A zener diode, having breakdown voltage equal to $15 \mathrm{~V}$, is used in a voltage regulator circuit…
- $10 \mathrm{~mA}$
- $15 \mathrm{~mA}$
- $20 \mathrm{~mA}$
- $5 \mathrm{~mA}$
Solution
For I $1 \mathrm{k} \Omega$
$i_1=\frac{15}{1}=15 \mathrm{~mA}$
For II $250 \Omega$
$\begin{aligned}
& i_{250 \Omega}=\frac{20-15}{250}=\frac{5}{250} \\
&=\frac{20}{1000}=20 \mathrm{~mA} \\
& \therefore \quad i_{\text {zener }}=20-15=5 \mathrm{~mA}
\end{aligned}$Asked in: NEET 2011 (Mains)