Question

(a) Six lead-acid type of secondary cells each of emf `2-0` V and internal resistance `0-015Omega` are jouned in series to provide a supply to a resistance of `8-5Omega`. What are the current drawn from the supply and its terminal voltage ?
(b) A secondary cells after long use has an emf of `1-9` V and a large internal resistance of `380 Omega`. What maximum current can be drawn from the cell ? Could the cell drive the starting motor of a car ?

Answer 1

(a) Here, `epsilon = 2.0V , n = 6, r = 0.015 Omega , R = 8.5 Omega`
Current, `I = (n epsilon)/(R+ nr) = (6xx2.0)/(8.5 + 6 xx 0.015) = 1.4A`
Terminal voltage, `V = I R = 1.4 xx 8.5 = 11.9 V`
(b) Here, `epsilon = 1.9 V , r =380 Omega`
`I_(max) = epsilon/r = 1.9/380 = 0.005A`
This amount of current cannot start a car because to start a car beacuse to start the motor, the current required is 100 A for few seconds.