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

Step 1: Apply the loop rule.
Use Kirchhoff's voltage law in each loop. The first loop gives $-28I_1-6-8=0$, so $I_1=-\frac{14}{28}=-0.5$ A.

Step 2: Second loop.
The other loop gives $54I_2-12+6=0$, so $I_2=\frac{6}{54}=\frac{1}{9}$ A.

Step 3: Add at the junction.
By the junction rule, $I=I_1+I_2=-\frac{1}{2}+\frac{1}{9}=\frac{-9+2}{18}=-\frac{7}{18}$ A.
\[ \boxed{-\frac{7}{18}\text{ A}} \]