Question

Consider the circuit shown in the figure below:


In the circuit the maximum power rating of the zener diode is equal to 400 mW, the maximum possible value of $R_L$ is equal to ________ $k\Omega$

• A. 1

Answer 1

The correct option is A 1

$\begin{array}{cc}\text{Applying KCL at node 'A' , we get,}& \\ I_S& =I_z+I_L\\ I_{z\left(max\right)}& =\frac{P_2\left(max\right)}{V_z}=\frac{400\times {10}^{-3}}{10}=40mA\\ now,I_s& =\frac{V_s-V_z}{R_s}=\frac{20-10}{200}=\frac{10}{200}=50mA\\ \therefore I_{L\left(nin\right)}& =I_s-I_{z\left(max\right)}\\ I_{L\left(min\right)}& =10\times {10}^{-3}=10mA\\ \therefore R_{L\left(max\right)}& =\frac{V_z}{I_{L\left(max\right)}}=\frac{10}{10\times {10}^{-3}}=1k\Omega \end{array}$