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

Explanation

Here `epsilon = 2.0 V, n = 6, r = 0.015 Omega, R = 8.5 Omega`
Current `I = (nE)/(R + nr) = (6 xx 2.0)/(8.5 + 6 xx 0.015) = 1.4 A`
Terminal voltage, `V = IR = 1.5 xx 8.5 = 11.9 V`