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MCQ Practice Test & Solutions: Test: Basic Concepts : PN Junction (10 Questions)

You can prepare effectively for Electrical Engineering (EE) GATE Electrical Engineering (EE) Mock Test Series 2027 with this dedicated MCQ Practice Test (available with solutions) on the important topic of "Test: Basic Concepts : PN Junction". These 10 questions have been designed by the experts with the latest curriculum of Electrical Engineering (EE) 2026, to help you master the concept.

Test Highlights:

  • - Format: Multiple Choice Questions (MCQ)
  • - Duration: 10 minutes
  • - Number of Questions: 10

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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: 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: 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: 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: 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: 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: 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: Question 8

Test: Basic Concepts : PN Junction - Question 9

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

Detailed Solution: 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: 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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