JEE Mains · Physics · STD 12 - 14. Semicondutor electronics
The circuit shown below is working as a \(8\; \mathrm{V}\;dc\) regulated voltage source. When \(12 \;\mathrm{V}\) is used as input, the power dissipated (in \(\mathrm{mW}\) ) in each diode is (considering both zener diodes are identical)
- A \(20\)
- B \(8\)
- C \(24\)
- D \(40\)
Answer & Solution
Correct Answer
(D) \(40\)
Step-by-step Solution
Detailed explanation
Current in circuit \(=\frac{4}{400}=\frac{1}{100} \mathrm{A}\) So power dissipited in each diode \(=\) \(\mathrm{VI}\) \(=4 \times \frac{1}{100} \;\mathrm{W}\) \(=40 \times 10^{-3}\; \mathrm{mW}\)
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