Answer to Question #280998 in Electric Circuits for Anonymous005

Question

A battery is connected in series with a variable resistor. When the resistor is 8.50 ȍ, the current is 1.00 A and when the resistor is 6.50 ȍ, the current is 1.25 A. What is the value of the internal resistance?

Answer 1

Answer

Given data

First case

Resistance R₁=8.5$\Omega$

Current I₁=1A

Now second case

Resistance R₂=6.50$\Omega$

Current I₂=1.25A

Now using the ohms law formula

$I=\frac{E}{r+R}$

Where r us internal resistance

Now write equation for both case

$1=\frac{E}{r+8.5}.........(1)$

$1.25=\frac{E}{r+6.50}.........(2)$

Divided equation 1 by 2 and solve this

We get internal resistance

r=1.5$\Omega$