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i1−iz=800Vz=8005.6=7mA
For loop ABCF,
9−200i1−Vz=0
i1=0.017or17mA
iz=(i1−7)mA=10mA
JEE Main 2019 — Physics Modern Physics
The reverse break down voltage of a Zener diode is 5.6V in the given circuit.
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The current Iz through the Zener is:
Held on 8 Apr 2019 · Verified 6 Jul 2026.
10mA
7mA
17mA
15mA
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