JEE Mains · Physics · STD 12 - Semicondutor electronics
- JEE Mains 2025
- Easy
- MCQ
A zener diode with 5 V zener voltage is used to regulate an unregulated dc voltage input of 25 V. For a \(400 \Omega\) resistor connected in series, the zener current is found to be 4 times load current. The load current \(\left(I_L\right)\) and load resistance \(\left(R_{\mathrm{L}}\right)\) are :
Pick an option to check your answer.
Answer & Solution
Correct Answer
(D) \(\mathrm{I}_{\mathrm{L}}=10 \mathrm{~mA} ; \mathrm{R}_{\mathrm{L}}=500 \Omega\)
Step-by-step Solution
Detailed explanation
From the circuit diagram,
\(\begin{aligned}
& 5 \mathrm{i}=\frac{20}{400}=\frac{1}{20} \mathrm{~A} \\
& \therefore \mathrm{i}=\frac{1}{100} \mathrm{~A}=10 \mathrm{~mA}=\text { Load current }
\end{aligned}\)
Also, \(\mathrm{V}_{\mathrm{L}}=5 \mathrm{~V}\)
\(\therefore \mathrm{R}_{\mathrm{L}}=\frac{5}{10 \times 10^{-3}} \Omega=500 \Omega\)
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