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Solution


Q.

The reverse breakdown voltage of a Zener diode is 5.6 V in the given circuit. The current Iz through the Zener is:
1 10 mA  
2 17 mA  
3 15 mA  
4 7 mA  

Ans.

(1)

P.D. across 800Ω resistors=5.6 V

So, I800Ω=5.6800 A=7 mA

Now, P.D. across 200Ω resistors

=9-5.6 V=3.4 V

So, I200Ω=9-5.6200=17 mA

So, current through zener diode =IZ=17-7=10 mA