JEE MAIN - Physics (2024 - 1st February Morning Shift - No. 20)
In the given circuit if the power rating of Zener diode is $10 \mathrm{~mW}$, the value of series resistance $R_s$ to regulate the input unregulated supply is :
_1st_February_Morning_Shift_en_20_1.png)
$10 \mathrm{k} \Omega$
$10 \Omega$
$1 \mathrm{k} \Omega$
$$ \frac{3}{7} \mathrm{k} \Omega<\mathrm{R}_{\mathrm{s}}<\frac{3}{5} \mathrm{k} \Omega$$
Explanation
_1st_February_Morning_Shift_en_20_2.png)
Pd across $R_s$
$$ \mathrm{V}_1=8-5=3 \mathrm{~V} $$
Current through the load resistor
$$ \mathrm{I}=\frac{5}{1 \times 10^3}=5 \mathrm{~mA} $$
Maximum current through Zener diode
$$ \mathrm{I}_{\mathrm{z} \max .}=\frac{10}{5}=2 \mathrm{~mA} $$
And minimum current through Zener diode
$$ \begin{gathered} \mathrm{I}_{\mathrm{z} \text { min. }}=0 \\\\ \therefore \mathrm{I}_{\mathrm{s} \text { max. }}=5+2=7 \mathrm{~mA} \end{gathered} $$
And $\mathrm{R}_{\mathrm{s} \text { min }}=\frac{\mathrm{V}_1}{\mathrm{I}_{\mathrm{s} \text { max }}}=\frac{3}{7} \mathrm{k} \Omega$
Similarly $$ \mathrm{I}_{\mathrm{s} \min .}=5 \mathrm{~mA} $$
And $\mathrm{R}_{\mathrm{s} \text { max. }}=\frac{\mathrm{V}_1}{\mathrm{I}_{\mathrm{s} \text { min. }}}=\frac{3}{5} \mathrm{k} \Omega$
$$ \therefore \frac{3}{7} \mathrm{k} \Omega<\mathrm{R}_{\mathrm{s}}<\frac{3}{5} \mathrm{k} \Omega $$
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