Question

In the given circuit, \(E_1 = E_2 = E_3 = 2V\) and \(R_1 = R_2 = 4\Omega\), then the current flowing through the branch AB is: 
 

• A. \(0\)
• B. \(2A\) from A to B
• C. \(2A\) from B to A
• D. \(5A\) from B to B

Hint:

Use Kirchhoff’s Laws to analyze complex circuits efficiently.

Answer 1

The Correct Option is B

Step 1: {Finding Equivalent EMF and Resistance}
Using Kirchhoff's Voltage Law (KVL): \[ E_{eq} = \frac{E_1 R_1 + E_2 R_2}{R_1 + R_2} = \frac{2 \times 4 + 2 \times 4}{4 + 4} = 2V \] \[ R_{eq} = \frac{R_1 R_2}{R_1 + R_2} = \frac{4 \times 4}{4+4} = 2\Omega \] Step 2: {Finding the Current}
\[ I = \frac{E'}{R_{eq}} = \frac{4V}{2\Omega} = 2A \] Thus, the current flows from A to B at \(2A\).