(a) Six lead-acid type of secondary cells each of emf `2.0 V` and internal resistacne `0.015Omega` are joined in series to provide a supply to a resistance of `8.5 Omega.` What are the current drawn from the suply 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 tghe starting motor of a car ?

Explanation

Here no. of rows, m = 1
No. of cells in one row, n = 6
emf of one cell, `epsilon ` =2 V
External resistance , R = 8.5 `Omega`
(a) Here, current passing above closed loop is,
I = `("mne")/(mR + nr)`
`therefore I = (1 xx 6 xx 2)/((1 xx 8.5 ) + (6 xx 0.015))`
`therefore I = (12)/(8.59)`
`therefore I = 1.3969 A = 1.4` A
Now, terminal voltage of given source will be,
V = IR
= (1.4) (8.5)
`therefore V = 11.9 ` Volt
(b) Maximum current that can be drawn from given source is ,
`I_("max") = (epsilon)/(r) = (1.9)/(380)`
` therefore I_(max) = 0.005 ` A
Motor of a car can not be started with above current because while starting the motor of a car (i.e. for triggering the engine of a car) minimum 100 A current is required.