Question #347998

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.

Expert's answer

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."

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