(a) 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 a resistance of 8.5 `Omega `. What are the current drawn from the supply and its terminal voltage ?
(b) A secondary cell 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?

Explanation

(a) Here emf of each cell `epsi`= 2.0 V, internal resistance of each cell r = 0.015 `Omega ` Number of cells n= 6 and external resistance R = 8.5 `Omega`
` therefore` In series arrangement total emf `epsi_(eq) = n epsi = 6 xx 2.0 = 12 V ` and total internal resistance ` r_(eq) = nr = 6 x 0.015 = 0.09 Omega`
Current drawn from the supply `I = (epsi_(eq))/(R + r_(eq)) = (12 )/(8.5 + 0.09) = (12)/(8.59) = 1.4 A`
`therefore ` Terminal voltage `V = epsi_(eq) - Ir_(eq) = 12 - 1.4 xx (0.09) = 11.9 V`
(b) Here emf `epsi.` = 1.9 V, internal resistance `r. = 380 Omega` Maximum current which may be drawn from the cell
`I. = (epsi.)/(r.) = (1.9)/(380) = 0.005A`
As this current is extremely small, it cannot be used to drive the starting motor of a car.