Question

The resistance \[R = 100\;{{k}}\Omega \] and the capacitance \[C = 1\;{{\mu F}}\] are connected in a series with a $12\;{{V}}$ battery. What is the maximum energy stored in the capacitor:
(A) \[72\;{{\mu J}}\]
(B) \[6\;{{\mu J}}\]
(C) \[24\;{{\mu J}}\]
(D) \[18\;{{\mu J}}\]

Answer 1

Hint: In this question we will use the concept of the energy store in the capacitor that is the energy stored in a capacitor is equal to the half of the product of the charge stored and the potential.

Complete step by step answer:
In this question, we are given that a resistance of \[R = 100\;{{k}}\Omega \], let's denote the resistance by \[R\]. And represent the capacitance by \[C\] of the capacitor. So, the capacitance is \[C = 1\;{{\mu F}}\].
Now, we convert the capacitance into Farad which is denoted by the symbol \[\left( {{F}} \right)\].
\[ \Rightarrow C = 1 \times {10^{ - 6\,}}\;{{F}}\]
The capacitance of the capacitor is \[C = 1 \times {10^{ - 6\,}}\;{{F}}\] and which is having a 12 volt battery, let denote the voltage by \[V\] so, \[V = 12\;{{Volt}}\].
 The energy stored in a capacitor is nothing but the electric potential energy and that is related to the voltage of the capacitor and the charge of the capacitor. Let denote the energy stored in the capacitor by the\[U\].
So, the energy stored in capacitor is equal to the half of the product of charge and square of the potential, So, energy stored is,
\[ \Rightarrow U = \dfrac{1}{2}C{V^2}\]
Now we substitute the values in the above expression as,
\[ \Rightarrow U = \dfrac{1}{2} \times 1 \times {10^{ - 6}} \times {12^2}\]
After simplifying the above expression, we get.
\[ \Rightarrow U = \dfrac{1}{2} \times 1 \times {10^{ - 6}} \times 144\]
We solve the expression for $U$ and get
\[\therefore U = 72\;{{\mu J}}\]
Therefore, the energy stored in the capacitor is \[7{{2}}\;{{\mu J}}\] which is having a resistance of\[100\;{{k}}\Omega \] and the capacitance is \[1\;{{\mu F}}\], and they are connected in a series with a $12\;{{V}}$ battery.

Thus, the correct option is (A).

Note: As we know that when the circuit connection is parallel then there will be the same voltage or potential across the circuit. As we know that, if the resistors are in parallel the voltage across each resistor will be the same and if the resistors are in series then the current flowing through each resistor will be the same and the voltage will be different.