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

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 :
  1. $\mathrm{I}_{\mathrm{L}}=20 \mathrm{~mA} ; \mathrm{R}_{\mathrm{L}}=250 \Omega$
  2. $\mathrm{I}_{\mathrm{L}}=10 \mathrm{~A} ; \mathrm{R}_{\mathrm{L}}=0.5 \Omega$
  3. $\mathrm{I}_{\mathrm{L}}=0.02 \mathrm{~mA} ; \mathrm{R}_{\mathrm{L}}=250 \Omega$
  4. $\mathrm{I}_{\mathrm{L}}=10 \mathrm{~mA} ; \mathrm{R}_{\mathrm{L}}=500 \Omega$

Solution


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, $\mathrm{V}_{\mathrm{L}}=5 \mathrm{~V}$
$\therefore \mathrm{R}_{\mathrm{L}}=\frac{5}{10 \times 10^{-3}} \Omega=500 \Omega$

Asked in: JEE Main 2025 (02 Apr Shift 1)

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