Question

In this circuit, the value of $I_2$, is

• A. 0.2 A
• B. 0.3 A
• C. 0.4 A
• D. 0.6 A

Answer 1

Answer: (C)

By current divider rule, we have,
$I_2=\frac{\frac{1}{R_2}}{\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}}I=\frac{\frac{1}{15}}{\frac{1}{10}+\frac{1}{15}+\frac{1}{30}}\times 1.2$
$=0.4A$