1

### JEE Main 2019 (Online) 10th January Morning Slot

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
2

### JEE Main 2019 (Online) 10th January Evening Slot

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
3

### JEE Main 2019 (Online) 11th January Morning Slot

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
4

### JEE Main 2019 (Online) 11th January Evening Slot

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)