# Determine the potential difference VA–VB between points A and B of circuit shown in fig. under what condition is it equal to zero?

Aug 14, 2018

${V}_{A} - {V}_{B} = \left[{C}_{2} / \left({C}_{1} + {C}_{2}\right) - {C}_{4} / \left({C}_{3} + {C}_{4}\right)\right] V$

For, ${V}_{A} - {V}_{B} = 0$ the required condition is ${C}_{2} / \left({C}_{1} + {C}_{2}\right) = {C}_{4} / \left({C}_{3} + {C}_{4}\right)$ or $V = 0$

#### Explanation:

Let's calculate the total capacitance of the circuit.

${C}_{1}$ and ${C}_{2}$ are in series,so their combined capacitance is $\frac{{C}_{1} {C}_{2}}{{C}_{1} + {C}_{2}}$

Similarly, ${C}_{3}$ and ${C}_{4}$ are in series,so their combined capacitance is $\frac{{C}_{3} {C}_{4}}{{C}_{3} + {C}_{4}}$

Again,their combination are in parallel,(because potential drop across them is the same)so net capacitance of the circuit is $\frac{{C}_{1} {C}_{2}}{{C}_{1} + {C}_{2}} + \frac{{C}_{3} {C}_{4}}{{C}_{3} + {C}_{4}} = C$ (let)

So,total charge flowing in the circuit is $Q = C V$

Now,let,charge flowing through capacitor ${C}_{1}$ and ${C}_{2}$ is ${Q}_{1}$ and that flowing through ${C}_{3}$ and ${C}_{4}$ is ${Q}_{2}$

So, ${Q}_{1} = \frac{{C}_{1} {C}_{2}}{{C}_{1} + {C}_{2}} V$ (as, $q = c v$)

and, ${Q}_{2} = \frac{{C}_{3} {C}_{4}}{{C}_{3} + {C}_{4}} V$ (as, $q = c v$)

Applying Kirchoff's law in the circuit,

${V}_{A} - \frac{{Q}_{1}}{{C}_{1}} + \frac{{Q}_{2}}{{C}_{3}} = {V}_{B}$

or, ${V}_{A} - {V}_{B} = \frac{{Q}_{1}}{{C}_{1}} - \frac{{Q}_{2}}{{C}_{3}}$

putting the value of ${Q}_{1}$ and ${Q}_{2}$ we get,

${V}_{A} - {V}_{B} = \frac{{C}_{1} {C}_{2}}{{C}_{1} + {C}_{2}} \left(\frac{V}{{C}_{1}}\right) - \frac{{C}_{3} {C}_{4}}{{C}_{3} + {C}_{4}} \left(\frac{V}{{C}_{3}}\right)$

$\implies {V}_{A} - {V}_{B} = \left[{C}_{2} / \left({C}_{1} + {C}_{2}\right) - {C}_{4} / \left({C}_{3} + {C}_{4}\right)\right] V$

If, ${V}_{A} - {V}_{B} = 0$

then,

$\left[{C}_{2} / \left({C}_{1} + {C}_{2}\right) - {C}_{4} / \left({C}_{3} + {C}_{4}\right)\right] V = 0$

Now, either,$V = 0$

Or, ${C}_{2} / \left({C}_{1} + {C}_{2}\right) - {C}_{4} / \left({C}_{3} + {C}_{4}\right) = 0$

i.e ${C}_{2} / \left({C}_{1} + {C}_{2}\right) = {C}_{4} / \left({C}_{3} + {C}_{4}\right)$

This is the condition for which ${V}_{A} - {V}_{B}$ will be $0$