Question

A circuit has a section $ABC$ as shown in figure.

If the potentials at points $A$, $B$ and $C$ are $V_1$, $V_2$ and $V_3$ respectively. The potential at point $O$ is

• A. $V_1+V_2+V_3$
• B. $\left[\frac{V_1}{R_1}+\frac{V_2}{R_2}+\frac{V_3}{R_3}\right]{\left[\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}\right]}^{-1}$
• C. Zero
• D. $\left[\frac{V_1}{R_1}+\frac{V_2}{R_2}+\frac{V_3}{R_3}\right]\left(R_1+R_2+R_3\right)$

Answer 1

Answer: (B)

Applying junction rule to $O$

$-I_1-I_2=I_3=0$
i,e., $I_1+I_2+I_3=0\dots \left(i\right)$
Let, $V_0$ be the potential at point $O$. By Ohm's law for resistances $R_1$, $R_2$ and $R_3$ respectively, we get
$\left(V_0-V_1\right)=I{R}_1$;
$\left(V_0-V_2\right)=I_2{R}_2$ and $\left(V_0-V_3\right)=I_3{R}_3$
or $I_1=\frac{\left(V_0-V_1\right)}{R_1}$;
$I_2=\frac{\left(V_0-V_2\right)}{R_2}$;
$I_3=\frac{\left(V_0-V_3\right)}{R_3}$
So substituting these values of $I_1$, $I_2$ and $I_3$ in eqn. $\left(i\right)$, we get
$V_0\left[\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}\right]-\left[\frac{V_1}{R_1}+\frac{V_2}{R_2}+\frac{V_3}{R_3}\right]=0$
$V_0=\left[\frac{V_1}{R_1}+\frac{V_2}{R_2}+\frac{V_3}{R_3}\right]{\left[\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}\right]}^{-1}$