Question

Correct option for the circuit shown is: 

• A. Current \(i_1 = \frac{15}{16}\,A\)
• B. Current \(i_2 = \frac{15}{8}\,A\)
• C. Current \(i_3 = \frac{5}{4}\,A\)
• D. Current \(i_1 = \frac{5}{8}\,A\)

Answer 1

The Correct Option is B

Concept: Use Kirchhoff’s current law (KCL) and Kirchhoff’s voltage law (KVL). Let the currents in the branches be \(i_1, i_2, i_3\). 

 Step 1: Apply KVL in first loop \[ -2(i_1-i_1-i_3) - 4(i_1-i_1) -10V + i_1 = 0 \] \[ 2i_1 - 6i_1 + 7i_3 = 10 \] \[ 2i_1 - 6i_1 + 7i_3 = 10 \quad ...(1) \] Step 2: Second loop \[ -2(i_2-i_1) + i_3 + 10 - 4i_2 = 0 \] \[ -2i_1 + 6i_1 + i_3 = 10 \quad ...(2) \] Step 3: Third loop \[ 5 - 4(i_1-i_3) - 4i_2 = 0 \] \[ 8i_2 - 4i_3 = 5 \quad ...(3) \] Step 4: Solve simultaneous equations Solving (1), (2) and (3): \[ i_3 = \frac{5}{2}\,A \] \[ i_2 = \frac{15}{8}\,A \] \[ i_1 = \frac{15}{8}\,A \] Thus the correct statement is \[ \boxed{i_2 = \frac{15}{8}\,A} \]