MHT CET · Physics · Semiconductors
A zener diode, having breakdown voltage 15 V is used in a voltage regulator circuit as shown. The current through the zener diode is
- A 20 mA
- B 5 mA
- C 10 mA
- D 15 mA
Answer & Solution
Correct Answer
(B) 5 mA
Step-by-step Solution
Detailed explanation
The voltage drop across \(1 \mathrm{k} \Omega\) is 15 V
\(\therefore \quad\) The current through it is \(I=\frac{15 \mathrm{~V}}{1 \times 10^3 \Omega}\)
\(=15 \times 10^{-3} \mathrm{~A}\)
The voltage drop across \(250 \Omega=20-15=5 \mathrm{~V}\)
\(\therefore \quad\) the current through it
\(I=\frac{V}{R}=\frac{5}{250}=20 \times 10^{-3} \mathrm{~A}\)
The current through zener diode is,
\(I_z=\left(20 \times 10^{-3}\right)-\left(15 \times 10^{-3}\right)=5 \mathrm{~mA}\)
\(\therefore \quad\) The current through it is \(I=\frac{15 \mathrm{~V}}{1 \times 10^3 \Omega}\)
\(=15 \times 10^{-3} \mathrm{~A}\)
The voltage drop across \(250 \Omega=20-15=5 \mathrm{~V}\)
\(\therefore \quad\) the current through it
\(I=\frac{V}{R}=\frac{5}{250}=20 \times 10^{-3} \mathrm{~A}\)
The current through zener diode is,
\(I_z=\left(20 \times 10^{-3}\right)-\left(15 \times 10^{-3}\right)=5 \mathrm{~mA}\)
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