A car has a fresh storage battery of emt `12 V` and internal resistance `5.0 xx 10^(-2) Omega`. If the starter moter draws a current of 90A, what is the terminal voltage of the battery when the starter is on?
(b) After long use, the internal resistance of the storange battery increase to `500 Omega`. What maximum current can be drawn from the battery? Assume the emt of the battery to remain unchanged.
(c) If the discharged battery is changed by an external emf source, is the terminal voltage of the battery during changing greater of less than is emf `12 V`?

Explanation

We have emf of the storage battery as `epsilon = 12 V`, internal resistance of the battery as `r = 0.5 xx 10^(2) Omega`, and current flowing through the circuit as `I = 90 A`.
(a) If `V` is the terminal voltage,
`V = epsilon - lrj =12 - 90 (5.0xx10^(-2)) = 7.5V`
b. We know that `I = epsilon //(R +r)`, where R is sthe external
resistance. For I to be maximum (i.e., `I_(max), R = 0 ` (i.e.,
the batter is to be shorted ). Thus,
`I_(max) = (epsilon)/(r ) = (12V)/(500Omega) = 24mA`
(as after long use `r = 500 Omega)`
Clearly, the battery can now no longer be used for
starting the car.
c. During charging, the current inside the battery flows in
a direction opposite to that when it is discharging.
Clearly, replacing I by -I we get `V = epsilon - (-I)r = epsilon +lr.`
Hence, V should be greater then `epsilon (=12 V)` during
charging.