JEE Mains · Physics · STD 12 - 14. Semicondutor electronics
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 :
- A \(\mathrm{I}_{\mathrm{L}}=20 \mathrm{~mA} ; \mathrm{R}_{\mathrm{L}}=250 \Omega\)
- B \(\mathrm{I}_{\mathrm{L}}=10 \mathrm{~A} ; \mathrm{R}_{\mathrm{L}}=0.5 \Omega\)
- C \(\mathrm{I}_{\mathrm{L}}=0.02 \mathrm{~mA} ; \mathrm{R}_{\mathrm{L}}=250 \Omega\)
- D \(\mathrm{I}_{\mathrm{L}}=10 \mathrm{~mA} ; \mathrm{R}_{\mathrm{L}}=500 \Omega\)
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, \(\mathrm{V}_{\mathrm{L}}=5 \mathrm{~V}\)…
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