A zener diode with 5 V zener voltage is used to regulate an unregulated dc voltage input of 25 V. For a 400 resistor connected in series, the zener current is found to be 4 times load current. The load current (I_L) and

Correct answer: D. I_ L =10 ~mA ; R_ L =500 From the circuit diagram, aligned & 5 i=(20)/(400)=(1)/(20) ~A & i=(1)/(100) ~A=10 ~mA= Load current aligned Also, V_ L =5 ~V R_ L = 5 10 × 10^ -3 =500

JEE Main 20252 AprilMorning Shift

Exam: JEE Main
Year: 2025
Date: 2 April
Shift: Morning Shift
Subject: Physics
Chapter: Semiconductors

JEE Main 2025 · 2 April · Morning Shift · Physics · Semiconductors

A zener diode with 5 V zener voltage is used to regulate an...

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 zener diode with 5 V zener voltage is used to regulate an unregulated dc voltage input of 25 V. For a 400 resistor

Options

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 $\mathrm{I}_{\mathrm{L}}=10 \mathrm{~mA} ; \mathrm{R}_{\mathrm{L}}=500 \Omega$

Correct answer

D. $\mathrm{I}_{\mathrm{L}}=10 \mathrm{~mA} ; \mathrm{R}_{\mathrm{L}}=500 \Omega$

Step-by-step solution

From the circuit diagram,

\begin{aligned}<br/>& 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 }<br/>\end{aligned}

Also,

\mathrm{V}_{\mathrm{L}}=5 \mathrm{~V}<br/>

\therefore

\mathrm{R}_{\mathrm{L}}=\frac{5}{10 \times 10^{-3}} \Omega=500 \Omega

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