For the given circuit, the power across the Z | vedclass

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

Question

(C) The circuit consists of a source voltage $V = 24 	ext{ V}$, a series resistor $R_s = 1 	ext{ k}\Omega$, a Zener diode with breakdown voltage $V_

Vedclass · Accepted Answer

$120$

Vedclass · Answer

(C) The circuit consists of a source voltage $V = 24 	ext{ V}$, a series resistor $R_s = 1 	ext{ k}\Omega$, a Zener diode with breakdown voltage $V_z = 10 	ext{ V}$, and a load resistor $R_L = 5 	ext{ k}\Omega$.$1$. The current through the series resistor $R_s$ is given by:$I_s = \frac{V - V_z}{R_s} = \frac{24 	ext{ V} - 10 	ext{ V}}{1 	imes 10^3 	ext{ }\Omega} = \frac{14 	ext{ V}}{1000 	ext{ }\Omega} = 14 	ext{ mA}$.$2$. The current through the load resistor $R_L$ is:$I_L = \frac{V_z}{R_L} = \frac{10 	ext{ V}}{5 	imes 10^3 	ext{ }\Omega} = 2 	ext{ mA}$.$3$. The current through the Zener diode $I_z$ is:$I_z = I_s - I_L = 14 	ext{ mA} - 2 	ext{ mA} = 12 	ext{ mA}$.$4$. The power dissipated across the Zener diode $P_z$ is:$P_z = V_z 	imes I_z = 10 	ext{ V} 	imes 12 	ext{ mA} = 120 	ext{ mW}$.

For the given circuit, the power across the Zener diode is $....... \text{mW}$.

Similar Questions

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

In the circuit diagram shown in the figure,the current through the zener diode is (in $mA$)

In a Zener diode,breakdown occurs in reverse bias due to

The Zener diode is used for

$A$ Zener diode of power rating $2 \, W$ is to be used as a voltage regulator. If the Zener diode has a breakdown voltage of $10 \, V$ and it has to regulate an unregulated input voltage that fluctuates between $6 \, V$ and $14 \, V$, the value of $R_s$ for safe operation should be $.... \Omega$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module