Question 47 of 209

A Zener diode of breakdown voltage $10V$ is used as a voltage regulator as shown in the figure. The current through the Zener diode is

A$50mA$B$0$C$30mA$D$20mA$

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Answer: (C) $30mA$

As Zener is in breakdown region, therefore potential drop across it would be $10V$. Now,
${I}_{3}=\dfrac{10}{500}=\dfrac{1}{50}A$ and
${I}_{1}=\dfrac{10}{200}=\dfrac{1}{20}A$
Also, ${I}_{2}={I}_{1}-{I}_{3}$
$\Rightarrow {I}_{2}=\left(\dfrac{1}{20}-\dfrac{1}{50}\right)=\left(\dfrac{3}{100}\right)A=30mA$

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