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
Answer & explanation
Correct answer: option 2
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:
$I = \frac{E_{\text{total}}}{R + r_{\text{total}}}$
$0.5 = \frac{11}{R + 5}$
$R + 5 = \frac{11}{0.5} = 22$
$R = 22 - 5 = 17\ \Omega$