Question

In the circuit shown in the figure, the current through the $10\Omega$ resistor is

• A. $\frac{1}{9}\text{A}$
• B. $\frac{4}{9}\text{A}$
• C. $\frac{2}{3}\text{A}$
• D. $\frac{5}{6}\text{A}$

Answer 1

The correct option is B $\frac{4}{9}\text{A}$

Applying Kirchoff's junction law at node $A$, we get
$\frac{12-v^{\prime }}{3}=\frac{v^{\prime }}{3}+\frac{v^{\prime }}{12}$
$\Rightarrow 4=v^{\prime }(\frac{1}{3}+\frac{1}{3}+\frac{1}{12})$
$\Rightarrow v^{\prime }=\frac{16}{3}\text{V}$
Now, current in $10\Omega$ resistor,
$I=\frac{\frac{16}{3}}{10+2}=\frac{4}{9}\text{A}$