Physics MCQs for NEET — Practice Questions with Answers

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.'

Figure 7.8(a) and the accompanying description clearly state: 'Here the current phasor I is $\pi/2$ ahead of the voltage phasor V as they rotate counterclockwise.' This aligns with the mathematical derivation that current leads voltage by $\pi/2$.

According to the provided text, 'We see that the current reaches its maximum value later than the voltage by one-fourth of a period $T/4 = \pi/(2\omega)$'.

The text states, 'So, the average power over a complete cycle is $(P_L) = \text{average of } [- (v_m i_m/2) \sin(2\omega t)]$ = 0, since the average of $\sin (2\omega t)$ over a complete cycle is zero. Thus, the average power supplied to an inductor over one complete cycle is zero'.

The example calculates inductive reactance as $X_L = 2\pi \nu L$. This formula clearly shows that inductive reactance is directly proportional to the frequency ($\nu$) of the AC source.

The text mentions, 'in the case of an inductor, the current lags the voltage by $\pi/2$'. This is equivalent to saying voltage leads current by $\pi/2$.

From Example 7.2: $X_L = 2\pi \nu L = 2 \times 3.14 \times 50 \times 25 \times 10^{-3} \Omega = 7.85 \Omega$.