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14.Semiconductor Electronics
hard
The reverse breakdown voltage of a Zener diode is $5.6\, V$ in the given circuit. The current $I_z$ through the Zener is......$mA$
A
$10$
B
$15$
C
$7$
D
$17$
(JEE MAIN-2019)
Solution
$9=V_{z}+V_{R_{1}}$
$V_{z}=5.6 v$
$V_{R_1}=9-5.6$
$V_{R_1}=3.4$
$\mathrm{I}_{\mathrm{R}_{1}}=\frac{\mathrm{V}_{\mathrm{R}_{1}}}{\mathrm{R}}=\frac{3.4}{200} ; \mathrm{I}_{\mathrm{R}_{1}}=17 \mathrm{mA}$
${{\text{V}}_{\text{z}}}={{\text{V}}_{{{\text{R}}_{2}}}}={{\text{I}}_{{{\text{R}}_{2}}}}\left( {{\text{R}}_{2}} \right)$
$\frac{5.6}{800}={{\text{l}}_{{{\text{R}}_{2}}}}$ ; ${{I}_{{{\text{R}}_{2}}}}=7\text{mA}$
${{I}_{\text{z}}}=(17-7)\,\text{mA}=10\,\text{mA}$
Standard 12
Physics
