Question

The current i in the given branch of circuit is

• A. 0.1 A
• B. 0.2 A
• C. 0.5 A
• D. 0.4 A

Answer 1

Answer: (B)

0.2 A

By applying KCL at the node

let the node potential be V

$L^{{}^{\prime }}=L_1^{{}^{\prime }}+L_2^{{}^{\prime }}$

$\frac{(10-V)}{10}=\frac{(V-6)}{20}+\frac{(V-5)}{30}$

$10-V=\left(\frac{V-6}{2}\right)+\frac{V-5}{3}$

$10-V=\frac{5V}{6}-3-\frac{5}{3}$

$\frac{44}{3}=\frac{11V}{6}\Rightarrow V=8$

$i=\frac{10-8}{10}=0.2A$