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प्रश्न
In the circuit shown in the figure, ε1 = 3 V, ε2 = 2 V, εa = 1 V and r1 = r2 = r3 = 1Ω. Find the potential difference between the points A and B and the current through each branch.

बेरीज
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उत्तर

Applying KVL in loop 1, we get:-
\[i_1 + \left( i_1 - i_2 \right) - 1 + 2 = 0\]
\[2 i_1 - i_2 = - 1 .............(1)\]
Similarly, for loop 2:-
\[i_2 - \left( i_1 - i_2 \right) - 2 + 3 = 0\]
\[2 i_2 - i_1 = - 1 ............(2)\]
Solving (1) and (2), we get:-
\[i_1 = i_2 = 1 A\]
Potential difference between A and B:-
\[V_A - V_B = E_2 - \left( i_1 - i_2 \right) \times 1 = 2 - \left( 1 - 1 \right) = 2 V\]
Current through the top branch = i2 = 1 A
Current through the middle branch = i1 - i2 = 0 A
Current through the bottom branch = i1 = 1 A
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Potential Difference and Emf of a Cell
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