Physics MCQs for NEET — Practice Questions with Answers

The text clearly states: 'This property is used to rectify alternating voltages and the circuit used for this purpose is called a rectifier.' Rectification means converting AC to DC, but specifically, it's a pulsating DC before filtering, which is what the diode achieves.

For a half-wave rectifier, output occurs during one half-cycle for every complete input cycle. So, if the input completes 'f' cycles per second, the output also creates 'f' pulses per second, meaning the fundamental ripple frequency is the same as the input frequency, $f$.

The NCERT text states: '(The reverse breakdown voltage of the diode must be sufficiently higher than the peak ac voltage at the secondary of the transformer to protect the diode from reverse breakdown.)' This implies $V_{BR} > V_{peak}$.

The text states: 'The forward bias resistance is low as compared to the reverse bias resistance. This property is used for rectification of ac voltages as discussed in the next section.'

The context mentions: 'This voltage is called the threshold voltage or cut-in voltage (~0.2V for germanium diode and ~0.7 V for silicon diode).'

The text defines dynamic resistance as: 'For diodes, we define a quantity called dynamic resistance as the ratio of small change in voltage $\Delta V$ to a small change in current $\Delta I$: $r_d = \Delta V / \Delta I$ (14.6)'.

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.'.