AP EAMCET · PHYSICS · Current Electricity
When a resistance \(R_1\) is connected across a cell, the current is \(I_1\) and if the resistance \(R_1\) is replaced by \(R_2\), the current is \(\mathrm{I}_2\). Then the internal resistance of the cell is
- A \(\frac{\mathrm{I}_1 \mathrm{R}_1+\mathrm{I}_2 \mathrm{R}_2}{\mathrm{I}_1+\mathrm{I}_2}\)
- B \(\frac{\mathrm{I}_1 \mathrm{R}_2-\mathrm{I}_2 \mathrm{R}_1}{\mathrm{I}_1-\mathrm{I}_2}\)
- C \(\frac{\mathrm{I}_1 \mathrm{R}_2-\mathrm{I}_2 \mathrm{R}_1}{\mathrm{I}_1-\mathrm{I}_2}\)
- D \(\frac{\mathrm{I}_2 \mathrm{R}_2-\mathrm{I}_1 \mathrm{R}_1}{\mathrm{I}_1-\mathrm{I}_2}\)
Answer & Solution
Correct Answer
(D) \(\frac{\mathrm{I}_2 \mathrm{R}_2-\mathrm{I}_1 \mathrm{R}_1}{\mathrm{I}_1-\mathrm{I}_2}\)
Step-by-step Solution
Detailed explanation
Current \(I_1=\frac{E}{R_1+r}\) and \(R_1\) is replaced by \(R_2\) then,…
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