In the given circuit, Zener breakdown voltage is 8 V. If power of Zener diode is 1.6 W . The value of $R$ is
In the given circuit, Zener breakdown voltage is 8 V. If power of Zener diode is 1.6 W . The value of $R$ is
$2 \Omega$
$4 \Omega$
$6 \Omega$
$10 \Omega$
Solution
$\begin{array}{ll}
& \mathrm{V}_{\mathrm{Z}}=8 \mathrm{~V} \\
& \mathrm{P}=\mathrm{V}_{\mathrm{Z}} \mathrm{I} \\
\therefore \quad & 1.6=8 \mathrm{I} \\
\therefore \quad & \mathrm{I}=0.2 \mathrm{~A}
\end{array}$ Voltage drop across Zener is 8 V . Hence the voltage drop across R will be $10-8=2 \mathrm{~V}$.
$\mathrm{R}=\frac{\mathrm{V}}{\mathrm{I}}=\frac{2}{0.2}=10 \Omega$
