JEE MAIN - Physics (2019 - 10th January Evening Slot - No. 8)
For the circuit shown below, the current through the Zener diode is -
_10th_January_Evening_Slot_en_8_1.png)
5 mA
zero
14 mA
9 mA
Explanation
Assuming zener diode doesnot undergo breakdown, current in circuit
= $${{120} \over {15000}}$$ = 8mA
$$ \therefore $$ Voltage drop across diode
= 80 V > 50 V.
The diode undergo breakdown.
_10th_January_Evening_Slot_en_8_2.png)
Current is R1 = $${{70} \over {5000}}$$ = 14mA
Current is R2 = $${{50} \over {10000}}$$ = 5mA
$$ \therefore $$ Current through diode = 9mA
= $${{120} \over {15000}}$$ = 8mA
$$ \therefore $$ Voltage drop across diode
= 80 V > 50 V.
The diode undergo breakdown.
Current is R1 = $${{70} \over {5000}}$$ = 14mA
Current is R2 = $${{50} \over {10000}}$$ = 5mA
$$ \therefore $$ Current through diode = 9mA
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