Question

In a series LCR circuit, resonance occurs at $105\;Hz$. At that time, the potential difference across the 100 resistance is $40\;V$, while the potential difference across the pure inductor is $30\;V$. The inductance $L$ of the inductor is equal to

& A.2.0\times {{10}^{-4}} \\\ & B.3.0\times {{10}^{-4}} \\\ & C.1.2\times {{10}^{-4}} \\\ & D.2.4\times {{10}^{-4}} \\\ \end{aligned}$$

Answer 1

We know that the source of an AC circuit is sinusoidal. Then there exists a phase difference between the voltage and the current in the circuit. If the phase difference between the current and voltage is zero, then both are said to be in phase, and if the phase difference is not equal to zero, then both are said to be out of phase.
Formula: $V_{L}=i\times X_{L}$

Complete answer:
When the natural frequency of the given system matches the applied frequency, then there is an increase in amplitude of the system, then the system is said to experience resonance.
Similarly, when an AC current is passed through a LCR circuit, then the circuit experiences resonance, when the reactance due to the capacitance and the inductance are equal and opposite to each other, thus cancelling each other.
Then, we can say $X_{L}=X_{C}$, where $X_{L}, X_{C}$ is the reactance due to the inductance and the capacitance respectively.
$2\pi fL=\dfrac{1}{2\pi fC}$
$\implies f^{2}=\dfrac{1}{4\pi^{2} LC}$
$\implies f=\sqrt{\dfrac{1}{{4\pi^{2} LC}}}$
$\implies f=\dfrac{1}{2\pi\sqrt{LC}}$
$\implies \omega=\dfrac{1}{\sqrt{LC}}$
Then the maximum angular frequency given as $\omega=\dfrac{1}{\sqrt{LC}}$, where $L$ is the inductor and $C$ is the capacitance.
Here it is given that, the potential difference across the 100 resistance is $40\;V$
Then we can say that $i=\dfrac{V}{R}=\dfrac{100}{40}=2.5A$
Also given that the potential difference across the pure inductor is $30\;V$.
Then we can say that, $V_{L}=i\times X_{L}$
$\implies X_{L}=\dfrac{V_{L}}{i}$
$\implies \omega L=\dfrac{V_{L}}{i}$
$\implies L=\dfrac{30}{2.5\times 2\times \pi\times105}=1.2\times 10^{-4}Hz$

Therefor, the answer is option $C.1.2\times {{10}^{-4}}$.

Note:
Clearly, resonance is observed when RLC is connected in series. Not when the following connections are seen RLC is in parallel or in RL, RC or LC circuit. In RLC connection the impedance is said to be purely real, as only R is active in the circuit and there is no impedance in the circuit. For frequency lesser than the resonant frequency, the impedance is capacitive in nature and for frequency higher than the resonant frequency; the impedance is inductive in nature.