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Kirchhoff's Junction Law

A circuit has...

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}

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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

Zero

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D

[\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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Solution

The correct option is A $V_{1} + V_{2} + V_{3}$
According to kirchhoff's first rule to $O$
$- l_{1} - l_{2} - l_{3} = 0$ or
$l_{1} + l_{2} + l_{3} = 0$ ...(i)
If $V_{0}$ is the electric potential at $O$
$V_{1}, V_{2}$ and $V_{3}$ be the potential at $A, B$ and $C$ respectively then $V_{0} - V_{1} = l_{1} R_{1}$ .
$V_{0} - V_{2} = l_{2} R_{2}$ and $V_{0} - V_{3} = l_{3} R_{3}$
This, $l_{1} = (\frac{V_{0} - V_{1}}{R_{1}}), l_{2} = (\frac{V_{0} - V_{2}}{R_{2}})$ , and $l_{3} = (\frac{V_{0} - V_{3}}{R_{3}})$

putting volume of $l_{1}, l_{2}$ and $l_{3}$ in (I) we get
$\frac{V_{0} - V_{1}}{R_{1}} + \frac{V_{0} - V_{2}}{R_{2}} + \frac{V_{0} - V_{3}}{R_{3}} = 0$

or $V_{0} [\frac{1}{R_{1}} + \frac{1}{R_{2}} + \frac{1}{R_{3}}] - ⎡ ⎢ ⎣ \frac{\frac{V_{1}}{R_{1}} + \frac{V_{2}}{R_{2}} + V_{3}}{R_{3}} ⎤ ⎥ ⎦ = 0$

or $V_{0} = ⎡ ⎢ ⎣ \frac{\frac{V_{1}}{R_{1}} + \frac{V_{2}}{R_{2}} + V_{3}}{R_{3}} ⎤ ⎥ ⎦ {[\frac{1}{R_{1}} + \frac{1}{R_{2}} + \frac{1}{R_{3}}]}^{- 1}$

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