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Find the emf (V) and internal resistance (r ) of a single battery which is equivalent to a parallel combination of two batteries of emfs V_(1) and V_(2) and internal resistances r_(1) and r_(2) respectively, with polarties as shown in figure |
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Answer» Solution :EMF of battery is equal to potential difference ACROSS the terminals, when no CURRENT is drawn from battery (for external circuit) [Here, all the elements in the circuit are in series] Current in internal circuit `=i` `therefore i=("Net emf")/("Toltal resistance") or i=(V_(1)+V_(2))/(r_(1)+r_(2))`  `therefore V_(A)-V_(B)=V_(1)-ir_(1)""[because V_(1)" cell is discharging"]` `or V_(1)-V_(B)=V_(1)-((V_(1)+V_(2))/(r_(1)+r_(2))) r_(1) or V_(A)-V_(B)=(V_(1)r_(2)-V_(2)r_(1))/(r_(1)+r_(2))` `therefore"Equivalent emf of the battery "=V=(V_(1)r_(2)-V_(2)r_(1))/(r_(1)+r_(2))` (ii) Internal resistance of equivalent battery. `r_(1) and r_(2)` are in parallel. `(1)/(r)=(1)/(r_(1))+(1)/(r_(2)) or r=(r_(1)r_(2))/(r_(1)+r_(2))` |
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