COMEDK · Physics · 19. Current Electricity
A cell of emf \(2 \mathrm{~V}\) is connected with a load of resistance \(1.5 \Omega\). The power delivered by the cell to the load is maximum, then power transferred to the load is
- A \(0.67 \mathrm{~W}\)
- B \(2.67 \mathrm{~W}\)
- C \(1.33 \mathrm{~W}\)
- D \(3.25 \mathrm{~W}\)
Answer & Solution
Correct Answer
(A) \(0.67 \mathrm{~W}\)
Step-by-step Solution
Detailed explanation
For maximum power transfer, internal resistance \(r\) = load resistance \(R\). Since \(R = 1.5\ \Omega\), we have \(r = 1.5\ \Omega\). Maximum power delivered to load: \(P_{max} = \dfrac{E^2}{4R} = \dfrac{(2)^2}{4 \times 1.5} = \dfrac{4}{6} = \dfrac{2}{3} \approx 0.67\) W
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