JEE MAIN - Physics (2016 - 9th April Morning Slot - No. 24)
An experiment is performed to determine the I - V characteristics of a Zener diode, which has a protective resistance of R = 100 $$\Omega $$, and a maximum power of dissipation rating of 1 W. The minimum voltage range of the DC source in the circuit is :
0 $$-$$ 5 V
0 $$-$$ 8 V
0 $$-$$ 12 V
0 $$-$$ 24 V
Explanation
_9th_April_Morning_Slot_en_24_1.png)
Potential drop accross zener diode.
Vz = V $$-$$ 100 I
$$ \therefore $$ Power dissiption = Vz I
= (V $$-$$ 100 I) I
Given that,
(V $$-$$ 100 I) I = 1
$$ \Rightarrow $$ VI $$-$$ 100 I2 = 1
$$ \Rightarrow $$ 100 I2 $$-$$ VI + 1 = 0
As As I is real,
So, b2 $$-$$ 4ac $$ \ge $$ far this quadratic equation.
$$ \therefore $$ V2 $$-$$ 4(100)1 $$ \ge $$ 0
$$ \Rightarrow $$ V $$ \ge $$ 20 V
$$ \therefore $$ Voltage range should be 0 $$-$$ 24 V
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