The circuit shown below is equivalent to:

    

1.  OR gate

2.  NOR gate

3.  AND gate

4.  NAND gate

An LED is constructed from a p-n junction diode using GaAsP. The energy gap is 1.9 eV. The wavelength of the light emitted will be equal to:

1. 10.4 $\times {10}^{-26}m$

2. 654 nm

3. 654 m

4. 654 $\times {10}^{-11}m$

The circuit diagram shown here corresponds to the logic gate:

1. NOR

2. AND

3. OR

4. NAND

The battery is shown in the figure, current drawn from the battery is:

1.  1 A

2.  2 A

3.  2.5 A

4.  4 A

The input signal is given to a CE amplifier having a voltage gain of 150 is ${\mathrm{V}}_{\mathrm{i}}=2\cos \left[15\mathrm{t}+\frac{{\displaystyle \mathrm{II}}}{{\displaystyle 3}}\right]$. The corresponding output signal will be:

1. $30 \cos \left[15t+\frac{\mathrm{\pi }}{3}\right]$

2. $75 \cos \left[15t+\frac{2\mathrm{\pi }}{3}\right]$

3. $2 \cos \left[15t+\frac{5\mathrm{\pi }}{3}\right]$

4. $300 \cos \left[15t+\frac{4\mathrm{\pi }}{3}\right]$

In the given figure, a diode D is connected to an external resistance R=100$\Omega$ and an e.m.f. of 3.5 V. If the barrier potential developed across the diode is 0.5 V, the current in the circuit will be:

1. 30 mA

2. 40 mA

3. 20 mA

4. 35 mA

If in a p-n junction diode, a square input signal of 10 V is applied as shown

Then the output signal across ${\mathrm{R}}_{\mathrm{L}}$ will be

1. 

2. 

3. 

4. 

A semiconductor device is connected in a series circuit with a battery and a resistance. A current is found to pass through the circuit. If the polarity of the battery is reversed, the current drops almost to zero. The device may be:

1. a p-type semiconductor

2. an n-type semiconductor

3. a p-n junction

4. an intrinsic semiconductor

The reverse bias in a junction diode is changed from 8 V to 13 V, then the value of the current changes from $40 \mathrm{\mu } \mathrm{A} \mathrm{to} 60 \mathrm{\mu } \mathrm{A}$. The resistance of the junction diode will be: 

1. $2\times {10}^5 \Omega$

2. $2.5\times {10}^5 \Omega$

3. $3\times {10}^5 \Omega$

4. $4\times {10}^5 \Omega$