WBJEE · Physics · Current Electricity
A cell of emf \(E\) is connected to a resistance \(R_{1}\) for time \(t\) and the amount of heat generated in it is \(H\). If the resistance \(R_{1}\) is replaced by another resistance \(R_{2}\) and is connected to the cell at the same time \(t,\) the amount of heat generated in \(R_{2}\) is \(4 \mathrm{H}\). Then internal resistance of the cell is
- A \(\frac{2 R_{1}+R_{2}}{2}\)
- B \(\sqrt{R_{1} R_{2}} \frac{2 \sqrt{R_{2}}-\sqrt{R_{1}}}{\sqrt{R_{2}}-2 \sqrt{R_{1}}}\)
- C \(\sqrt{R_{1} R_{2}} \frac{\sqrt{R_{2}}-2 \sqrt{R_{1}}}{2 \sqrt{R_{2}}-\sqrt{R_{1}}}\)
- D \(\sqrt{R_{1} R_{2}} \frac{\sqrt{R_{2}}-\sqrt{R_{1}}}{\sqrt{R_{2}}+\sqrt{R_{1}}}\)
Answer & Solution
Correct Answer
(B) \(\sqrt{R_{1} R_{2}} \frac{2 \sqrt{R_{2}}-\sqrt{R_{1}}}{\sqrt{R_{2}}-2 \sqrt{R_{1}}}\)
Step-by-step Solution
Detailed explanation
We have, \(H=I^{2} R\) According to given condition, \(I_{1}^{2} R_{1}=H\) and \[ I_{2}^{2} R_{2}=4 H \] \[ \Rightarrow \quad \frac{E^{2}}{\left(R_{1}+r\right)^{2}} R_{1}=H \text { and } \frac{E^{2}}{\left(R_{2}+r\right)^{2}} R_{2}=4 H \]…
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