Six lead-acid type of secondary cells each of emf 2.0 V and internal resistance 0.015Ω are joined in series to provide a supply to a resistance of 8.5Ω. What are the current drawn from the supply and its terminal voltage?

Question

Philoid · Accepted Answer

Number of secondary cells n = 6

Emf of each secondary cell E = 2Volt

Internal resistance of each cell r = 0.015 Ω

Resistance of resistor R = 8.5 Ω

Let current drawn from supply = I

⇒ I = 1.39Ampere

Hence current drawn from supply is 1.39Ampere