Question

Three identical capacitors \[{C_1},{C_2},{C_3}\] of capacitance \[6\mu F\] each are connected to a \[12\] V battery as shown. Find the charge on each capacitor.

Answer 1

Hint: To solve this problem we have to know what capacitor is and what is series connection and parallel connection. So, we know that a capacitor is a device that can store electrical energy in an electric field. It contains two terminals. It can store two energies one is positive and another one is negative energy on two separate plates which is separated by an insulator. We know that the capacitance value of a capacitor is measured by Farads.

Complete step by step answer:
We know that from the figure that capacitor \[{C_1}\] and capacitor \[{C_2}\] are in series. And they potential difference of \[12\] V at \[{C_3}\].Now, according to the formula of effective capacitance of \[{C_1},{C_2},{C_3}\] is,
\[{C_{12}} = \dfrac{{{C_1} \times {C_2}}}{{{C_1} + {C_2}}} \\
\Rightarrow{C_{12}} = \dfrac{{6 \times 6}}{{6 + 6}} \\
\Rightarrow{C_{12}} = 3\mu F\]
We know the two capacitors \[{C_1},{C_2}\]are in series so we can say, the charge of each capacitor is the same. So,
\[{q_1} = {q_2} = {C_{12}}V = (3\mu F)(12V) = 36\mu C\]
Now, we have to calculate the charge on \[{C_3}\] which is ,
\[{q_3} = {C_3}V = (6\mu F)(12V) = 72\mu C\]

So the right answer is for \[{C_1}\] and \[{C_2}\] the charge is \[36\mu C\]. And for \[{C_3}\] the charge is \[72\mu C\].

Note:We can get confused by the series connection and the parallel connection of capacitors. We have to know that, A number of capacitors are said to be in series if they are joined end to end and each condenser gets equal charge. Now, we can say when the number of condensers is said to be in a parallel connection when they are connected between two common points and their potential difference across each condenser is equal.