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 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:

For a diode connected in parallel with a resistor, Choose the correct graph between \((I)\) and \((V)\) for the given circuit:

1.		2.
3.		4.

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:

\({C}\) and \({Si}\) both have the same lattice structure, having \(4\) bonding electrons in each. However, \(C\) is an insulator whereas \(Si\) is an intrinsic semiconductor. This is because:

The current through an ideal p-n junction diode shown in the circuit will be:

1.  5 A

2.  0.2 A 

3.  0.6 A

4.  Zero