Alternating Current MCQs for NEET — Physics Questions with Answers

At resonance, the peak current $i_m = v_m/R$. Given $v_m = 100 \text{ V}$ and $R = 100 \Omega$. So, $i_m = 100 \text{ V} / 100 \Omega = 1 \text{ A}$. The provided text also specifically mentions this for the given values: 'Since $i_m = v_m/R$ at resonance, the current amplitude for case (i) is twice to that for case (ii).'

At resonance, $X_L = X_C$, which means $(X_L - X_C) = 0$. From $\tan \phi = (X_L - X_C)/R$, we get $\tan \phi = 0$, so $\phi = 0$. This implies current and voltage are in phase.

From the provided context, specifically Equation 7.16, for an AC voltage applied to a capacitor, the current is given by $i = i_m \sin(\omega t + \pi/2)$. This indicates that the current leads the voltage by a phase of $\pi/2$ (or 90 degrees).

As per the context, 'It (capacitive reactance) is inversely proportional to the frequency and the capacitance.' (page 185) and Equation 7.17 states $X_C = 1/\omega C$. Therefore, it is inversely proportional to capacitance, not directly.

The context states, 'A comparison of Eq. (7.16) with the equation of source voltage, Eq. (7.1) shows that the current is $\pi/2$ ahead of voltage.' This means the current leads the voltage by $\pi/2$ or 90 degrees.

The context clearly states, 'So, as in the case of an inductor, the average power $P_C = \frac{1}{2} i_m v_m \langle \sin(2\omega t) \rangle = 0$ since $\langle \sin(2\omega t) \rangle = 0$ over a complete cycle.' (Equation 7.19 and accompanying text). Also, point 4 under summary confirms 'the average power supplied to a capacitor over one complete cycle is zero.'.

The text explains, 'When a capacitor is connected to a voltage source in a dc circuit, current will flow for the short time required to charge the capacitor. As charge accumulates on the capacitor plates, the voltage across them increases, opposing the current. ... When the capacitor is fully charged, the current in the circuit falls to zero.' (page 184).

Example 7.3 discusses this scenario: 'Reducing C will increase reactance and the lamp will shine less brightly than before.' This is because $X_C = 1/\omega C$. If C decreases, $X_C$ increases, which reduces the current ($i_m = v_m / X_C$), leading to less brightness.

The context states, 'where the amplitude of the oscillating current is $i_m = \omega C v_m$.' (page 184).

Point 7 under 'POINTS TO PONDER' (page 199) clarifies: 'There are no power losses associated with pure capacitances and pure inductances in an ac circuit. The only element that dissipates energy in an ac circuit is the resistive element.'