Question

The reverse breakdown voltage of a Zener diode is 5.6 V in the given circuit .
The current Iz through the Zener is:
(A) 10 mA (B) 15 mA
(C) 7 mA (D) 17 mA

Answer 1

Answer:

Current through diode is given as 10 mA

Explanation:

Let the current will flow through the two resistors in series

So the current through it is given as

$i = \frac{9}{1000}$

now the voltage across 800 ohm is given as

$V = \frac{9}{1000}800$

$V = 7.2 V$

so here the diode will break down and voltage across it is 5.6 V

now the voltage across 200 ohm is

$V = 9 - 5.6 = 3.4 V$

current in 200 ohm resistance is given as

$i = \frac{3.4}{200} = 17 mA$

similarly current in 800 ohm

$i_1 = \frac{5.6}{800}$

$i_1 = 7 mA$

so current through diode is given as

$i_2 = i - i_1$

$i_2 = 10 mA$

#Learn

Topic : Zener Diode

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