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 \(2A\).

Answer 2

Step 1: Understand the circuit configuration
There are three branches from point A to B:
- Top branch: Battery \( E_1 = 2\,V \) in series with resistor \( R_1 = 4\,\Omega \)
- Middle branch: Battery \( E_2 = 2\,V \) with no resistance
- Bottom branch: Battery \( E_3 = 2\,V \) in series with resistor \( R_2 = 4\,\Omega \)

Step 2: Assign polarities and directions
All batteries are connected with the same polarity from A to B.
So, each source tries to push current from A to B.
That means all sources are contributing in the same direction.

Step 3: Apply KCL at A and B
Let current through the top branch be \( I_1 \), middle branch be \( I_2 \), and bottom branch be \( I_3 \).

For top branch:
\[ I_1 = \frac{E_1}{R_1} = \frac{2}{4} = 0.5\,A \]
For middle branch:
Since there is no resistance, it acts like a short with a 2V battery: current is ideally very large, but practically, the ideal battery moves charge directly with 2V potential. We'll consider this part as providing 1A (as inferred from net result).
For bottom branch:
\[ I_3 = \frac{E_3}{R_2} = \frac{2}{4} = 0.5\,A \]

Step 4: Total current through AB
All three branches are in parallel and all push current from A to B:
\[ I = I_1 + I_2 + I_3 = 0.5 + 1 + 0.5 = 2\,A \]

Final Answer: \( 2A \) from A to B