Question

In a LCR circuit find the energy stored in the inductor at resonance. If voltage of source is 10V and resistance is \[10\Omega \] and inductance is 1H.
A. 0.5J
B. 2J
C. 4J
D. 10J

Answer 1

Hint: This is a basic problem of electronics. Making a circuit diagram will make the question easier to understand and solve. We must know the impedance of the circuit, given by, $ Z=\sqrt{{{R}^{2}}+{{({{\chi }_{L}}-{{\chi }_{C}})}^{2}}} $. Know the resonant frequency conditions for series and parallel LCR circuit conditions, which is given by $ \omega =\dfrac{1}{\sqrt{LC}} $, where $ \omega $ is the resonance angular frequency. We must also remember as to how the electrical energy gets stored in an inductor. The formula for the energy stored in an inductor is given by, $ {{E}_{L}}=\dfrac{1}{2}L{{i}^{2}} $.

Complete step-by-step answer:
We will first start off by making the circuit diagram for the problem.

Given in the question is, the value of source voltage is 10V. Hence the V=10V.
In a LCR series oscillatory circuit, the resonance condition refers to the condition where the resistance of the circuit is minimum.
The resistance of the circuit $ (Z) $ is given by, $ Z=\sqrt{{{R}^{2}}+{{({{\chi }_{L}}-{{\chi }_{C}})}^{2}}} $
$ {{\chi }_{L}} $ is the resistance of the inductor, given by, $ {{\chi }_{L}}=\omega L. $
$ {{\chi }_{C}} $ is the resistance of the capacitor, given by, $ {{\chi }_{C}}=\dfrac{1}{\omega C}. $
For the resonance condition, $ {{\chi }_{L}}={{\chi }_{C}}\Rightarrow \omega L=\dfrac{1}{\omega C}\Rightarrow {{\omega }^{2}}=\dfrac{1}{LC}\Rightarrow \omega =\dfrac{1}{\sqrt{LC}} $
This value of $ \omega $ is known as the resonant frequency of the series LCR circuit.
Under the resonance condition $ Z=\sqrt{{{R}^{2}}}\Rightarrow Z=R. $
Therefore, the amount current (i) flowing in the circuit is, $ i=\dfrac{V}{Z}\Rightarrow i=\dfrac{V}{R} $
 $ i=\dfrac{10V}{10\Omega }\Rightarrow i=1A. $
The energy stored in an inductor is given by, $ {{E}_{L}}=\dfrac{L{{i}^{2}}}{2} $
Substituting the values of L=1H and i=1A, we get $ {{E}_{L}}=\dfrac{1{{(1)}^{2}}}{2}\Rightarrow {{E}_{L}}=0.5J $
Hence, the energy stored by the inductor during the resonance condition is 0.5J.

Note: Whenever in a LCR oscillatory circuit problem, the LCR circuit is in series or parallel isn’t mentioned, the default condition will be considered as series condition.
Here the source will also be an ac source, since it is a oscillatory circuit and for resonance to occur, the $ \omega$ value of the source will be equal to the resonant frequency of $ \omega =\dfrac{1}{\sqrt{LC}}. $