Test: Real Time Applications


10 Questions MCQ Test Electromagnetic Theory | Test: Real Time Applications


Description
This mock test of Test: Real Time Applications for Electrical Engineering (EE) helps you for every Electrical Engineering (EE) entrance exam. This contains 10 Multiple Choice Questions for Electrical Engineering (EE) Test: Real Time Applications (mcq) to study with solutions a complete question bank. The solved questions answers in this Test: Real Time Applications quiz give you a good mix of easy questions and tough questions. Electrical Engineering (EE) students definitely take this Test: Real Time Applications exercise for a better result in the exam. You can find other Test: Real Time Applications extra questions, long questions & short questions for Electrical Engineering (EE) on EduRev as well by searching above.
QUESTION: 1

Calculate the capacitance of a material in air with area 20 units and distance between plates is 5m.

Solution:

Answer: a
Explanation: The capacitance of any material is given by, C = εA/d, where ε = εoεr is the permittivity in air and the material respectively. Thus C = 1 X 8.854 X 10-12 X 20/5 = 35.36pF

QUESTION: 2

The resistance of a material with conductivity 2millimho/m2, length 10m and area 50m is 

Solution:

Answer: c
Explanation: The resistance is given by, R = ρL/A, where ρ is the resistivity, the inverse of conductivity. R = 10/(0.002 X 50) = 100 ohm.

QUESTION: 3

Find the inductance of a coil with permeability 3.5, turns 100 and length 2m. Assume the area to be thrice the length.

Solution:

Answer: a
Explanation: The inductance is given by L = μ N2A/l, where μ= μoμr is the permeability of air and the material respectively. N = 100 and Area = 3 X 2 = 6. L = 4π X 10-7 X 1002 X 6/2 = 131.94mH

QUESTION: 4

Find the current density of a material with resistivity 20 units and electric field intensity 2000 units.

Solution:

Answer: d
Explanation: The current density is given by J = σ E, where σ is the conductivity. Thus resistivity ρ = 1/σ. J = E/ρ = 2000/20 = 100 units.

QUESTION: 5

Find the current in a conductor with resistance 2 ohm, electric field 2 units and distance 100cm.

Solution:

Answer: a
Explanation: We know that E = V/d. To get potential, V = E X d = 2 X 1 = 2 volts. From Ohm’s law, V = IR and current I = V/R = 2/2 = 1A.

QUESTION: 6

In electric fields, D= ε E. The correct expression which is analogous in magnetic fields will be

Solution:

Answer: b
Explanation: In electric fields, the flux density is a product of permittivity and field intensity. Similarly, for magnetic fields, the magnetic flux density is the product of permeability and magnetic field intensity, given by B= μ H.

QUESTION: 7

Find the force on a conductor of length 12m and magnetic flux density 20 units when a current of 0.5A is flowing through it.

Solution:

Answer: b
Explanation: The force on a conductor is given by F = BIL, where B = 20, I = 0.5 and L = 12. Force F = 20 X 0.5 x 12 = 120 N.

QUESTION: 8

From the formula F = qE, can prove that work done is a product of force and displacement. State True/False 

Solution:

Answer: a
Explanation: We know that F = qE = qV/d and W = qV. Thus it is clear that qV = W and qV = Fd. On equating both, we get W = Fd, which is the required proof.

QUESTION: 9

Calculate the power of a material with electric field 100 units at a distance of 10cm with a current of 2A flowing through it.

Solution:

Answer: b
Explanation: Power is defined as the product of voltage and current.
P = V X I, where V = E X d. Thus P = E X d X I = 100 X 0.1 X 2 = 20 units.

QUESTION: 10

Compute the power consumed by a material with current density 15 units in an area of 100 units. The potential measured across the material is 20V.

Solution:

Answer: c
Explanation: Power is given by, P= V X I, where I = J X A is the current.
Thus power P = V X J X A = 20 X 15 X 100 = 30,000 joule = 30kJ.

Related tests