Question

Two cells of each emf \(1.5\,{V}\) and internal resistance \(1\,\Omega\) are first connected in series to an external resistance \(R\), and then the cells are connected in parallel to the same resistance. If the ratio of the potential differences across \(R\) in the two cases is \(4:3\), then the value of \(R\) is:

• A. \(5.5\,\Omega\)
• B. \(4.5\,\Omega\)
• C. \(3.5\,\Omega\)
• D. \(2.5\,\Omega\)

Hint:

Use Ohm's law \(V = IR\) and analyze series and parallel setups separately for such circuit comparisons.

Answer 1

The Correct Option is D

Series case: - Total emf \(= 3\,{V}\), total internal resistance \(= 2\,\Omega\) Current: \[ I_1 = \frac{3}{R + 2} \Rightarrow V_1 = I_1 R = \frac{3R}{R + 2} \] Parallel case: - Equivalent emf = \(1.5\,{V}\), equivalent internal resistance = \(0.5\,\Omega\) Current: \[ I_2 = \frac{1.5}{R + 0.5} \Rightarrow V_2 = I_2 R = \frac{1.5R}{R + 0.5} \] Given: \[ \frac{V_1}{V_2} = \frac{4}{3} \Rightarrow \frac{3R}{R + 2} \cdot \frac{R + 0.5}{1.5R} = \frac{4}{3} \] Solving: \[ \frac{3(R + 0.5)}{1.5(R + 2)} = \frac{4}{3} \Rightarrow \frac{2(R + 0.5)}{(R + 2)} = \frac{4}{3} \Rightarrow 6(R + 0.5) = 4(R + 2) \Rightarrow 6R + 3 = 4R + 8 \Rightarrow R = 2.5\,\Omega \]