Question

Use Kirchhoff’s rules to find the current through \(3\,\Omega\) resistor in the circuit shown in the figure.

Answer 1

Kirchhoff’s rules application

Step 1: Assign node potentials
Take bottom node \(E = D\) as reference: \[ V_E = 0 \] Let:
• \(A = B = X\)
• \(C = W\)
• \(F = Z\)

Step 2: Battery relations
From the 3V battery: \[ V_X - V_Z = 3 \] From the 5V battery: \[ V_X - V_W = 5 \] So: \[ Z = X - 3,\quad W = X - 5 \]

Step 3: Apply KCL at supernode
Currents from supernode \((X, W, Z)\) to ground: Through \(3\Omega\): \[ I_1 = \frac{X}{3} \] Through \(2\Omega\): \[ I_2 = \frac{W}{2} \] Through \(4\Omega\): \[ I_3 = \frac{Z}{4} \] KCL: \[ \frac{X}{3} + \frac{W}{2} + \frac{Z}{4} = 0 \] Substitute \(W = X-5,\ Z = X-3\): \[ \frac{X}{3} + \frac{X-5}{2} + \frac{X-3}{4} = 0 \]

Step 4: Solve equation
Multiply by 12: \[ 4X + 6(X-5) + 3(X-3) = 0 \] \[ 4X + 6X - 30 + 3X - 9 = 0 \] \[ 13X - 39 = 0 \] \[ X = 3\,\text{V} \]

Step 5: Current through \(3\Omega\)
\[ I = \frac{X}{3} = \frac{3}{3} = 1\,\text{A} \] Final Answer: \[ \boxed{I_{3\Omega} = 1\,\text{A}} \]