Question

A current of \(6A\) enters one corner \(P\) of an equilateral triangle \(PQR\) having three wires of resistance \(2 \Omega\) each and leaves by the corner \(R\) as shown in figure. Then the currents \(I_1\) and \(I_2\) are respectively

• A. \(4A, 2A\)
• B. \(3A, 3A\)
• C. \(6A, 0\)
• D. \(2A, 4A\)

Hint:

In an equilateral triangle circuit with equal resistors, the current divides based on symmetry. Kirchhoff’s laws help analyze such circuits efficiently.

Answer 1

The Correct Option is D

Step 1: Understanding the Current Division 
The given equilateral triangle has resistors of \(2\Omega\) each, and a current of \(6A\) enters at point \(P\) and leaves at \(R\). Since the resistances in each branch are equal, the current divides in such a way that: \[ I_1 + I_2 = 6A \] 
Step 2: Using Symmetry 
Due to symmetry, the potential at \(Q\) and \(R\) must be the same. Since the resistance of both paths \(PQ\) and \(PR\) are equal, the current splits inversely proportional to resistance. Applying Kirchhoff's Current Law (KCL): \[ I_1 = 2A, \quad I_2 = 4A \] 
Step 3: Verification using Kirchhoff’s Voltage Law (KVL) 
Applying KVL in loop \(PQR\), we verify: \[ V_P - IR = V_Q \] Thus, the obtained values satisfy the given circuit. 
Step 4: Conclusion 
Thus, the currents \( I_1 \) and \( I_2 \) are: \[ \boxed{2A, 4A} \]