A cell of emf ε and internal resistance r is connected to an external …

Question

A cell of emf ε and internal resistance r is connected to an external resistance R. The maximum power delivered to R is

Accepted Answer

B. {'text': 'ε²/(4r)', 'label': 'B'}

1. Current in the loop: I = ε / (R + r).
2. Power delivered to R: P = I² R = ε² R / (R + r)².
3. Differentiate with respect to R and set dP/dR = 0.
4. This gives R = r, and substituting back yields P_max = ε² / (4 r).
_Source: NCERT Class 12 Physics, Ch 3 "Current Electricity", §3 EMF and internal resistance_

Answer

{'text': 'ε²/(2r)', 'label': 'A'}

Answer

{'text': 'ε²/r', 'label': 'C'}

Answer

{'text': 'ε²R/(R+r)²', 'label': 'D'}