Six cell emf 2V and the internal resistance 0.015 ohm are joined in a series provide supply to a resistance of 8.5 ohm what are the current drawn from the supply on terminal resistance

Six cell emf 2V and the internal resistance 0.015 ohm are joined in a

1.	Six cell emf 2V and the internal resistance 0.015 ohm are joined in a series provide supply to a resistance of 8.5 ohm what are the current drawn from the supply on terminal resistance
Answer» udent,◆ Answer -I = 1.397 AV = 11.87 V◆ EXPLAINATION -# Given -n = 6E = 2 Vr = 0.015 ohmR = 8.5 ohm# Solution -For battery of 6 CELLS -E' = nE = 6×2 = 12 Vr' = NR = 6×0.015 = 0.09 ohmCurrent drawn from the supply is -I = E' / (r' + R)I = 12 / (0.09 + 8.5)I = 12 / 8.59I = 1.397 AWe can also calculate voltage across terminal resistance as -V = I.RV = 1.397 × 8.5V = 11.87 VHope this HELPS you...

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Six cell emf 2V and the internal resistance 0.015 ohm are joined in a series provide supply to a resistance of 8.5 ohm what are the current drawn from the supply on terminal resistance

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