1

### 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
2

### 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)
3

### JEE Main 2019 (Online) 12th January Morning Slot

The output of the given logic circuit is :

A
$\overline A B$
B
$AB + \overline {AB}$
C
$A\overline B + \overline A B$
D
$A\overline B$

## Explanation

Y $= \overline {\left( {\overline A + \overline B } \right)\overline A }$

$= \overline {\overline A + \overline A B}$

$= A\left( {\overline {\overline A B} } \right)$

$= A\left( {A + \overline B } \right)$

$= A + A\overline B = A\overline B$
4

### JEE Main 2019 (Online) 12th January Evening Slot

In the figure, given that VBB supply can vary from 0 to 5.0 V, VCC = 5V, $\beta$dc = 200, RB = 100 k$\Omega$, RC = 1 k$\Omega$ and VBE = 1.0 V. The minimum base current and the input voltage at which the transistor will go to saturation, will be respectively :

A
20 $\mu$A and 2.8 V
B
25 $\mu$A and 2.8 V
C
20 $\mu$A and 3.5 V
D
25 $\mu$A and 3.5 V

## Explanation

At saturation, VCE = 0

VCE = VCC $-$ ICRC

$\Rightarrow$  IC = ${{{V_{CC}}} \over {{R_C}}} = 5 \times {10^{ - 3}}\,A$

Given

$\beta$dc = ${{{{\rm I}_C}} \over {{{\rm I}_B}}}$

IB = ${{5 \times {{10}^{ - 3}}} \over {200}}$

IB = 25$\mu$A

At input side

VBB = IBRB + VBE

= (25mA) (100k$\Omega$) + 1V

VBB = 3.5 V