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

0%(a) internal resistance of the cell is 2 Ω

0%(b) emf of the cell is 12 V

100%(c) maximum power that can be dissipated in the external resistance is 18 watt ( 1 voter )

0%(d) short circuit current from the cell is infinite

Based On 1 Vote

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).

Leave an answer