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/m^{2}, 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 N^{2}A/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 100^{2} 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

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: 7

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: 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.

### Top Real Time Big data Applications

Doc | 3 Pages

### Logarithms - Real Life Applications

Video | 05:03 min

### Real time Opearting systm

Doc | 8 Pages

### Real time state machines-II

Doc | 6 Pages

- Test: Real Time Applications
Test | 10 questions | 10 min

- Test: Real Time Applications
Test | 10 questions | 10 min

- Test: Bernoulliâ€™s Equation For Real Fluids & Applications Of Bernoulliâ€™s Equation
Test | 10 questions | 20 min

- Test: Applications Of Probability
Test | 25 questions | 25 min

- Test: Real Analysis- 8
Test | 20 questions | 60 min