Question

In the following circuit diagram, the potential of points $A,B$ and $C$ are $10\text{V},20\text{V}$ and $20\text{V}$ respectively. The potential of the junction $O$ is


$[$0.77 Mark$]$

• A. $100\text{V}$
• B. $\frac{160}{11}\text{V}$
• C. $20\text{V}$
• D. $10\text{V}$

Answer 1

The correct option is B $\frac{160}{11}\text{V}$

As per Junction law, net current at $O$ is equal to zero.
$\Rightarrow \frac{V_B-V_O}{R_2}+\frac{V_A-V_O}{R_1}+\frac{V_C-V_O}{R_3}=0$
$\Rightarrow \frac{20-V_O}{20}+\frac{10-V_O}{10}+\frac{20-V_O}{30}=0$
$1-\frac{V_O}{20}+1-\frac{V_O}{10}+\frac{2}{3}-\frac{V_O}{30}=0$
$2+\frac{2}{3}=V_O(\frac{1}{20}+\frac{1}{10}+\frac{1}{30})$
$\Rightarrow \frac{8}{3}=V_O\left(\frac{3+6+2}{60}\right)$
$\Rightarrow \frac{8}{3}=V_O\left(\frac{11}{60}\right)\Rightarrow V_O=\frac{160}{11}\text{V}$