Question

Obtain the equivalent capacitance of the network shown in the figure for a 300 V supply. Determine the charge and voltage across each capacitor.

Answer 1

Hint: Capacitors connected in series have the same charge on them and those connected parallel have the same voltage across them. Also remember the formula for series and parallel combination.

Formula used: The equivalent capacitance for capacitors connected in series is given by

$\dfrac{1}{{{C_{eff}}}} = \dfrac{1}{{{C_1}}} + \dfrac{1}{{{C_2}}} + \dfrac{1}{{{C_3}}}....\dfrac{1}{{{C_n}}}$

The equivalent capacitance for capacitors connected in parallel is given by

${C_{eff}} = {C_1} + {C_2} + {C_3} + \cdot \cdot \cdot + {C_n}$.

Complete step by step solution :

From the diagram we can see that the capacitors ${C_2}$ and ${C_3}$ are connected in series whereas ${C_1}$, ${C_4}$ and the combined capacitance of ${C_2}$ and ${C_3}$ are connected in parallel.

Given ${C_1} = 100pF,{C_2} = 200pF,{C_3} = 200pF,{C_4} = 100pF.$

${C_{eff}} = {C_1} + \dfrac{1}{{\dfrac{1}{{{C_2}}} + \dfrac{1}{{{C_3}}}}} + {C_4}$.

Substituting the values of ${C_1},{C_2},{C_3},{C_4}$ gives:

${C_{eff}} = 100 + \dfrac{1}{{\dfrac{1}{{200}} + \dfrac{1}{{200}}}} + 100$
${C_{eff}} = 100 + \dfrac{1}{{\dfrac{1}{{100}}}} + 100$
${C_{eff}} = 100 + 100 + 100 = 300pF$

Given $V = 300V$

Since $Q = C \times V$, We can find the total charge on all capacitors combined as:

$Q = 300 \times 300 \times {10^{ - 12}}$
$Q = 9 \times {10^{ - 8}}$C

We know that the effective capacitance of ${C_2}$ and ${C_3}$ combined is $100pF$ and this is equal to the individual capacitances of ${C_1}$ and ${C_4}$.

So the given circuit consists of three capacitors , each of $100pF$ connected parallel.

From the figure, we see that The potential difference across each of these three capacitors is $300V$.

Since the Voltages and the capacitances of these capacitors are equal. The charge present in them should also be equal since $V = \dfrac{Q}{C}$

Thus, ${V_2} = {V_3} = 300V$
Now ${V_2} + {V_3} = 300$

Since ${C_2}$ and ${C_3}$ are in series, the charge on both of them are the same. Now since $V = \dfrac{Q}{C}$,

$300 = \dfrac{Q}{{{C_2}}} + \dfrac{Q}{{{C_3}}}$
$300 = 2 \times \dfrac{Q}{C}$
$\dfrac{Q}{C} = {V_2} = {V_3}$
$\dfrac{Q}{C} = \dfrac{{300}}{2}$
$= 150V$

Additional Information:

Series

When the capacitances are connected in series ,the charge across each capacitor remain constant whereas their voltages vary. If ${V_1},{V_2},{V_3}$are the respective voltages across each capacitors then ${V_{net}}$ is given by

${V_{net}} = {V_1} + {V_2} + {V_3}...... + {V_n}.$

By substituting the equation $V = \dfrac{Q}{C}$

$\dfrac{Q}{{{C_{eff}}}} = \dfrac{Q}{{{C_1}}} + \dfrac{Q}{{{C_2}}} + \dfrac{Q}{{{C_3}}}....\dfrac{Q}{{{C_n}}}$
$\dfrac{1}{{{C_{eff}}}} = \dfrac{1}{{{C_1}}} + \dfrac{1}{{{C_2}}} + \dfrac{1}{{{C_3}}}....\dfrac{1}{{{C_n}}}$
${C_{eff}} = \dfrac{1}{{\dfrac{1}{{{C_1}}} + \dfrac{1}{{{C_2}}} + \dfrac{1}{{{C_3}}}....\dfrac{1}{{{C_n}}}}}$

Parallel

In parallel connection the voltage across the capacitors remains constant and the charge varies. If ${Q_1},{Q_2},{Q_3}$ are the respective charges across each capacitor, then ${Q_{net}}$ is given by

${Q_{net}} = {Q_1} + {Q_2} + {Q_3}...... + {Q_n}.$

Since $Q = C \times V$, let’s replace charges in the above equation.

${C_{eff}} \times V = {C_1} \times V + {C_2} \times V + {C_3} \times V...... + {C_n} \times V$.
${C_{eff}} = {C_1} + {C_2} + {C_3}...... + {C_n}$.

Note:
(i)When $n$ capacitors of equal capacitance are connected in series the effective capacitance becomes $\dfrac{1}{n} \times C$
(ii) When $n$ capacitors of equal capacitance are connected in parallel the effective capacitance becomes $n \times C$ .