Question:
In the given circuit, current passing through \( R \) is 1 ampere. Now polarity of one cell is reversed, then current through \( R \) becomes \( \frac{\alpha}{5} \) Amp, then find \( \alpha \).
In the given circuit, current passing through \( R \) is 1 ampere. Now polarity of one cell is reversed, then current through \( R \) becomes \( \frac{\alpha}{5} \) Amp, then find \( \alpha \).
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When the polarity of a cell is reversed in a circuit, the net voltage changes, which affects the current. Use Kirchhoff’s law and Ohm’s law to find the current before and after the polarity reversal.
Updated On: Apr 7, 2026
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Correct Answer: 3
Solution and Explanation
First, for the equivalent emf (\( \epsilon_{\text{eq}} \)):
\[
\epsilon_{\text{eq}} = \frac{2 \times 2 + 1 \times 1}{3} = \frac{5}{3} \, \text{V}
\]
Next, the equivalent resistance (\( r_{\text{eq}} \)):
\[
r_{\text{eq}} = \frac{2}{3} \, \Omega
\]
The current \( I \) is given by:
\[
I = \frac{5}{3} \left[ R + \frac{2}{3} \right]
\]
Solving for \( R \):
\[
3R + 2 = 5
\]
\[
R = 1 \, \Omega
\]
Now, for the second part of the circuit, we are given:
\[
\epsilon_{\text{eq}} = 4 - 1 \, \text{V} = 1 \, \text{V}
\]
The equivalent resistance for the second part is:
\[
r_{\text{eq}} = \frac{2}{3} \, \Omega
\]
The current \( I \) is:
\[
I = \frac{1}{1 + \frac{2}{3}} = \frac{3}{5}
\]
Finally, the value of \( \alpha \):
\[
\alpha = 3
\]
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