Question

Determine the potential difference ${\phi }_A-{\phi }_B$ between points $A$ and $B$ of the circuit shown in Fig. Under what condition is it equal to zero?

Answer 1

Make the charge distribution, as shown in the figure. In the circuit, $12561$.

$-△\phi =0yields$

$\frac{q_1}{C_4}+\frac{q_1}{C_3}-\xi =0$ or, $q_1=\frac{\xi C_3{C}_4}{C_3+C_4}$

and in the circuit $13461$,

$\frac{q_2}{C_2}+\frac{q_2}{C_1}-\xi =0$ or, $q_2=\frac{\xi C_1{C}_2}{C_1+C_2}$

Now ${\phi }_A-{\phi }_B=\frac{q_2}{C_1}-\frac{q_1}{C_3}$

$=\xi [\frac{C_2}{C_1+C_2}-\frac{C_4}{C_3+C_4}]=\xi \left[\frac{C_2{C}_3-C_1{C}_4}{(C_1+C_2)(C_3+C_4)}\right]$

It becomes zero, when

$(C_2{C}_3-C_1{C}_4)=0$. or $\frac{C_1}{C_2}=\frac{C_3}{C_4}$.