two batteries A and B each of emf 2volt are connected in series to an external resistance R=1 ohm. if the internal resistance of battery A is 1.9 ohm and that of B is 0.9 ohm.what is the potential difference between the terminals of battery A?

Asked by  | 13th Sep, 2012, 02:11: AM

Expert Answer:

The circuit diagram can be drawn as:
using the Kirschoff's Loop law:
I + 0.9I -2 + 1.9I - 2 = 0
I = 4/3.8 A
thus potential drop across 1.9 ohm resistor: 1.9 * 4/3.8 = 2 V
hence the potential difference of battery A in the circuit will  be: 0 V

Answered by  | 16th Sep, 2012, 01:21: AM

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