Sum

In the circuit shown in the figure, ε_{1} = 3 V, ε_{2} = 2 V, ε_{a} = 1 V and r_{1} = r_{2}_{ }= r_{3} = 1Ω. Find the potential difference between the points A and B and the current through each branch.

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#### Solution

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 = i_{2} = 1 A

Current through the middle branch = i_{1} - i_{2} = 0 A

Current through the bottom branch = i_{1} = 1 A

Concept: Potential Difference and Emf of a Cell

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