Ncert Exercises & Solutions

In the circuit shown in the figure given below, if the diode forward voltage drop is 0.3 V, the voltage difference between A and B is:

1. 1.3 V

2. 2.3 V

3. 0

4. 0.5 V

(b) Hint: The potential difference between A and B depends on the biasing of the diode.

Step 1: Find the potential difference between A and B.

Consider the fig. (b).

r_{1} = 5 k Ω and r_{2} = 5 k Ω

are resistance in series connection.

Then,

V - 0.3 = [(r_{1} + r_{2}) 10^{3}] \times (0.2 \times 10^{- 3})]

= [(5 + 5) 10^{3}] \times (0.2 \times 10^{- 3})

= 10 \times 10^{3} \times 0.2 \times 10^{- 3} = 2

\Rightarrow V = 2 + 0.3 = 2.3 V