Question

A pn junction diode is connected to a battery of e.m.f 5.5V and external

resistance 5.1 kΩ. The barrier potential in the diode is 0.4 V. The current in

the circuit is: 1

a. 1A b. 1mA c.2mA d.0.08mA​

Answer 1

ANSWER

b) 2 mA

EXPLAINATION

emf = 5.5V barrier potential = 0.4 V

potential difference= emf - barrier potential

= 5.5 - 0.4

= 5.1 V

V = IR

5.1= I × 5.1 × 1000

I = 5.1 /5.1 × 1000

I = 0.001 A

I = 1 mA

Answer 2

Answer:

The correct answer is  b. 1mA.

Explanation:

• p-n junction: A p-n junction is created where a p-type semiconductor is joined to an n-type semiconductor.
• A p-type semiconductor has extra holes and an $\mathrm{n}$ -type semiconductor has excess free electrons.
• Due to this difference in concentration of charge carriers, when the junction is formed, the electrons diffuse from the n side to the p side, and holes diffuse from the p side to the n side.

Resistance, $R=5.1 \mathrm{k} \Omega=5.1 \times 10^{3} \Omega$

From the provided circuit diagram, it is inferred that the p-n junction exist forward-biased.

Due to forward biasing, the barrier potential generated exists opposite to the external voltage source.

Thus, the net voltage drop across the resistor,

$=5.5 \mathrm{~V}-0.5 \mathrm{~V}$

$=5 \mathrm{~V}$

Using Ohm's law,

$&\Rightarrow V=I R$

$&\Rightarrow V=I \times 5.1 \times 10^{3}$

$&\Rightarrow I=\frac{5}{5.1 \times 10^{3}}=0.98 \times 10^{-3} A$

$\approx 1 m A$

Hence, The correct answer is  b. 1mA.

The circuit is given below,

#SPJ2