Description

This mock test of Test: Basic Concepts : PN Junction for Electrical Engineering (EE) helps you for every Electrical Engineering (EE) entrance exam.
This contains 10 Multiple Choice Questions for Electrical Engineering (EE) Test: Basic Concepts : PN Junction (mcq) to study with solutions a complete question bank.
The solved questions answers in this Test: Basic Concepts : PN Junction quiz give you a good mix of easy questions and tough questions. Electrical Engineering (EE)
students definitely take this Test: Basic Concepts : PN Junction exercise for a better result in the exam. You can find other Test: Basic Concepts : PN Junction extra questions,
long questions & short questions for Electrical Engineering (EE) on EduRev as well by searching above.

QUESTION: 1

A solid copper sphere, 10 cm in diameter is deprived of 10^{20} electrons by a charging scheme. The charge on the sphere is

Solution:

n 10^{20}, Q = ne = e 10^{20} = 16.02 C.

Charge on sphere will be positive.

QUESTION: 2

A lightning bolt carrying 30,000 A lasts for 50 microseconds. If the lightning strikes an airplane flying at 20,000 feet, what is the charge deposited on the plane?

Solution:

QUESTION: 3

If 120 C of charge passes through an electric conductor in 60 sec, the current in the conductor is

Solution:

i = dQ/dt = 120/60 = 2A.

QUESTION: 4

The energy required to move 120 coulomb through 3 V is

Solution:

W = Qv = 360 J.

QUESTION: 5

Consider the circuit graph shown in figure below. Each branch of circuit graph represent a circuit element. The value of voltage V1 is

Solution:

100 = 65 + V2 => V2 = 35 V

V3 – 30 = V2 => V3 = 65 V

105 – V3 + V4 – 65 = 0 => V4 = 25 V

V4 + 15 – 55 + V1 = 0 => V1 = 15 V.

QUESTION: 6

Req = ?

Solution:

Req – 5 = 10(Req + 5)/(10 + 5 +Req). Solving for Req we have

Req = 11.18 ohm.

QUESTION: 7

The superposition theorem requires as many circuits to be solved as there are

Solution:

QUESTION: 8

Twelve 6 resistor are used as edge to form a cube. The resistance between two diagonally opposite corner of the cube is (in ohm)

Solution:

QUESTION: 9

The energy required to charge a 10 µF capacitor to 100 V is

Solution:

Energy provided is equal to 0.5 CVxV= 0.5x10^{-6} x 100 x 100 = 0.05 J

QUESTION: 10

A capacitor is charged by a constant current of 2 mA and results in a voltage increase of 12 V in a 10 sec interval. The value of capacitance is ________

Solution:

The capacitor current is given as i=C*(dv/dt), where dv/dt is the derivative of voltage, dt=t2-t1 given as 10 sec and dv is the change in voltage which is given as 12V.

So, we have C=i/(dv/dt)

=> C = 2mA/(12/10) = 2mA/(1.2).

Hence C = 1.67mF.

### Pn Junction Diode

Doc | 3 Pages

### PN-JUNCTION-DIODE

Doc | 27 Pages

### PPT - PN Junction Diode

Doc | 27 Pages

### Characteristics of PN Junction Diode

Video | 12:51 min

- Test: Basic Concepts : PN Junction
Test | 10 questions | 10 min

- The PN - Junction, MCQ Test
Test | 20 questions | 60 min

- Test: Qualitative Theory Of The PN Junction
Test | 10 questions | 10 min

- Test: The PN Junction As A Diode
Test | 10 questions | 10 min

- Test: Basic Concepts - 2
Test | 10 questions | 30 min