The reverse breakdown voltage of a Zener diode is 5.6 V in the given c

1.	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» 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 brainly.in/question/15673253