Question

In the given circuit the current through Zener Diode is close to :

• A. $6.0mA$
• B. $4.0mA$
• C. $6.7mA$
• D. $0.0mA$

Answer 1

The correct option is D $0.0mA$

From the given diagram,

$R_1=500\Omega$

$R_2=1500\Omega$

$V_2=10V$

As the resistors are in parallel, the potential difference across the $1500\Omega$ will also be 10 V.

The potential difference across the $500\Omega$ will be

$V_1=12-10=2V$

According to ohm’s law,

$V=IR$

$\Rightarrow I=\frac{V}{R}$

Current across the $500\Omega$ is

$\Rightarrow I_1=\frac{V_1}{R_1}=\frac{2}{500}=4\times {10}^{-3}A$

Current across the $1500\Omega$ is

$\Rightarrow I_2=\frac{V_2}{R_2}=\frac{10}{1500}=6.66\times {10}^{-3}A$

From above calculations, $I_1<I_2$ (Not possible case)

So the potential difference across Zener diode will never reach to breakdown voltage.

The current flows in the Zener diode only when the applied reverse voltage reaches breakdown voltage. Therefore, the current flowing through the Zener diode is 0 mA

Hence, Option (D) is the correct answer