2 Marks Questions

Question 12 Marks

What are the readings of the ammeters $A_1$ and $A_2$ shown in figure. Neglect the resistance of the meters.

Answer

The diode is reverse biased.
Hence the resistance is infinite.
So, current through $A_1$ is zero.

For $A_2,$ current $=\frac{2}{10}=0.2\ \text{Amp}.$

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Question 22 Marks

Each of the resistances shown in figure. has a value of $20\Omega.$ Find the equivalent resistance between A and B. Does it depend on whether the point A or B is at higher potential?

Answer

From the figure.

According to wheat stone bridge principle, there is no current through the diode.
Hence net resistance of the circuit is $\frac{40}{2}=20\Omega.$

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Question 32 Marks

Calculate the current through the circuit and the potential difference across the diode shown in figure. The drift current for the diode is $20\mu\text{A}.$

Answer

Current in the circuit = Drift current
(Since, the diode is reverse biased $=20\mu\text{A})$
Voltage across the diode $=5-\big(20\times20\times10^{-6}\big)$
$=5-\big(4\times10^{-4}\big)=5\text{V}.$

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Question 42 Marks

The drift current in a p-n junction is $20.0\mu\text{A}.$ Estimate the number of electrons crossing a cross-section per second in the depletion region.

Answer

Drift current $=20\mu\text{A}=20\times10^{-6}\text{A}$
Both holes and electrons are moving
So, no.of electrons $=\frac{20\times10^{-6}}{2\times1.6\times10^{-19}}=6.25\times10^{13}.$

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Question 52 Marks

Consider an amplifier circuit using a transistdf. The output power is several times greater than the input power. Where does the extra power come from?

Answer

The amplifier takes this energy from the power supply. Amplifiers take the energy from the power supply and control the output to match with the input signal but with greater amplitude.

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Question 62 Marks

Find the maximum wavelength of electromagnetic radiation which can create a hole-electron pair in germanium. The band gap in germanium is 0.65eV.

Answer

Given,
Band gap of germanium, E = 0.65eV
Wavelength of the incident radiation, $\lambda=?$
For the electron‒hole pair creation, the energy of the incident radiation should be at least equal to the band gap of the material.
So,
$\text{E}=\frac{\text{hc}}{\lambda}$
$\Rightarrow\lambda=\frac{\text{hc}}{\text{E}}=\frac{1242\text{eV}-\text{nm}}{0.65\text{eV}}$
$\Rightarrow\lambda=1910.7\times10^{-9}\text{m}$
$\Rightarrow\lambda=1.9\times10^{-5}\text{m}$

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Question 72 Marks

Suppose the energy liberated in the recombination of a hole-electron pair is converted into electromagnetic radiation. If the maximum wavelength emitted is 820nm, what is the band gap?

Answer

Given,
Wavelength, $\lambda = 820\text{nm}$
The minimum energy released in the recombination of a conduction band electron with a valence band hole is equal to the band gap of the material.
Band gap, $\text{E}=\frac{\text{hc}}{\lambda}$
$\Rightarrow\text{E}=\frac{1240}{820}\frac{\text{eV}-\text{nm}}{\text{nm}}$
$\Rightarrow\text{E}=1.5\text{eV}$

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Question 82 Marks

What is the resistance of an intrinsic semiconductor at 0K?

Answer

At 0K, the valence band is full and the conduction band is empty. As no electron is available for conduction in an intrinsic semiconductor, the intrinsic semiconductor at 0K acts as an insulator and hence offers infinite resistance.

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