1
JEE Main 2019 (Online) 8th April Morning Slot
+4
-1
The reverse breakdown voltage of a Zener diode is 5.6 V in the given circuit.

The current IZ through the Zener is :
A
7 mA
B
17 mA
C
15mA
D
10 mA
2
JEE Main 2019 (Online) 12th January Evening Slot
+4
-1
Out of Syllabus
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
3
JEE Main 2019 (Online) 12th January Morning Slot
+4
-1
The output of the given logic circuit is :

A
$$\overline A B$$
B
$$AB + \overline {AB}$$
C
$$A\overline B + \overline A B$$
D
$$A\overline B$$
4
JEE Main 2019 (Online) 11th January Evening Slot
+4
-1
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
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