The reverse breakdown voltage of Zener diode is 5.6V in the given circuit. The current Iz through the Zener is:-

Question

The reverse breakdown voltage of Zener diode is 5.6V in the given circuit. The current Iz through the Zener is:- <img class="img-fluid question-image" alt="Question" src="https://cdn.tardigrade.in/q/nta/p-c7u17lsrjmbexuv2.jpg" /> (A) 7mA (B) 17mA (C) 10mA (D) 15mA

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="Solution" src="https://cdn.tardigrade.in/q/nta/p-c7u17lsrjmbeerqq.jpg" /> In first loop 9V=200I1+5.6V I1=(3 . 4/200)=17mA In second loop I2=(5 . 6/800)=7mA IZ=I1-I2 =(.17-7.)mA=10mA

Q. The reverse breakdown voltage of Zener diode is $5.6 V$ in the given circuit. The current $I_{z}$ through the Zener is:-

$7 m A$

$17 m A$

$10 m A$

$15 m A$

In first loop
$9 V = 200 I_{1} + 5.6 V$
$I_{1} = \frac{3.4}{200} = 17 m A$
In second loop
$I_{2} = \frac{5.6}{800} = 7 m A$
$I_{Z} = I_{1} - I_{2}$
$= (17 - 7) m A = 10 m A$

Q. The reverse breakdown voltage of Zener diode is 5.6V in the given circuit. The current Iz​ through the Zener is:-

7mA

17mA

10mA

15mA