Ncert Exercises & Solutions

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

1.	\(1.3~\text V\)	2.	\(2.3~\text V\)
3.	\(0\)	4.	\(0.5~\text V\)

(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

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