The storage battery of a car has an emf of 12 V. If the internal resistance of the battery is 0.4 `Omega` , what is the maximum current that can be drawn from the battery ?

Explanation

Consider a battery with emf e and internal resistance r, connected with external resistance R. Applying KVL in above closed loop, we get
- IR - Ir = -` epsilon`
`therefore IR + Ir = epsilon`
`therefore I(R + r) = epsilon `
`therefore I = (epsilon)/(R + r) " "` .... (1)
For a given battery, values of £ and r are constants. Here as R decreases I increases. Hence , when R = minimum = 0,we get
I = maximum = `I_(max)`,
thus ,
`I_(max) = (epsilon)/(r)" " `... (2)
`therefore I_(max) = (12)/(0.4) = 30` A