JEE MAIN - Physics (2025 - 2nd April Morning Shift - No. 18)
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 $\left(R_L\right)$ are :
$\mathrm{I}_{\mathrm{L}}=0.02 \mathrm{~mA} ; \mathrm{R}_{\mathrm{L}}=250 \Omega$
$\mathrm{I}_{\mathrm{L}}=10 \mathrm{~A} ; \mathrm{R}_{\mathrm{L}}=0.5 \Omega$
$\mathrm{I}_{\mathrm{L}}=10 \mathrm{~mA} ; \mathrm{R}_{\mathrm{L}}=500 \Omega$
$\mathrm{I}_{\mathrm{L}}=20 \mathrm{~mA} ; \mathrm{R}_{\mathrm{L}}=250 \Omega$
Explanation
_2nd_April_Morning_Shift_en_18_1.png)
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, $\mathrm{V}_{\mathrm{L}}=5 \mathrm{~V}$
$$\therefore \mathrm{R}_{\mathrm{L}}=\frac{5}{10 \times 10^{-3}} \Omega=500 \Omega$$
Comments (0)
