(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 currents drawn from the supply, and its terminal voltage?
(b) A secondary cell after long use has an emf 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) The total emf of a number of cells connected in series combination will be equal to the sum of their individual emfs.
`:.` Total emf of 6 cells, `E= 6 xx 2 = 12V`
Total internal resistance in the circuit, `r= 6 xx 0.015 = 0.09Omega`
If R is the external resistance in the circuit, then the current in the circuit is given by `I= (E )/( R +r) = (12)/(8.5 + 0.09) = (12)/(8.59) = 1.4A`
The terminal voltage across the supply is given by: `V= E- Ir`
`rArr V= 12-1.4 xx 0.09`
`=11.87V ~~ 11.9V`
(b) The maximum current will be drawn from the cell when external resistance is zero. The current drawn will be given by `I = (E )/(r )`
Here, E= 1.9V, `r= 380 Omega`
`:. I = (1.9)/(380) = 0.005A`
Since the value of current is very small, motor of a car cannot be driven with this cell because it requires large current in a very short interval (few second) of time.