Question

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

Answer 1

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

For, `V_A-V_B=0` the required condition is `C_2/(C_1 +C_2) =C_4/(C_3 +C_4)` 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 `(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`