NEET LR, RC and LCR Circuits MCQs & PYQs

In an inductive-resistive (RL) series circuit, the phase angle \( \phi \) is given by \( \tan \phi = \frac{X_L}{R} \). Since \( \phi = 60^\circ \), \( \tan 60^\circ = \frac{X_L}{R} \implies X_L = \sqrt{3}R \).

The resonant frequency of a series LCR circuit is given by \(f_0 = \frac{1}{2\pi\sqrt{LC}}\). Substituting values: \(f_0 = \frac{1}{2\pi\sqrt{10^{-4} \times 10^{-6}}} = \frac{10^5}{2\pi}\text{ Hz}\).

Inserting an iron rod increases magnetic flux \( \Phi \). By Lenz's law, this induces an opposing \( \text{EMF} \), causing current \( \text{I} \) to decrease during insertion.

Inserting a soft iron core increases solenoid inductance \(L\), thus increasing inductive reactance \(X_L = omega L\). This increases circuit impedance \(Z\), reducing current \(I = V/Z\) and making the bulb dimmer. Both Assertion (A) and Reason (R) are false.

At resonance, \(V_L = V_C\) but they are \(180^\circ\) out of phase. The applied voltage is \(V = IR\). Due to voltage magnification (high \(Q\) factor), \(V_L\) or \(V_C\) can be much greater than \(V\). Reason is true, but it doesn't explain *why* they can be larger than applied voltage, it explains why they cancel out to make \(V=IR\).

Peak voltage across resistor is \(V_R = I_0 R\). Peak source voltage is \(V_0 = I_0 Z\). Since \(Z \ge R\), \(V_R \le V_0\). So A is false. Peak voltage across inductor is \(V_L = I_0 X_L\). At resonance, if \(X_L > R\), then \(V_L\) can be greater than \(V_0\) (voltage magnification). So R is true. Thus, A is false and R is true.

Metal detectors use \(LC\) resonant circuits. When a metal object enters the coil's magnetic field, it changes the coil's inductance, altering the circuit's resonant frequency and triggering detection.

Smaller bandwidth implies a higher quality factor \(Q\) and sharper resonance, leading to better frequency selectivity (easier tuning). Resonant frequency \(\omega_0\) is the *geometric mean* \(sqrt{\omega_1 \omega_2}\,\) not the arithmetic mean of half-power frequencies.

At resonance, the inductive reactance \(X_L\) equals the capacitive reactance \(X_C\), making the total impedance purely resistive \(Z=R\). This results in the current and emf being in phase. Since impedance is minimal \(Z=R\), the current is maximal. Therefore, R is the correct explanation of A.

For frequencies greater than resonant frequency, inductive reactance \(X_L = \omega L\) becomes greater than capacitive reactance \(X_C = 1/(\omega C)\), making the circuit inductive. The reciprocal of reactance is defined as susceptance. Both statements are true, but R does not explain A.