A Zener diode,having a breakdown voltage equa | vedclass

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

Question

(B) The voltage across the Zener diode is constant at $15 \, V$ because it is in the breakdown region.First,calculate the current through the load res

Vedclass · Accepted Answer

$5$

Vedclass · Answer

(B) The voltage across the Zener diode is constant at $15 \, V$ because it is in the breakdown region.First,calculate the current through the load resistor $(R_L = 1 \, k\Omega)$:$I_L = \frac{V_Z}{R_L} = \frac{15 \, V}{1 	imes 10^3 \, \Omega} = 15 \, mA$.Next,calculate the total current $(I)$ flowing through the series resistor $(R_S = 250 \, \Omega)$:The voltage drop across the series resistor is $V_S = V_{in} - V_Z = 20 \, V - 15 \, V = 5 \, V$.$I = \frac{V_S}{R_S} = \frac{5 \, V}{250 \, \Omega} = 0.02 \, A = 20 \, mA$.Finally,apply Kirchhoff's Current Law at the node:$I = I_Z + I_L$$I_Z = I - I_L = 20 \, mA - 15 \, mA = 5 \, mA$.Therefore,the current through the Zener diode is $5 \, mA$.

$X = x$	$0$	$1$	$2$	$3$
$P(X = x)$	$\frac{1}{3}$	$\frac{1}{2}$	$0$	$\frac{1}{6}$

$A$ Zener diode,having a breakdown voltage equal to $15 \, V$,is used in a voltage regulator circuit as shown in the figure. The current through the diode is.....$mA$.

Similar Questions

If $m$ and $\sigma^2$ are the mean and variance of the random variable $X$,whose distribution is given by:
$X = x$ $0$ $1$ $2$ $3$
$P(X = x)$ $\frac{1}{3}$ $\frac{1}{2}$ $0$ $\frac{1}{6}$

Then:

The slope of the tangent to the curve $x=t^{2}+3t-8, y=2t^{2}-2t-5$ at the point $(2,-1)$ is

$\int \frac{d x}{7+5 \cos x}$ is equal to

Among the following,which is water insoluble?

The set $\{x \in R: \cos 2x + 2 \cos^2 x = 2\}$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module