Question

A circuit has a section ABC as shown in figure. If the potentials at points A, B and C are V1, V2 and V3 respectively. The potential at point O is

• A. $\frac{2\sigma_{1}\sigma_{2}}{\sigma_{1} + \sigma_{2}}$
• B. $\left[ \frac { \mathrm { V } _ { 1 } } { \mathrm { R } _ { 1 } } + \frac { \mathrm { V } _ { 2 } } { \mathrm { R } _ { 2 } } + \frac { \mathrm { V } _ { 3 } } { \mathrm { R } _ { 3 } } \right] \left[ \frac { 1 } { \mathrm { R } _ { 1 } } + \frac { 1 } { \mathrm { R } _ { 2 } } + \frac { 1 } { \mathrm { R } _ { 3 } } \right]$ $\overset{o}{A}$
• C. Zero
• D. $9.4 \times 10^{18}$

Answer 1

Answer: (B)

\left[ \frac { \mathrm { V } _ { 1 } } { \mathrm { R } _ { 1 } } + \frac { \mathrm { V } _ { 2 } } { \mathrm { R } _ { 2 } } + \frac { \mathrm { V } _ { 3 } } { \mathrm { R } _ { 3 } } \right] \left[ \frac { 1 } { \mathrm { R } _ { 1 } } + \frac { 1 } { \mathrm { R } _ { 2 } } + \frac { 1 } { \mathrm { R } _ { 3 } } \right]$$\overset{o}{A}

Answer 2

: Applying junction rule to O

$- I_{1} - I_{2} - I_{3} = 0$

i.e., $I_{1} + I_{2} + I_{3} = 0$ …(i)

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

$(V_{0} - V_{1}) = I_{1}R_{1};(V_{0} - V_{2}) = I_{2}R_{2}$

and,$(V_{0} - V_{3}) = I_{3}R_{3}$

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

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

$V_{0}\left\lbrack \frac{1}{R_{1}} + \frac{1}{R_{2}} + \frac{1}{R_{3}} \right\rbrack - \left\lbrack \frac{V_{1}}{R_{1}} + \frac{V_{2}}{R_{2}} + \frac{V_{3}}{R_{3}} \right\rbrack = 0$

$V_{0} = \left\lbrack \frac{V_{1}}{R_{1}} + \frac{V_{2}}{R_{2}} + \frac{V_{3}}{R_{3}} \right\rbrack ⥂ \left\lbrack \frac{1}{R_{1}} + \frac{1}{R_{2}} + \frac{1}{R_{3}} \right\rbrack^{- 1}$