Question

When a potential difference of ${10^3}$ volt is applied between A and B , a charge of $0.75mC$ is stored in the system of capacitor as shown . The value of C is (in $\mu F$ ) is:

A) $\dfrac{1}{2}$
B) $2$
C) $2.5$
D) $3$

Answer 1

First we can calculate the equivalent capacitance of system of capacitors by using given information that the charge stored in system of capacitors is $0.75mC$ and voltage across the terminal is ${10^3}$ volt. And we can calculate equivalent capacitance of circuit by using circuit diagram ,by using combination formula for series and parallel combination of capacitor

Complete step by step solution:
By the definition of capacitance we know charge stored in capacitor per unit potential difference is given by
$\Rightarrow C = \dfrac{Q}{V}$
In question it is given the charge stored in a system of given capacitors or charge stored in equivalent capacitor $Q = 0.75mC$ or $Q = 0.75 \times {10^{ - 3}}C$.
$V$ is given potential across the terminal is ${10^3}$ volt
So the equivalent capacitance of the given circuit is ${C_{eq}} = \dfrac{Q}{V}$
$\Rightarrow {C_{eq}} = \dfrac{{0.75 \times {{10}^{ - 3}}}}{{{{10}^3}}}$
Solving this
$\Rightarrow {C_{eq}} = 0.75 \times {10^{ - 6}}F$
$\Rightarrow {C_{eq}} = 0.75\mu F$ .................. (1)
So the equivalent capacitance of given system of capacitors is $0.75\mu F$

Step 2

Now look at the circuit it is clear from circuit capacitor C1 and C2 are connected in series so series combination equivalent capacitance of C1 and C2 is C12 can calculate as
$\Rightarrow \dfrac{1}{{{C_{12}}}} = \dfrac{1}{{{C_1}}} + \dfrac{1}{{{C_2}}}$
Put the value of C1 and C2
$\Rightarrow \dfrac{1}{{{C_{12}}}} = \dfrac{1}{2} + \dfrac{1}{2}$
$\Rightarrow \dfrac{1}{{{C_{12}}}} = \dfrac{1}{1}$
Hence ${C_{12}} = 1\mu F$
And $C_3$ and $C_4$ are also connected in series so equivalent capacitance C34 can be calculate as:
$\Rightarrow \dfrac{1}{{{C_{34}}}} = \dfrac{1}{{{C_3}}} + \dfrac{1}{{{C_4}}}$
Put value of ${C_3},{C_4}$
$\Rightarrow \dfrac{1}{{{C_{34}}}} = \dfrac{1}{2} + \dfrac{1}{2}$
$\Rightarrow \dfrac{1}{{{C_{34}}}} = \dfrac{1}{1}$
$\Rightarrow {C_{34}} = 1\mu C$
Hence we can replace ${C_1},{C_2}$ by their resultant ${C_{12}}$ and ${C_3},{C_4}$ by their resultant ${C_{34}}$ in circuit then circuit become.

Now in this circuit ${C_{12}}$ and $C$ connected in parallel so the resultant capacitance can calculate by
$\Rightarrow {C_{12c}} = {C_{12}} + C$
$\Rightarrow {C_{12c}} = 1 + C$
Now circuit become

Now ${C_{34}}$ and ${C_{12c}}$ is in series so
Equivalent capacitance of whole circuit
$\Rightarrow \dfrac{1}{{{C_{eq}}}} = \dfrac{1}{{{C_{34}}}} + \dfrac{1}{{{C_{12c}}}}$
Put values
$\Rightarrow \dfrac{1}{{{C_{eq}}}} = \dfrac{1}{1} + \dfrac{1}{{\left( {1 + C} \right)}}$
Solving this
$\Rightarrow \dfrac{1}{{{C_{eq}}}} = \dfrac{{1 + C + 1}}{{\left( {1 + C} \right)}}$
$\Rightarrow \dfrac{1}{{{C_{eq}}}} = \dfrac{{\left( {2 + C} \right)}}{{\left( {1 + C} \right)}}$
So the equivalent capacitance of given whole circuit is
$\Rightarrow {C_{eq}} = \dfrac{{\left( {1 + C} \right)}}{{\left( {2 + C} \right)}}$ ...................... (2)
Equation (1) and (2) both are representing the equivalent capacitance of given circuit
So (1) = (2)
$\Rightarrow \dfrac{{\left( {1 + C} \right)}}{{\left( {2 + C} \right)}} = 0.75$
$\Rightarrow C + 1 = 0.75\left( {C + 2} \right)$
$\Rightarrow C + 1 = 0.75\left( C \right) + 1.50$
Again solving this
$\Rightarrow C - 0.75\left( C \right) = 1.50 - 1$
$\Rightarrow C\left( {1 - 0.75} \right) = 0.50$
Further solving
$\Rightarrow C = \dfrac{{0.50}}{{0.25}}$
$\therefore C = 2\mu F$
Hence in given circuit the value of $C$ is $2\mu F$.

So option B is correct.

Note: We use here series combination formula for calculating equivalent capacitance let’s talk about this. If capacitors connected in series then the charge stored in every capacitor same but the voltage across every capacitor different let take three capacitor connected in series voltage across capacitors ${C_1},{C_2}$ and ${C_3}$ is ${V_1},{V_2}$ and ${V_3}$ respectively
${V_1} = \dfrac{q}{{{C_1}}},{V_2} = \dfrac{q}{{{C_2}}},{V_3} = \dfrac{q}{{{C_3}}}$
Net voltage $V = {V_1} + {V_2} + {V_3}$
$\Rightarrow V = \dfrac{q}{{{C_1}}} + \dfrac{q}{{{C_2}}} + \dfrac{q}{{{C_3}}}$
If equivalent capacitance is ${C_{eq}}$ then $V = \dfrac{q}{{{C_{eq}}}}$
$\Rightarrow \dfrac{q}{{{C_{eq}}}} = \dfrac{q}{{{C_1}}} + \dfrac{q}{{{C_2}}} + \dfrac{q}{{{C_3}}}$
So we get
$\therefore \dfrac{1}{{{C_{eq}}}} = \dfrac{1}{{{C_1}}} + \dfrac{1}{{{C_2}}} + \dfrac{1}{{{C_3}}}$.