Question

The potential difference between point B and C of the circuit is-
(A)\[\dfrac{{\left( {{C_2} - {C_1}} \right)}}{V}\]
(B)\[\dfrac{{\left( {{C_4} - {C_3}} \right)}}{V}\]
(C) \[\dfrac{{\left( {{C_2}{C_3} - {C_1}{C_4}} \right)}}{{\left( {{C_1} + {C_2} + {C_3} + {C_4}} \right)}}.V\]
(D) \[\dfrac{{\left( {{C_1}{C_4} - {C_2}{C_3}} \right)}}{{\left( {{C_1} + {C_2}} \right) \times \left( {{C_3} + {C_4}} \right)}}.V\]

Answer 1

Hint: As We know when capacitors are connected in series then the total capacitance will be the reciprocal of the individual capacitors that are connected. When the capacitors are connected in parallel combination then the total capacitance is the direct addition of the individual capacitances.
\[\dfrac{1}{{{C_S}}} = \dfrac{1}{{{C_1}}} + \dfrac{1}{{{C_2}}} + \dfrac{1}{{{C_3}}} + - - \]
\[{C_P} = {C_1} + {C_2} + {C_3} + - - - \]

Complete step by step answer:
In the arm ABD, the net voltage drop is equal to \[v\] volts and $C_1$ and $C_2$ are connected in series. So, the value of capacitance is-
\[\dfrac{1}{C} = \dfrac{1}{{{C_1}}} + \dfrac{1}{{{C_2}}}\]
\[C = \dfrac{{{C_1}{C_2}}}{{\left( {{C_1} + {C_2}} \right)}}\]
Charge will also be the same in series combination. So, \[Q = CV\]
Voltage across $C_2$ will be\[ = \Delta V\]\[ = \dfrac{Q}{{{C_2}}}\]
Here put the value of Q,
We get-\[\Delta V = \dfrac{{CV}}{{{C_2}}}\]
Substitute the value of C in this equation.
\[\Delta V = \dfrac{{{C_1}V}}{{{C_1} + {C_2}}}\]
\[{V_B} - {V_D} = \dfrac{{{C_1}V}}{{{C_1} + {C_2}}}\]
 As we know potential at point D is \[0\]. So
\[{V_B} - 0 = \dfrac{{{C_1}V}}{{{C_1} + {C_2}}}\]
Similarly we can write, \[{V_c} - 0 = \dfrac{{{C_3}V}}{{{C_3} + {C_4}}}\]
Now we will calculate \[{V_B} - {V_C}\] ,
\[{V_B} - {V_C}\]\[ = \dfrac{{{C_1}V}}{{{C_1} + {C_2}}} - \dfrac{{{C_3}V}}{{{C_3} + {C_4}}}\]
\[ = \dfrac{{V({C_1}{C_4} - {C_3}{C_2})}}{{\left( {{C_3} + {C_4}} \right)\left( {{C_1} + {C_2}} \right)}}\]
So, potential difference between point B and C is\[ = \dfrac{{V({C_1}{C_4} - {C_3}{C_2})}}{{\left( {{C_3} + {C_4}} \right)\left( {{C_1} + {C_2}} \right)}}\].

So, the correct answer is “Option D”.

Note:
A capacitor is a device which is used to store electrical energy and charge which are available in many shapes and sizes. The amount of storage in a capacitor is calculated by a property named capacitance. There are two types of capacitance- Self capacitance and mutual conductance
The potential difference between two points is defined as the change in potential energy of a charge\[q\] moved from one point to another. Units of potential difference are \[joules\] per \[Coulomb\].