A battery of emf E produces currents I1 and I2, when connected to external resistances R1 and R2 respectively. The internal resistance of the battery is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \( \displaystyle\frac{I_1R_1 - I_2R_2}{I_2 - I_1} \)
The emf of the battery is the same in both cases. By Ohm’s law for the complete circuit, the current is the emf divided by the total resistance, external plus internal.
\( E = I_1(R_1 + r) \)
\( E = I_2(R_2 + r) \)
\( I_1R_1 + I_1r = I_2R_2 + I_2r \)
\( r(I_1 - I_2) = I_2R_2 - I_1R_1 \)
\( \displaystyle r = \frac{I_2R_2 - I_1R_1}{I_1 - I_2} \)
Multiplying numerator and denominator by −1:
\( \displaystyle r = \frac{I_1R_1 - I_2R_2}{I_2 - I_1} \)