In the following circuit,the power dissipated | vedclass

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

Question

(C) Given: Source voltage $V_s = 12 \ V$,series resistance $R_s = 100 \ \Omega$,Zener voltage $V_z = 8 \ V$,load resistance $R_L = 800 \ \Omega$.The c

Vedclass · Accepted Answer

$0.24$

Vedclass · Answer

(C) Given: Source voltage $V_s = 12 \ V$,series resistance $R_s = 100 \ \Omega$,Zener voltage $V_z = 8 \ V$,load resistance $R_L = 800 \ \Omega$.The current through the series resistor $R_s$ is given by:$I_s = \frac{V_s - V_z}{R_s} = \frac{12 \ V - 8 \ V}{100 \ \Omega} = \frac{4 \ V}{100 \ \Omega} = 0.04 \ A = 40 \ mA$.The current through the load resistor $R_L$ is:$I_L = \frac{V_z}{R_L} = \frac{8 \ V}{800 \ \Omega} = 0.01 \ A = 10 \ mA$.The current through the Zener diode $I_z$ is:$I_z = I_s - I_L = 40 \ mA - 10 \ mA = 30 \ mA = 0.03 \ A$.The power dissipated in the Zener diode is:$P_z = V_z 	imes I_z = 8 \ V 	imes 0.03 \ A = 0.24 \ W$.

In the following circuit,the power dissipated in the Zener diode is (in $W$)

Similar Questions

An ideal Zener diode with a breakdown voltage of $3 \text{ V}$ is reverse-biased with an input voltage $V_i = 5 \text{ V}$. The magnitude of the voltage difference between points $B$ and $A$ is: (in $\text{ V}$)

If the voltage between the terminals $A$ and $B$ is $17\ V$ and the Zener breakdown voltage is $9\ V$,then the potential voltage across $R$ is.......$V$.

In the given circuit,the value of current $I_{L}$ will be in $mA$. (When $R_{L} = 1 \, k\Omega$)

$A$ Zener diode having a breakdown voltage equal to $5 \, V$ is used in the voltage regulator circuit shown in the figure. The current through the diode is ........ $mA$.

What happens during the regulation action of a Zener diode?

Vedclass Test Series

Exam Paper Generator

Online Exam Module