NEET Alternating Current MCQs & PYQs

The root-mean-square (rms) voltage for a sinusoidal alternating voltage is given by $V_{rms} = \frac{V_0}{\sqrt{2}}$. Rearranging for peak voltage $V_0$, we get $V_0 = \sqrt{2} V_{rms}$. Hence, the peak voltage is $\sqrt{2}$ times the rms value of the ac source.

For an alternating current $I = I_0 \sin \omega t$, the rms current is defined as $I_{rms} = \sqrt{\frac{1}{T} \int_0^T I^2 dt}$. Evaluating this integral over one complete cycle gives $I_{rms} = \frac{I_0}{\sqrt{2}}$. Therefore, the relation between rms and peak current is $I_{rms} = I_0/\sqrt{2}$.

The displacement current between the capacitor plates is equal to the conduction current, $I_d = \frac{dq}{dt}$. Since charge on the capacitor is $q = C V = C V_0 \sin \omega t$, we differentiate with respect to time. $I_d = \frac{d}{dt}(C V_0 \sin \omega t) = V_0 \omega C \cos \omega t$.

Capacitive reactance is $X_C = \frac{1}{\omega C} = \frac{1}{2\pi f C}$. RMS current is given by $I_{rms} = \frac{V_{rms}}{X_C} = V_{rms} \times 2\pi f C$. Substituting the values: $I_{rms} = 200 \times 2\pi \times 50 \times 40 \times 10^{-6} = 0.8\pi \approx 2.51 A \approx 2.5 A$.

For an ideal capacitor, the phase difference between current and voltage is $\phi = 90^\circ$ (current leads voltage). The average power consumed over a full cycle is $P_{avg} = V_{rms} I_{rms} \cos(90^\circ) = 0$. Since average power is zero, the capacitor does not consume any energy from the voltage source over a complete cycle.

Peak voltage is $E_0 = 200\sqrt{2} V$, so $E_{rms} = \frac{E_0}{\sqrt{2}} = 200 V$. Capacitive reactance is $X_C = \frac{1}{\omega C} = \frac{1}{100 \times 10^{-6}} = 10^4 \Omega$. The rms current is $I_{rms} = \frac{E_{rms}}{X_C} = \frac{200}{10^4} = 20 \times 10^{-3} A = 20 mA$.

Capacitive reactance is $X = \frac{1}{2\pi f C}$. When capacitance becomes $2C$ and frequency becomes $2f$, the new reactance is $X' = \frac{1}{2\pi(2f)(2C)} = \frac{1}{4(2\pi f C)} = \frac{X}{4}$.

Initially, power dissipated in resistance is $P = \frac{V_{rms}^2}{R}$. When inductance is connected in series, current is $I_{rms} = \frac{V_{rms}}{Z}$ and power factor is $\cos\phi = \frac{R}{Z}$. New power drawn is $P' = V_{rms} I_{rms} \cos\phi = V_{rms} \left(\frac{V_{rms}}{Z}\right) \left(\frac{R}{Z}\right) = \frac{V_{rms}^2 R}{Z^2} = P \left(\frac{R}{Z}\right)^2$.

Inserting an iron rod into the coil increases its self-inductance $L$. This increases the inductive reactance $X_L = \omega L$ and hence total circuit impedance $Z = \sqrt{R^2 + X_L^2}$. With increased impedance, the circuit current decreases, leading to reduced brightness of the bulb.

Inductive reactance is $X_L = 2\pi f L \propto f$. At $100 Hz$, $X_L' = 20 \times \frac{100}{50} = 40 \Omega$. Circuit impedance is $Z = \sqrt{R^2 + X_L'^2} = \sqrt{30^2 + 40^2} = 50 \Omega$. Current in the coil is $I = \frac{V}{Z} = \frac{200}{50} = 4.0 A$.