Question:
A zener diode having zener voltage $8 \mathrm{~V}$ and power dissipation rating of $0.5 \mathrm{~W}$ is connected across a potential divider arranged with maximum potential drop across zener diode is as shown in the diagram. The value of protective resistance $R_{p}$ is............$\Omega$
Solution:
$\mathrm{P}=\mathrm{Vi}$
$0.5=8 \mathrm{i}$
$\mathrm{i}=\frac{1}{16} \mathrm{~A}$
$E=20=8+i R_{P}$
$\mathrm{R}_{\mathrm{P}}=12 \times 16=192 \Omega$
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