Question

A battery of emf E and internal resistance r sends current I1 and I2, when connected to external resistance R1 and R2 respectively. Determine the emf and internal resistance of the battery.

Answer 1

The emf and internal resistance of the battery is E = I1I2(R2−R1) I1−I2

Explanation:

• When resistor R1 is connected with the cell of internal resistance r.From kirchoffs loop rule current in circuit is

I1 = E / r+R1    -----(1)

• When resistor R2 is connected with the cell of internal resistance r.From kirchoffs loop rule current in circuit is

I2 = Er+R2 -----(2)

Equating E from both equation:

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

solving for r

r=I2R2 − I1R1 / I1−I2

Putting value of r in equation ..1

r+R1 = E / I1

I2R2−I1R1 / I1−I2+R1 =E / I1

E = I1(I2R2 − I1R1 / I1−I2 + R1)

E = I1I2(R2−R1) I1−I2

Thus the emf and internal resistance of the battery is E = I1I2(R2−R1) I1−I2

 

Answer 2

Explanation:

A battery of emf E and internal resistance r sends current I1 and I2, when connected to external resistance is I1. I2(R2-R1)/I1-I2