Physics 2 Marks Questions

For a half-wave rectifier, the output frequency is equal to the input frequency.

$\therefore$ Output frequency = 50 Hz

For a full-wave rectifier, the output frequency is twice the input frequency.

$\therefore$ Output frequency = 2 × 50 = 100 Hz

(If X and Y are correctly identified but the identification of biasing is incorrect or  not given at all, award.)

$(\text{X}\rightarrow\text{p} \text{Type})$

$(\text{Y}\rightarrow\text{n} \text{Type})$

1. Reverse biased.
2. Graph:

1. Intrinsic semiconductor does not contain any impurity while p-type semiconductor is doped with a trivalent impurity.

Alternate Answer

For an intrinsic semiconductor, $\text{n}_{e} = \text{n}_{h}$ while for a p-type semiconductor $\text{n}_{h}>>\text{n}_{e}$or any other difference.

Alternate Answer

Shows the distinction by band diagram.

(HOTS)

2. p-type semiconductor is made by doping the neutral intrinsic semicondustor with trivalent impurity atoms.

Current in the circuit = Drift current

(Since, the diode is reverse biased $=20\mu\text{A})$

Voltage across the diode $=5-\big(20\times20\times10^{-6}\big)$

$=5-\big(4\times10^{-4}\big)=5\text{V}.$

1. The doped semiconductor is n-type.

2. $\text{n}_\text{e}\text{n}_\text{h}=\text{n}^2_\text{i}$

$\Rightarrow\text{n}_\text{h}=\frac{\text{n}^2_\text{i}}{\text{n}_\text{e}}$

$=\frac{(6\times10^8)^2}{9\times10^{12}}=4\times10^4\text{per m}^3$

1. The amplitude modulated waveform is given by superimposition of modulating signal on carrier wave and is shown below:

2. We have,

A_m = 1V, A_c = 2V.

Therefore, modulation index is given by,

$\mu=\frac{\text{A}_{\text{m}}}{\text{V}_\text{C}}=\frac{1}{2}=0.5$