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Two batteries each of emf E and internal resistance r are connected in series and in parallel, and are used to find current in an external resistance R. If the current in series is equal to that in parallel, the internal resistance of each battery is |
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Answer» <P> Solution : `I_(s)=(2E)/(2r+R) and I_(P)=(E)/((r)/(2)+R)=(2E)/(r+2R)` If `I_(S)+I_(P)`, (2E)/(2R+R)=(2E)/(r+2R), 2r+R=r+2R RARR r=R` |
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