Question

Two cells of emfs \(1\,\text{V}\) and \(2\,\text{V}\) and internal resistances \(2\,\Omega\) and \(1\,\Omega\) respectively connected in parallel, gave a current of \(1\,\text{A}\) through an external resistance. If the polarity of one cell is reversed, the current through the external resistance will be \( \frac{\alpha}{5} \) A. The value of \( \alpha \) is _____.

Answer 1

Correct Answer: 3

Step 1: Understanding the Question:
We first find the external resistance \(R\) from the initial parallel connection, then find the new current after reversing one cell's polarity.
Step 2: Key Formula or Approach:
Equivalent emf (\(E_{eq}\)) and resistance (\(r_{eq}\)) for cells in parallel:
\[ E_{eq} = \frac{E_1 r_2 + E_2 r_1}{r_1 + r_2}, \quad r_{eq} = \frac{r_1 r_2}{r_1 + r_2} \]
Step 3: Detailed Explanation:
1. Given: \(E_1 = 1 \text{ V}, r_1 = 2 \text{ }\Omega, E_2 = 2 \text{ V}, r_2 = 1 \text{ }\Omega\).
Initial Case: Polarity supports each other.
\[ E_{eq} = \frac{1 \times 1 + 2 \times 2}{2 + 1} = \frac{5}{3} \text{ V}, \quad r_{eq} = \frac{2 \times 1}{3} = \frac{2}{3} \text{ }\Omega \]
Current \(I = \frac{E_{eq}}{r_{eq} + R} \Rightarrow 1 = \frac{5/3}{2/3 + R} \Rightarrow \frac{2}{3} + R = \frac{5}{3} \Rightarrow R = 1 \text{ }\Omega\).
2. Case 2: Polarity of one cell is reversed.
\[ E'_{eq} = \frac{E_2 r_1 - E_1 r_2}{r_1 + r_2} = \frac{2 \times 2 - 1 \times 1}{3} = 1 \text{ V} \]
New Current \(I' = \frac{E'_{eq}}{r_{eq} + R} = \frac{1}{2/3 + 1} = \frac{1}{5/3} = \frac{3}{5} \text{ A}\).
3. Given \(I' = \alpha/5 \Rightarrow \alpha = 3\).
Step 4: Final Answer:
The value of \(\alpha\) is 3.