In the given circuit,the value of current I_{ | 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 $R_{L}$.Since the Zener diode is in the breakdown region,the voltage across the lo

Vedclass · Accepted Answer

$5$

Vedclass · Answer

(A) The Zener diode is connected in parallel with the load resistor $R_{L}$.Since the Zener diode is in the breakdown region,the voltage across the load resistor $R_{L}$ is equal to the Zener breakdown voltage,which is $V_{Z} = 5 \, V$.The load current $I_{L}$ is given by Ohm's law:$I_{L} = \frac{V_{Z}}{R_{L}}$Given $V_{Z} = 5 \, V$ and $R_{L} = 1 \, k\Omega = 1000 \, \Omega$.$I_{L} = \frac{5 \, V}{1000 \, \Omega} = 0.005 \, A = 5 \, mA$.

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

Similar Questions

$A$ Zener diode having a Zener voltage of $8\, V$ and a power dissipation rating of $0.5\, W$ is connected across a potential divider circuit as shown in the diagram. The value of the protective resistance $R_p$ is $....\, \Omega$.

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.......$V$.

The $I-V$ characteristics of an electronic device are shown in the figure. The device 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