A 79.0 ? and a 40.0 ? resistor are connected in parallel. When this combination

Question

A 79.0 ? and a 40.0 ? resistor are connected in parallel. When this combination is connected across a battery, the current delivered by the battery is 0.350 A. When the 40.0 ? resistor is disconnected, the current from the battery drops to 0.195 A. Determine the emf of the battery.

Explanation / Answer

A 79.0 ? and a 40.0 ? resistor are connected in parallel. When thiscombination is connected across a battery, the current delivered bythe battery is 0.350 A. When the 40.0 ? resistor is disconnected,the current from the battery drops to 0.195 A. Determine the emf ofthe battery. effective resistance of resistors as they are connected inparallel=79*40/79+40=26.55ohm now current=E/R=>0.350A=E/26.55+r......1 where r is the internal resistance of the battery when When the 40.0 ohm resistor is disconnected, the current fromthe battery =0.195 now 0.195A=E/79+r   ....2 solving equations 1 and 2 we will get the value of E emf of thebattery. internal resistance of the battery=r=39.43ohm now E=23.09V

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