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Electronics and Communication Engineering (ECE) Exam  >  Electronics and Communication Engineering (ECE) Tests  >  GATE ECE (Electronics) Mock Test Series 2025  >  Test: Basic Concepts : PN Junction - Electronics and Communication Engineering (ECE) MCQ

Test: Basic Concepts : PN Junction - Electronics and Communication Engineering (ECE) MCQ


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10 Questions MCQ Test GATE ECE (Electronics) Mock Test Series 2025 - Test: Basic Concepts : PN Junction

Test: Basic Concepts : PN Junction for Electronics and Communication Engineering (ECE) 2024 is part of GATE ECE (Electronics) Mock Test Series 2025 preparation. The Test: Basic Concepts : PN Junction questions and answers have been prepared according to the Electronics and Communication Engineering (ECE) exam syllabus.The Test: Basic Concepts : PN Junction MCQs are made for Electronics and Communication Engineering (ECE) 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Basic Concepts : PN Junction below.
Solutions of Test: Basic Concepts : PN Junction questions in English are available as part of our GATE ECE (Electronics) Mock Test Series 2025 for Electronics and Communication Engineering (ECE) & Test: Basic Concepts : PN Junction solutions in Hindi for GATE ECE (Electronics) Mock Test Series 2025 course. Download more important topics, notes, lectures and mock test series for Electronics and Communication Engineering (ECE) Exam by signing up for free. Attempt Test: Basic Concepts : PN Junction | 10 questions in 10 minutes | Mock test for Electronics and Communication Engineering (ECE) preparation | Free important questions MCQ to study GATE ECE (Electronics) Mock Test Series 2025 for Electronics and Communication Engineering (ECE) Exam | Download free PDF with solutions
Test: Basic Concepts : PN Junction - Question 1
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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
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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.

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Test: Basic Concepts : PN Junction - Question 3
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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
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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
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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
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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
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The superposition theorem requires as many circuits to be solved as there are:
Test: Basic Concepts : PN Junction - Question 8
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Twelve 6 resistor are used as edge to form a cube. The resistance between two diagonally opposite corner of the cube is: (in ohm)
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Test: Basic Concepts : PN Junction - Question 9
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 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 CV2 = 0.5x10-6 x 100 x 100 = 0.05 J

Test: Basic Concepts : PN Junction - Question 10
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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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