Question

Three cells (EMF =9V) are connected in series to a 4 ohms light bulb. If the resulting current is 0.65 A, what is the internal resistance of each cell, assuming they are identical and neglecting the wires?

Answer 1

$$r=\frac{\frac{E}{I}-R}{3}=\frac{\frac{9V}{0.65A}-4\Omega}{3}\approxeq3.28 \Omega$$

Answer: 3.28 ohms