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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 ? |
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Answer» Solution :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 |
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