Two batteries of emf 5 V with internal resistance 2 Ω and 6 V with internal resistance 3 Ω are connected in series with an external resistor. If the current in the circuit is 0.5 A, then the resistance of the resistor is

17 Ω

The correct answer is Option (2) → 17 Ω

Given:

EMF₁ = 5 V, r₁ = 2 Ω

EMF₂ = 6 V, r₂ = 3 Ω

Current, I = 0.5 A

Let external resistance be R.

Total EMF = 5 + 6 = 11 V

Total internal resistance = 2 + 3 = 5 Ω

Using Ohm's law:

$I = \frac{E_{\text{total}}}{R + r_{\text{total}}}$

$0.5 = \frac{11}{R + 5}$

$R + 5 = \frac{11}{0.5} = 22$

$R = 22 - 5 = 17\ \Omega$