NEET Alternating Current MCQs & PYQs

Inserting mica increases capacitance ($C_b > C_a$), so reactance decreases ($X_{Cb} < X_{Ca}$). Total impedance $Z = \sqrt{R^2 + X_C^2}$ decreases, causing current to increase ($i_b > i_a$). Since $V_C = \frac{V}{\sqrt{1 + (R/X_C)^2}}$, as $X_C$ decreases, $V_C$ decreases, hence $V_a > V_b$.

In a series capacitive-resistive (C-R) circuit, charge grows or decays according to $q(t) = q_0 e^{-t/\tau}$. The time constant $\tau$ is defined as the product of capacitance and resistance. Therefore, the time constant is $\tau = CR$.

For the AC source $V = 200 \sin(100t)$, angular frequency $\omega = 100 rad/s$.
Source frequency $\nu = \frac{\omega}{2\pi} = \frac{100}{2\pi} = \frac{50}{\pi} Hz$.
Resonant frequency $\nu_0 = \frac{1}{2\pi\sqrt{LC}} = \frac{1}{2\pi\sqrt{10 \times 10 \times 10^{-6}}} = \frac{1}{2\pi\times 10^{-2}} = \frac{50}{\pi} Hz$.
Thus, $\nu_0 = \nu = \frac{50}{\pi} Hz$.

When L is removed (RC circuit): $\tan(\pi/3) = \frac{X_C}{R}$.
When C is removed (RL circuit): $\tan(\pi/3) = \frac{X_L}{R}$.
This implies $X_C = X_L$, meaning the circuit is in resonance and total impedance $Z = R$.
Power factor is $\cos\phi = \frac{R}{Z} = \frac{R}{R} = 1.0$.

For better tuning of an L-C-R circuit, the quality factor $Q = \frac{1}{R}\sqrt{\frac{L}{C}}$ must be as high as possible.
For option a, $Q = \frac{1}{15}\sqrt{\frac{3.5}{30 \times 10^{-6}}} \approx 22.7$.
For options b, c, and d, $Q$ is approximately $7.3$, $10.4$, and $9.4$ respectively.
Therefore, the combination in option a has the highest Q-factor.

Phase difference when L is removed is $\tan(\pi/3) = \frac{X_C}{R}$, and when C is removed is $\tan(\pi/3) = \frac{X_L}{R}$.
Equating these gives $X_C = X_L$, which is the condition for resonance.
At resonance, the net reactance is zero, and the circuit is purely resistive ($Z = R$).
The power factor is $\cos\phi = \frac{R}{Z} = \frac{R}{R} = 1$.

Current in a series LCR circuit is maximum at resonance, where $X_L = X_C$.
This gives $\omega L = \frac{1}{\omega C} \implies L = \frac{1}{\omega^2 C}$.
Substituting the given values: $L = \frac{1}{(1000)^2 \times 10 \times 10^{-6}} = \frac{1}{10^6 \times 10^{-5}} = \frac{1}{10} H$.
Therefore, $L = 0.1 H = 100 mH$.

The quality factor (Q-factor) of a resonant series LCR circuit measures the sharpness of resonance.
It is defined as the ratio of voltage across the inductor (or capacitor) to the applied voltage at resonance.
$Q = \frac{V_L}{V_R} = \frac{I \cdot X_L}{I \cdot R} = \frac{X_L}{R}$.
Substituting $X_L = \omega L$, we get the quality factor as $\frac{\omega L}{R}$.

At resonance in a series L-C-R circuit, the inductive and capacitive reactances cancel out ($X_L = X_C$).
The total impedance of the circuit is purely resistive ($Z = R$).
Since the circuit is purely resistive, the current and applied voltage are in phase.
Therefore, the phase difference between the applied voltage and current is zero.

By conservation of energy, the decrease in electrical energy of the capacitor equals the magnetic energy gained by the inductor.
$\frac{1}{2} C V_1^2 - \frac{1}{2} C V_2^2 = \frac{1}{2} L I^2$.
$C(V_1^2 - V_2^2) = L I^2 \implies I^2 = \frac{C(V_1^2 - V_2^2)}{L}$.
Therefore, the current is $I = \left[ \frac{C(V_1^2 - V_2^2)}{L} \right]^{\frac{1}{2}}$.