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1

JEE Main 2019 (Online) 11th January Evening Slot

MCQ (Single Correct Answer)
The circuit shown below contains two ideal diodes, each with a forward resistance of 50 $$\Omega $$. If the battery voltage is 6 V, the current through the 100 $$\Omega $$ resistance (in Amperes) is :

A
0.027
B
0.030
C
0.036
D
0.020

Explanation

I = $${6 \over {300}}$$ = 0.02 (D2 is in reverse bias)
2

JEE Main 2019 (Online) 11th January Morning Slot

MCQ (Single Correct Answer)
In the given circuit the current through Zener Diode is close to:

A
0.0 mA
B
6.7 mA
C
4.0 mA
D
6.0 mA

Explanation

Since voltage across zener diode must be less than 10V therefore it will not work in breakdown region, & its resistance will be infinite & current through it = 0
3

JEE Main 2019 (Online) 10th January Evening Slot

MCQ (Single Correct Answer)
For the circuit shown below, the current through the Zener diode is -

A
5 mA
B
zero
C
14 mA
D
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.


Current is R1 = $${{70} \over {5000}}$$ = 14mA

Current is R2 = $${{50} \over {10000}}$$ = 5mA

$$ \therefore $$  Current through diode = 9mA
4

JEE Main 2019 (Online) 10th January Morning Slot

MCQ (Single Correct Answer)
To get output 1 at R, for the given logic gate circuit the input values must be

A
x = 0,  y = 0
B
x = 1,  y = 0
C
x = 0,  y = 1
D
x = 1,  y = 1

Explanation

P = $$\overline x + $$ y

Q = $$\overline {\overline y .x} $$ = y + $$\overline x $$

Output, R = $$\overline {P + Q} $$

To make R = 1

$$\overline {P + Q} $$ should be = 1

$$ \therefore $$ P + Q must be 0

$$ \therefore $$ ($$\overline x + $$ y) + (y + $$\overline x $$) = 0

Now by checking each option you can see

When y = 0 and x = 1 then

$$\overline x + $$ y = 0

$$ \therefore $$ ($$\overline x + $$ y) + (y + $$\overline x $$) = 0

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