Question 48 of 209

In the given circuit, the breakdown voltage of the Zener diode is $3.0V$. What is the value of ${I}_{z}$ ?

A$3.3mA$B$5.5mA$C$10mA$D$7mA$

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Answer: (B) $5.5mA$

Given that, ${V}_{z}=3V$
Let potential at $B=0V$
Potential at $E\left({V}_{E}\right)=10V$
Now, ${V}_{C}={V}_{A}=3V$ and,
${I}_{z}+{I}_{1}=I...\left(1\right)$
The current through $1k\Omega$ resistor is given by
$\begin{matrix}I & = & \dfrac{10-3}{1000}A \\ & = & \dfrac{7}{1000}A\end{matrix}$
From the diagram, it follows that
${I}_{1}=\dfrac{3}{2000}A$
Therefore, from equation (1), it follows that
$\begin{matrix}{I}_{z} & = & \dfrac{7-1.5}{1000}A \\ & = & 5.5mA\end{matrix}$

JEE Main Physics practice question on Semiconductors (Diodes), with the answer and a step-by-step solution. Scan a question from your own book or paper to get a whiteboard video.