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Test: Basic Concepts : PN Junction - Electrical Engineering (EE) MCQ


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10 Questions MCQ Test GATE Electrical Engineering (EE) 2024 Mock Test Series - Test: Basic Concepts : PN Junction

Test: Basic Concepts : PN Junction for Electrical Engineering (EE) 2024 is part of GATE Electrical Engineering (EE) 2024 Mock Test Series preparation. The Test: Basic Concepts : PN Junction questions and answers have been prepared according to the Electrical Engineering (EE) exam syllabus.The Test: Basic Concepts : PN Junction MCQs are made for Electrical Engineering (EE) 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Basic Concepts : PN Junction below.
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Test: Basic Concepts : PN Junction - Question 1

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

Detailed Solution for Test: Basic Concepts : PN Junction - Question 1

 i = dQ/dt = 120/60 = 2A.

Test: Basic Concepts : PN Junction - Question 2

A solid copper sphere, 10 cm in diameter is deprived of 1020 electrons by a charging scheme. The charge on the sphere is:

Detailed Solution for Test: Basic Concepts : PN Junction - Question 2

n = 1020
Q = ne = e * 1020 = 16.02 C.
The charge on the sphere will be positive.

Test: Basic Concepts : PN Junction - Question 3

The energy required to move 120 coulomb through 3 V is:

Detailed Solution for Test: Basic Concepts : PN Junction - Question 3

W = QV = 360 J.

Test: Basic Concepts : PN Junction - Question 4

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?

Detailed Solution for Test: Basic Concepts : PN Junction - Question 4

Test: Basic Concepts : PN Junction - 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​:

Detailed Solution for Test: Basic Concepts : PN Junction - Question 5
  • 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.
Test: Basic Concepts : PN Junction - Question 6

What is the value of Req = ?

Detailed Solution for Test: Basic Concepts : PN Junction - Question 6
  • Req – 5 = 10(Req + 5)/(10 + 5 +Req).
  • Solving for Req we have
    Req = 11.18 ohm.
Test: Basic Concepts : PN Junction - Question 7

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

Test: Basic Concepts : PN Junction - 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)

Detailed Solution for Test: Basic Concepts : PN Junction - Question 8

Test: Basic Concepts : PN Junction - Question 9

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

Detailed Solution for Test: Basic Concepts : PN Junction - Question 9

Energy provided is equal to 0.5 CV= 0.5x10-6 x 100 x 100 = 0.05 J

Test: Basic Concepts : PN Junction - 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 ________.

Detailed Solution for Test: Basic Concepts : PN Junction - Question 10
  • 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 12 V.
  • So, we have C = i / (dv/dt)
    C = 2mA/(12/10) = 2mA/(1.2).
  • Hence C = 1.67mF
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