Question

A Zener diode is connected to a battery and a load as shown below. The currents $I,I_z$ and $I_L$ are respectively

• A. $12.5\text{mA},5\text{mA},7.5\text{mA}$
• B. $15\text{mA},7.5\text{mA},7.5\text{mA}$
• C. $15\text{mA},5\text{mA},10\text{mA}$
• D. $12.5\text{mA},7.5\text{mA},5\text{mA}$

Answer 1

The correct option is D $12.5\text{mA},7.5\text{mA},5\text{mA}$

Zener diode is reverse biased and $V_z=10\text{V}$

So, $I=\frac{60-10}{4}$

$=12.5\text{mA}$

Appling Ohm's law to the load resistance,

$\Rightarrow I_L=\frac{10}{2}=5\text{mA}$

Applying KCL, we get,

$I=I_Z+I_L$

$\Rightarrow 12.5=I_Z+5=7.5\text{mA}$

Hence, $\left(D\right)$ is the correct answer.