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Equivalent Resistance in Series Circuit

Three resista...

Question

Three resistances are joined together to form a letter $Y$ as shown in the figure.If the potentials of the terminals $A, B$ and $C are V_{1}, V_{2} and V_{3}$ respectively, then determine the potential of the node $O$

A

[\frac{V_{1}}{R_{1}^{2}} + \frac{V_{2}}{R_{2}^{2}} + \frac{V_{3}}{R_{3}^{2}}] {[\frac{1}{R_{1}} + \frac{1}{R_{2}} + \frac{1}{R_{3}}]}^{- 2}

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B

[\frac{V_{1}}{R_{1}} + \frac{V_{2}}{R_{2}} + \frac{V_{3}}{R_{3}}] {[\frac{1}{R_{1}} + \frac{1}{R_{2}} + \frac{1}{R_{3}}]}^{- 1}

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C

[\frac{V_{1}}{R_{1}} + \frac{V_{2}}{R_{2}} + \frac{V_{3}}{R_{3}}] [R_{1} + R_{2} + R_{3}]

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D

[\frac{V_{1}}{R_{1}^{2}} + \frac{V_{2}}{R_{2}^{2}} + \frac{V_{3}}{R_{3}^{2}}] [R_{1}^{2} + R_{2}^{2} + R_{3}^{2}]

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Solution

The correct option is B $[\frac{V_{1}}{R_{1}} + \frac{V_{2}}{R_{2}} + \frac{V_{3}}{R_{3}}] {[\frac{1}{R_{1}} + \frac{1}{R_{2}} + \frac{1}{R_{3}}]}^{- 1}$
From the figure,
Applying the junction rule to $O$ , we get

$- I_{1} - I_{2} - I_{3} = 0 \Rightarrow I_{1} + I_{2} + I_{3} = 0$

Now if $V_{0}$ is the potential at point $O$ , then

$I_{1} = \frac{(V_{0} - V_{1})}{R_{1}}; I_{2} = \frac{(V_{0} - V_{2})}{R_{2}}$ and $I_{3} = \frac{(V_{0} - V_{3})}{R_{3}}$ ;

So substituting these values of $I_{1}, I_{2}$ and $I_{3}$ in eq. (i), we get

$V_{0} [\frac{1}{R_{1}} + \frac{1}{R_{2}} + \frac{1}{R_{3}}] - [\frac{V_{1}}{R_{1}} + \frac{V_{2}}{R_{2}} + \frac{V_{3}}{R_{3}}] = 0$

$V_{0} = [\frac{V_{1}}{R_{1}} + \frac{V_{2}}{R_{2}} + \frac{V_{3}}{R_{3}}] {[\frac{1}{R_{1}} + \frac{1}{R_{2}} + \frac{1}{R_{3}}]}^{- 1}$

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