AP EAMCET · PHYSICS · Current Electricity
Two cells with same emf \(\mathrm{E}\) but different inteınal resistances, \(r_1\), and \(r_2\) are connected in series to an external resistance \(\mathrm{R}\). If the potential difference across the first cell is zero then the value of \(\mathrm{R}\) is
- A \(\frac{r_1-r_2}{2}\)
- B \(\frac{r_1+r_2}{2}\)
- C \(r_1-r_2\)
- D \(\left(r_1+r_2\right)\)
Answer & Solution
Correct Answer
(A) \(\frac{r_1-r_2}{2}\)
Step-by-step Solution
Detailed explanation
Equivalent emf, \(\mathrm{E}_{\mathrm{eq}}=\mathrm{E}+\mathrm{E}=2 \mathrm{E}\) Equivalent resistance, \(R_{e q}=r_1+r_2+R\) Current flowing through the circuit, \[ i=\frac{2 E}{r_1+r_2+R} \] Potential drop across the first cell, \(\mathrm{V}_1=\mathrm{E}-\mathrm{ir}_1\)…
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