The reverse breakdown voltage of a Zener diod | vedclass

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

Question

(A) The total voltage supplied is $V = 9\, V$. The Zener diode is in parallel with the load resistor $R_L = 800\, \Omega$,so the voltage across the lo

Vedclass · Accepted Answer

$10$

Vedclass · Answer

(A) The total voltage supplied is $V = 9\, V$. The Zener diode is in parallel with the load resistor $R_L = 800\, \Omega$,so the voltage across the load is $V_z = 5.6\, V$.The voltage drop across the series resistor $R_s = 200\, \Omega$ is $V_{R_s} = V - V_z = 9 - 5.6 = 3.4\, V$.The total current flowing through the series resistor $R_s$ is $I = \frac{V_{R_s}}{R_s} = \frac{3.4}{200} = 0.017\, A = 17\, mA$.The current flowing through the load resistor $R_L$ is $I_L = \frac{V_z}{R_L} = \frac{5.6}{800} = 0.007\, A = 7\, mA$.Applying Kirchhoff's current law at the junction,the current through the Zener diode is $I_z = I - I_L = 17\, mA - 7\, mA = 10\, mA$.

The reverse breakdown voltage of a Zener diode is $5.6\, V$ in the given circuit. The current $I_z$ through the Zener is......$mA$

Similar Questions

In the following circuit,the current flowing through the $1\,k\Omega$ resistor is......$mA$.

In the given circuit,the Zener breakdown voltage is $8 \ V$. If the power of the Zener diode is $1.6 \ W$,the value of $R$ is (in $\Omega$)

In the given circuit, if the power rating of the Zener diode is $10 \, mW$, the value of series resistance $R_s$ to regulate the input unregulated supply is:

$A$ Zener diode acts as a/an

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module