Question

A battery of EMF E and internal resistance r is connected across a variable resistor. When the resistor is set at 2lΩ the current through it is 0.48A; when it is set at 36Ω. the current is 0.30A. Find E and r.

Answer 1

Given:

$I1=0.48A; I2=0.3A; R1=21\Omega; R2=36\Omega.$

Find:

$E$ -EMF and $r$ - internal resistance .

Solution:

Let's use Ohm's law for whole circuit:

$I=E/(R+r);$ In our case:

$I1=E/(R1+r); I2=E/(R2+r);$

So,

$E=I1(R1+r); E=I2((R2+r);$

$I1(R1+r)=I2(R2+r);$

$r(I1-I2)=I2R2-I1R1;$

$r=(I2R2-I1R1)/(I1-I2)= (0.3A*36\Omega-0.48A*21\Omega)/(0.48A-0.3A)=4\Omega.$

$E=I1(R1+r)=0.3A*(21\Omega+4\Omega)=7.5V.$

Answer:

$E=7.5V; r=4\Omega.$