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For the circuit shown below, the current through the Zener diode is:
(1) 9 mA
(2) 5 mA
(3) Zero
(4) 14 mA
$I_{5k\Omega} = 70/5000 = 14$ mA. $I_{10k\Omega} = 50/10000 = 5$ mA. $I_{zener} = 14 - 5 = 9$ mA.
BETA
