The reverse breakdown voltage of a Zener diode is $5.6 V$ in the given circuit.
The current $I_{Z}$ through the Zener is

7 m A

No worries! We‘ve got your back. Try BYJU‘S free classes today!

17 m A

No worries! We‘ve got your back. Try BYJU‘S free classes today!

10 m A

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

15 m A

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C $10 m A$

$9 = V_{Z} + V_{R_{1}}$
$V_{Z} = 5.6 V$
$V_{R_{1}} = 9 - 5.6$
$= 3.4 V$
$I_{R_{1}} = \frac{V_{R_{1}}}{R} = \frac{3.4}{200}$
$= 17 m A$
$V_{Z} = V_{R_{2}} = I_{R_{2}} (R_{2})$
$\frac{5.6}{800} = I_{R_{2}}$
$⟹ I_{R_{2}} = 7 m A$
$I_{Z} = (17 - 7) m A$
$= 10 m A$

Suggest Corrections

Similar questions

5.6 V

I_{Z}

15 V,

15 V

5.6 V

10 V

100 Ω

Join BYJU'S Learning Program