Question

Three condensers of capacities 3 mF, 6 mF, 12 mF are connected in series with a battery. If the charge on a 12 mF condenser is 24 mC, the P.D. across the battery is
A. 2 V
B. 4 V
C. 8 V
D. 14 V

Answer 1

Hint: In order to answer this question, we will use the idea that each capacitor has the same charge when the same potential difference is applied across capacitors that are wired in series, and that the applied potential is equal to the total of the potential differences on each capacitor.

Formula used:
$V = \dfrac{Q}{C}$

Complete answer:
Let's define the capacitor's capacitance as the relationship between the charge on a plate and the applied potential difference. Charge and capacitance are inversely correlated; that is, when the charge increases, capacitance increases as well.

Let us assume potential difference across $3mF = {V_1}$
Potential difference across $6mF = {V_2}$
Potential difference across $12mF = {V_3}$
As given in question charge on 12mF is 24mC
$ \Rightarrow {V_3} = \dfrac{{24}}{{12}}V = 2V$

Similarly, since in series we have same charge across each we can say that
${V_2} = \dfrac{{24}}{6}V = 4V$ and ${V_1} = \dfrac{{24}}{3}V = 8V$

So total P.D will be $V = {V_1} + {V_2} + {V_3}$
So substituting the value we get $V = 2 + 4 + 8 = 14V$

Therefore, option D is correct.

Note: There are a few things to keep in mind while connecting capacitors in series, such as how the charge moves from one capacitor to the next; it can only flow in one direction; if it does not, then there is no series connection. Another point is that the circuit's battery can only supply power to the capacitor that is electrically connected to the battery directly. Charges on other capacitors are caused by the current charge moving. The only way to create new charges is to redistribute existing ones.