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

Let's calculate the total capacitance of the circuit.

#C_1# and #C_2# are in series,so their combined capacitance is #(C_1 C_2)/(C_1+C_2)#

Similarly, #C_3# and #C_4# are in series,so their combined capacitance is #(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 #(C_1 C_2)/(C_1+C_2) +(C_3 C_4)/(C_3 +C_4)=C# (let)

So,total charge flowing in the circuit is #Q=CV#

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 = (C_1 C_2)/(C_1+C_2) V# (as, #q=cv#)

and, #Q_2 = (C_3 C_4)/(C_3 +C_4) V# (as, #q=cv#)

Applying Kirchoff's law in the circuit,

#V_A-(Q_1)/(C_1) +(Q_2)/(C_3)=V_B#

or, #V_A -V_B= (Q_1)/(C_1) -(Q_2)/(C_3)#

putting the value of #Q_1 # and #Q_2# we get,

#V_A -V_B=(C_1 C_2)/(C_1+C_2) (V/(C_1)) -(C_3 C_4)/(C_3 +C_4) (V/(C_3))#

#=> V_A -V_B= [C_2/(C_1 +C_2) -C_4/(C_3 +C_4)] V#

If, #V_A - V_B =0#

then,

# [C_2/(C_1 +C_2) -C_4/(C_3 +C_4)] V=0#

Now, either,#V=0#

Or, #C_2/(C_1 +C_2) -C_4/(C_3 +C_4)=0#

i.e #C_2/(C_1 +C_2) =C_4/(C_3 +C_4)#

This is the condition for which #V_A -V_B# will be #0#