JEE MAIN - Physics (2023 - 10th April Morning Shift - No. 18)
A zener diode of power rating 1.6 W is to be used as voltage regulator. If the zener diode has a breakdown of 8V and it has to regulate voltage fluctuating between 3 V and 10 V. The value of resistance Rs for safe operation of diode will be
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13 $$\Omega$$
13.3 $$\Omega$$
10 $$\Omega$$
12 $$\Omega$$
Explanation
_10th_April_Morning_Shift_en_18_2.png)
Power rating of zener $=1.6 \mathrm{~W}$
Maximum voltage across zener $=8 \mathrm{~V}$
Using $P=V I$, for zenner diode,
$$ I=\frac{P}{V}=\frac{1.6}{8}=0.2 \mathrm{~A} $$
Hence, for safe operation, $2 \mathrm{~V}$ (excess voltage of $10 \mathrm{~V}$ from $8 \mathrm{~V}$ ), potential drop must occur in series resistor $R_S$.
For $R_S, V=2 \mathrm{~V}$ and $I=0.2 \mathrm{~A}$
Using $V=I R$,
$$ R=\frac{2}{0.2}=10 \Omega $$
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