Question

A network of four capacitors of capacity equal to ${C_1} = C,{C_2} = 2C,{C_3} = 3C,{C_4} = 4C$ are connected to a battery as shown in the figure. The ratio of the charges on ${C_2}$ and ${C_4}$ is:

Answer 1

As we can see in the figure, 1st, 2nd and 3rd capacitors are connected in series. So find the resultant capacitance of these capacitors and then find the charge on these capacitors. As they are connected in series the charge will be equal. The 4th capacitor is connected parallel to these three capacitors, so the charge will be different for it. Calculate the ratio after finding the charges present on 2nd and 4th capacitors.

Complete step by step answer:

We are given that a network of four capacitors of capacity equal to ${C_1} = C,{C_2} = 2C,{C_3} = 3C,{C_4} = 4C$ are connected to a battery as shown in the above figure.
We have to calculate the ratio of the charges on ${C_2}$ and ${C_4}$
As we can see in the diagram, ${C_1},{C_2},{C_3}$ capacitors are connected in series configuration.

So the equivalent resultant capacitance of ${C_1},{C_2},{C_3}$ will be
$\dfrac{1}{{{C_{eq}}}} = \dfrac{1}{{{C_1}}} + \dfrac{1}{{{C_2}}} + \dfrac{1}{{{C_3}}} \\\ {C_1} = C,{C_2} = 2C,{C_3} = 3C \\\ \dfrac{1}{{{C_{eq}}}} = \dfrac{1}{C} + \dfrac{1}{{2C}} + \dfrac{1}{{3C}} \\\ \implies \dfrac{1}{{{C_{eq}}}} = \dfrac{{6 + 3 + 2}}{{6C}} \\\ \implies {C_{eq}} = \dfrac{6}{{11}}C \\\$
Capacitors which are connected in series will have equal charges.
Charge (Q) present on ${C_{eq}} = {C_1} = {C_2} = {C_3}$
Charge present on a capacitor is equal to the product of capacitance and the voltage flowing across the battery.
$Charg{e_{eq}} = C \times V \\\ C = \dfrac{6}{{11}}C,V = V \\\ Charg{e_{eq}} = \dfrac{6}{{11}}C \times V = \dfrac{{6CV}}{{11}} \\\$
Charge on ${C_1} = {C_2} = {C_3}$ is $\dfrac{{6CV}}{{11}}$
Capacitor 4 is connected in parallel with the remaining 3 capacitors so the charge is different for the 4th capacitor from the remaining three capacitors.
Charge on ${C_4} = 4C \times V = 4CV$
The ratio of the charges on ${C_2}$ and ${C_4}$ is
$\dfrac{{Charg{e_2}}}{{Charg{e_4}}} = \dfrac{{\dfrac{{6CV}}{{11}}}}{{4CV}} \\\ = \dfrac{{6CV}}{{4CV \times 11}} \\\ = \dfrac{6}{{44}} \\\ = \dfrac{3}{{22}} \\\$
The ratio of charges present on 2nd and 4th capacitors is 3:22.
Note:
When the capacitors are connected in series then the reciprocal of the resultant capacitance will be the sum of the reciprocals of all the capacitances and when the capacitors are connected in parallel then the resultant capacitance will be the sum of all the capacitances; whereas when it comes to resistors it is vice versa. When the resistors are connected in parallel then the reciprocal of the resultant resistance will be the sum of the reciprocals of all the resistances and when the resistors are connected in parallel then the resultant resistance will be the sum of all the resistances. Do not confuse the series and parallel configurations among resistors and capacitors.