MHT CET · Physics · Current Electricity
Two cells \(E_1\) and \(E_2\) having equal e.m.f ' \(E\) ' and internal resistances \(r_1\) and \(r_2\left(r_1>r_2\right)\) respectively are connected in series. This combination is connected to an external resistance ' \(R\) '. It is observed that the potential difference across the cell \(\mathrm{E}_1\) becomes zero. The value of \(R\) will be
- A \(r_1-r_2\)
- B \(\mathrm{r}_1+\mathrm{r}_2\)
- C \(\frac{r_1-r_2}{2}\)
- D \(\frac{r_1+r_2}{2}\)
Answer & Solution
Correct Answer
(A) \(r_1-r_2\)
Step-by-step Solution
Detailed explanation
\(I = \frac{E+E}{R+r_1+r_2} = \frac{2E}{R+r_1+r_2}\) \(V_1 = E - Ir_1 = 0 \implies E = Ir_1\)
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