Question

Find the current flowing through the resistance $R_1$ of the circuit shown in figure, if the resistances are equal to $R_1=10\Omega$, $R_2=10\Omega$ and $R_3=10\Omega$, and the potential of points $1,2$ and $3$ are equal to $V_1=10\text{V}$, $V_2=6\text{V}$ and $V_3=5\text{V}$.

• A. $0.3\text{A}$
• B. $0.1\text{A}$
• C. $0.4\text{A}$
• D. $0.2\text{A}$

Answer 1

The correct option is A $0.3\text{A}$


Assume potential at the junction to be $x$ and apply KCL.

$i_1+i_2+i_3=0$

$\Rightarrow \frac{x-10}{10}+\frac{x-6}{10}+\frac{x-5}{10}=0$

$\Rightarrow x=7\text{V}$

So,

$i_1=\frac{x-10}{10}=\frac{7-10}{10}=-0.3\text{A}$

So, the direction of current will be opposite to the assumption.

Therefore, option $\left(c\right)$ is correct.