Question

A circuit is shown in figure. The ratio of currents $I_1/I_2$ is

• A. 5/3
• B. 4/3
• C. 3/2
• D. 2/3

Answer 1

Answer: (B)

The figure can be reduced as:

Current though $8\Omega =\frac{6}{8}=\frac{3}{4}A$
Current through $(3+1)\Omega =\frac{8-6}{4}=\frac{2}{4}=\frac{1}{2}A$
$\therefore I_1=\frac{3}{4}-\frac{1}{2}=\frac{1}{4}A$
Now, using current divider rule, current $I_2$ can be found out.
$\therefore I_2=\frac{3}{4}\times \frac{1}{2}\times \frac{1}{2}=\frac{3}{16}A$
$\therefore \frac{I_1}{I_2}=\frac{4}{3}$