MHT CET · Physics · Semiconductors
In the given circuit, Zener breakdown voltage is 8 V. If power of Zener diode is 1.6 W . The value of \(R\) is
- A \(2 \Omega\)
- B \(4 \Omega\)
- C \(6 \Omega\)
- D \(10 \Omega\)
Answer & Solution
Correct Answer
(D) \(10 \Omega\)
Step-by-step Solution
Detailed explanation
\(\begin{array}{ll}
& \mathrm{V}_{\mathrm{Z}}=8 \mathrm{~V} \\
& \mathrm{P}=\mathrm{V}_{\mathrm{Z}} \mathrm{I} \\
\therefore \quad & 1.6=8 \mathrm{I} \\
\therefore \quad & \mathrm{I}=0.2 \mathrm{~A}
\end{array}\)
Voltage drop across Zener is 8 V . Hence the voltage drop across R will be \(10-8=2 \mathrm{~V}\).
\(\mathrm{R}=\frac{\mathrm{V}}{\mathrm{I}}=\frac{2}{0.2}=10 \Omega\)
& \mathrm{V}_{\mathrm{Z}}=8 \mathrm{~V} \\
& \mathrm{P}=\mathrm{V}_{\mathrm{Z}} \mathrm{I} \\
\therefore \quad & 1.6=8 \mathrm{I} \\
\therefore \quad & \mathrm{I}=0.2 \mathrm{~A}
\end{array}\)
Voltage drop across Zener is 8 V . Hence the voltage drop across R will be \(10-8=2 \mathrm{~V}\).
\(\mathrm{R}=\frac{\mathrm{V}}{\mathrm{I}}=\frac{2}{0.2}=10 \Omega\)
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