A Zener diode is connected to a battery and a | vedclass

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

Question

(D) The circuit is given as shown in the figure.Potential drop across the load resistance $R_L$ is $V_L = V_Z = 10 	ext{ V}$.Potential drop across th

Vedclass · Accepted Answer

$12.5 	ext{ mA}, 7.5 	ext{ mA}, 5 	ext{ mA}$

Vedclass · Answer

(D) The circuit is given as shown in the figure.Potential drop across the load resistance $R_L$ is $V_L = V_Z = 10 	ext{ V}$.Potential drop across the $4 	ext{ k}\Omega$ series resistance is $V_R = 60 	ext{ V} - 10 	ext{ V} = 50 	ext{ V}$.Current through the $4 	ext{ k}\Omega$ resistance is $I = \frac{V_R}{R} = \frac{50 	ext{ V}}{4 	imes 10^3 \Omega} = 12.5 	imes 10^{-3} 	ext{ A} = 12.5 	ext{ mA}$.Current through the load resistance $I_L$ is $I_L = \frac{V_L}{R_L} = \frac{10 	ext{ V}}{2 	imes 10^3 \Omega} = 5 	ext{ mA}$.Using Kirchhoff's current law at the node,$I = I_Z + I_L$,therefore $I_Z = I - I_L = 12.5 	ext{ mA} - 5 	ext{ mA} = 7.5 	ext{ mA}$.Thus,the currents are $I = 12.5 	ext{ mA}$,$I_Z = 7.5 	ext{ mA}$,and $I_L = 5 	ext{ mA}$.

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

Similar Questions

Write a short note on a Zener diode.

For the circuit shown below,the current through the Zener diode is......$mA$

In the given circuit diagram,a $5 \,V$ Zener diode along with a series resistance $R$ is connected across a $50 \,V$ power supply. If the maximum Zener current is $90 \,mA$,the minimum value of the resistance $R$ required will be $..... \,\Omega$.

In a Zener regulated power supply, a Zener diode with $V_{z} = 6.0 \; V$ is used for regulation. The load current is to be $4.0 \; mA$ and the unregulated input is $10.0 \; V$. What should be the value of the series resistor $R_{S}$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module