JEE Mains · Physics · STD 12 - 3. current electricity
A cell of internal resistance \(r\) drives current through an external resistance \(R\) . The power delivered by the cell to the external resistance will be maximum when:
- A \(R = 0.001\,r\)
- B \(R = 1000\,r\)
- C \(R = 2\,r\)
- D \(R = r\)
Answer & Solution
Correct Answer
(D) \(R = r\)
Step-by-step Solution
Detailed explanation
Current \(i=\frac{E}{r+R}\) Power generated in \(R\) \(P=i^{2} R\) \(P=\frac{E^{2} R}{(r+R)^{2}}\) For maximum power \(\frac{\mathrm{dP}}{\mathrm{dR}}=0\) \(E^{2}\left[\frac{(r+R)^{2} \times 1-R \times 2(r+R)}{(r+R)^{4}}\right]=0\) \(\Rightarrow r=R\)
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