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The reverse breakdown voltage of a Zener diode is 5.6 V in the given circuit. The current $I_Z$ through the Zener is:
(1) 10 mA
(2) 17 mA
(3) 15 mA
(4) 7 mA
$I_{800\Omega} = 5.6/800 = 7$ mA. $I_{200\Omega} = (9-5.6)/200 = 17$ mA. $I_Z = 17 - 7 = 10$ mA.
BETA
