MHT CET · Physics · Electrostatics
Two cells \(\mathrm{E}_1\) and \(\mathrm{E}_2\) having equal \(\mathrm{EMF}\) ' \(\mathrm{E}\) ' 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 \(E_1\) becomes zero. The value of ' \(R\) ' will be
- A \(r_1-r_2\)
- B \(\mathrm{r}_1+\mathrm{r}_2\)
- C \(\frac{\mathrm{r}_1-\mathrm{r}_2}{2}\)
- D \(\frac{r_1+r_2}{2}\)
Answer & Solution
Correct Answer
(A) \(r_1-r_2\)
Step-by-step Solution
Detailed explanation
The total current in the circuit is
\(
I=\frac{2 E}{r_1+r_2+R}
\)
....(Given cells are in series, \(\mathrm{E}+\mathrm{E}=2 \mathrm{E}\) )
Now the potential drop across the first cell is
\(\mathrm{V}_1=\mathrm{E}-\mathrm{Ir}_1=0 \)
\( \therefore \quad \mathrm{E}-\left(\frac{2 \mathrm{E}}{\mathrm{r}_1+\mathrm{r}_2+\mathrm{R}}\right) \times \mathrm{r}_1=0 \)
\( \frac{2 \mathrm{E}}{\mathrm{r}_1+\mathrm{r}_2+\mathrm{R}}=\frac{\mathrm{E}}{\mathrm{r}_1} \Rightarrow 2 \mathrm{r}_1=\mathrm{r}_1+\mathrm{r}_2+\mathrm{R} \)
\( \therefore \mathrm{R}=\mathrm{r}_1-\mathrm{r}_2\)
\(
I=\frac{2 E}{r_1+r_2+R}
\)
....(Given cells are in series, \(\mathrm{E}+\mathrm{E}=2 \mathrm{E}\) )
Now the potential drop across the first cell is
\(\mathrm{V}_1=\mathrm{E}-\mathrm{Ir}_1=0 \)
\( \therefore \quad \mathrm{E}-\left(\frac{2 \mathrm{E}}{\mathrm{r}_1+\mathrm{r}_2+\mathrm{R}}\right) \times \mathrm{r}_1=0 \)
\( \frac{2 \mathrm{E}}{\mathrm{r}_1+\mathrm{r}_2+\mathrm{R}}=\frac{\mathrm{E}}{\mathrm{r}_1} \Rightarrow 2 \mathrm{r}_1=\mathrm{r}_1+\mathrm{r}_2+\mathrm{R} \)
\( \therefore \mathrm{R}=\mathrm{r}_1-\mathrm{r}_2\)
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