In the given circuit,the value of I_{L} (in m | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) The Zener diode is connected in parallel with the load resistor $R_{L} = 2 \ k\Omega = 2000 \ \Omega$.Since the Zener diode is in the breakdown re

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(A) The Zener diode is connected in parallel with the load resistor $R_{L} = 2 \ k\Omega = 2000 \ \Omega$.Since the Zener diode is in the breakdown region,the voltage across the load resistor $R_{L}$ is equal to the Zener voltage $V_{Z} = 5 \ V$.Using Ohm's law,the current through the load resistor $I_{L}$ is given by:$I_{L} = \frac{V_{Z}}{R_{L}}$$I_{L} = \frac{5 \ V}{2000 \ \Omega} = 0.0025 \ A$Converting to milliamperes $(mA)$:$I_{L} = 0.0025 	imes 1000 \ mA = 2.5 \ mA$.

In the given circuit,the value of $I_{L}$ (in $mA$) will be: (in $.5$)

Similar Questions

$A$ Zener diode with voltage $V_z = 6 \ V$ is used to maintain a constant voltage across a load resistance $R_L = 1000 \ \Omega$. $A$ series resistor $R_s = 100 \ \Omega$ is used. If the source $e.m.f$ is $E = 9 \ V$,then the power dissipated in the Zener diode is .......... $W$.

The device used for voltage regulation is

What is the value of output voltage $V_0$ in $V$ for the circuit shown in the figure?

In the given circuit, the breakdown voltage of the Zener diode is $3.0 \, V$. What is the value of $I_z$ (in $ \, mA$)?

Vedclass Test Series

Exam Paper Generator

Online Exam Module