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., $I_{1}+I_{2}+I_{3}=0\quad\ldots\left(i\right)$
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
$\left(V_{0}-V_{1}\right)=IR_{1}$;
$\left(V_{0}-V_{2}\right)=I_{2}R_{2}$ 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
$V_{0}\left[\frac{1}{R_{1}}+\frac{1}{R_{2}}+\frac{1}{R_{3}}\right]-\left[\frac{V_{1}}{R_{1}}+\frac{V_{2}}{R_{2}}+\frac{V_{3}}{R_{3}}\right]=0$
$V_{0}=\left[\frac{V_{1}}{R_{1}}+\frac{V_{2}}{R_{2}}+\frac{V_{3}}{R_{3}}\right]\left[\frac{1}{R_{1}}+\frac{1}{R_{2}}+\frac{1}{R_{3}}\right]^{-1}$