In the figure, the reverse breakdown voltage of a Zener diode is 5.6V , then the current Iz through the diode is

Question

In the figure, the reverse breakdown voltage of a Zener diode is 5.6V , then the current Iz through the diode is <img class="img-fluid question-image" alt="Question" src="https://cdn.tardigrade.in/q/nta/p-waszwkihe8zp5f2o.jpg" /> (A) 17mA (B) 15mA (C) 10mA (D) 7mA

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="Solution" src="https://cdn.tardigrade.in/q/nta/p-waszwkihe8zpatg4.jpg" /> For zener break down potential difference across 800 Ω resistor will be 5.6 V Vz=5.6 V i2=(V2/800)=(5 . 6/800)=7 mA ÎV across 200 Ω=9-5.6=3.4 V i1=(3 . 4/200)=17 mA i1=i2+iz iz=17 mA-7 mA=10 mA

Q. In the figure, the reverse breakdown voltage of a Zener diode is $5.6 V$ , then the current $I_{z}$ through the diode is

$17 m A$

$15 m A$

$10 m A$

$7 m A$

For zener break down potential difference across $800 Ω$ resistor will be $5.6 V$
$V_{z} = 5.6 V$
$i_{2} = \frac{V _{2}}{800} = \frac{5.6}{800} = 7 m A$
$Δ V$ across $200 Ω = 9 - 5.6 = 3.4 V$
$i_{1} = \frac{3.4}{200} = 17 m A$
$i_{1} = i_{2} + i_{z}$
$i_{z} = 17 m A - 7 m A = 10 m A$

Q. In the figure, the reverse breakdown voltage of a Zener diode is 5.6V , then the current Iz​ through the diode is

17mA

15mA

10mA

7mA