this problem, the initial current is I = 8 A. Then the resistance is doubled:

R ' = 2R

So the new current is

$I'=\frac{V}{R'}=\frac{V}{2R}=\frac{1}{2}(\frac{V}{R})=\frac{I}{2}=4 A$

so the current is halved.