If the reverse bias in a junction diode is changed from \(5\) V to \(15\) V then the value of current changes from \(38\) \(\mu \text{A}\) to \(88\) \(\mu \text{A}.\) The resistance of junction diode will be:
1. \(4\times10^{5}\) 
2. \(3\times10^{5}\)
3. \(2\times10^{5}\)
4. \(10^{6}\)

The energy band diagrams for semiconductor samples of silicon are as shown. We can assert that:

In a given circuit as shown the two input waveforms A and B are applied simultaneously. The resultant waveform Y is:

1.		2.
3.		4.

In the circuit shown in the figure, the maximum output voltage V_o is:

1. 0 V

2. 5 V

3. 10 V

4. $\frac{5}{\sqrt{2}} V$

In the circuit given below, if V(t) is the sinusoidal voltage source, then the voltage drop V_AB(t) across the resistance R:

A 2 V battery is connected across the points A and B as shown in the figure given below. Assuming that the resistance of each diode is zero in forward bias and infinity in reverse bias, the current supplied by the battery when its positive terminal is connected to A is:

The logic behind the 'NOR' gate is that it gives:

Zener breakdown will occur if:
1. impurity level is low.
2. impurity level is high.
3. impurity is less on the n-side.
4. impurity is less on the p-side.

An LED is constructed from a \(\mathrm{p\text{-}n}\) junction diode using \(\mathrm{GaAsP}\). The energy gap is \(1.9~\text{eV}\). The wavelength of the light emitted will be equal to:
1. \(10.4 \times 10^{-26} \text{m}\)
2. \(654~ \text{nm}\)
3. \(654~ \text{m}\)
4. \(654\times 10^{-11}~\text{m}\)

A potential barrier of 0.50 V exists across a p-n junction. If the depletion region is $5.0\times {10}^{-7}$ m wide, the intensity of the electric field in this region is: