Test: Real Time Applications - Electrical Engineering (EE) MCQ

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

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.

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

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.

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.

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.

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.

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.

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.

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.