Question

Consider the circuit shown in the figure. The value of current 'I' is

• A. $-\frac{7}{18}$ A
• B. 5 A
• C. 3 A
• D. $-3$ A

Hint:

Keep an eye on the sign conventions when setting up KVL loops. A negative final value simply means the actual physical current flows in the opposite direction of the arrow drawn on the diagram!

Answer 1

The Correct Option is A

Let's apply Kirchhoff's Voltage Law (KVL) to the independent structural loops of the circuit network: 1. In the first branch loop containing current component $I_1$: $$-28I_1 - 6 - 8 = 0 \implies -28I_1 = 14 \implies I_1 = -\frac{14}{28} = -0.5\ \text{A}$$ 2. In the secondary branch loop containing current component $I_2$: $$54I_2 - 12 + 6 = 0 \implies 54I_2 = 6 \implies I_2 = \frac{6}{54} = \frac{1}{9}\ \text{A}$$ By Kirchhoff's Current Law (KCL), the combined net target current $I$ equals the sum of these branch currents: $$I = I_1 + I_2 = -\frac{1}{2} + \frac{1}{9} = \frac{-9 + 2}{18} = -\frac{7}{18}\ \text{A}$$
Final Answer:
The value of the current $I$ is $-\frac{7}{18}$ A, which corresponds to option (A).