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A resistance of 4 Ω is connected across a cell. Then it is replaced by another resistance of 1 Ω. It is found that power dissipated in resistance in both the cases is 16 watt. Then
Using P = V²R/(R+r)² , equating power for
R = 4Ω and R = 1Ω, solving gives r = 2Ω and
E = 12V. Maximum power = 18W.
Answer: (c).
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