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By method of images, the problem can be easily calculated by replacing the boundary with which polygon?
d Explanation: When any field or potential needs to be calculated for either line charge or coaxial cable or concentric cylinder, the method of images uses a triangle which converts the three dimensional problem to one dimensional analysis. From this, the result can be calculated.
d
See lessExplanation: When any field or potential needs to be calculated for either line charge or
coaxial cable or concentric cylinder, the method of images uses a triangle which
converts the three dimensional problem to one dimensional analysis. From this, the
result can be calculated.
Choose the best definition of dielectric loss.
a Explanation: In the scenario of an AC field, the absorption of electrical energy by a dielectric material is called as dielectric loss. This will result in dissipation of energy in the form of heat.
a
See lessExplanation: In the scenario of an AC field, the absorption of electrical energy by a
dielectric material is called as dielectric loss. This will result in dissipation of energy in
the form of heat.
Calculate the charge density for the current density given 20sin x i + ycos z j at the origin.
b Explanation: Using continuity equation, the problem can be solved. Div(J) = – dρ/dt. Div(J) = 20cos x + cos z. At origin, we get 20cos 0 + cos 0 = 21. To get ρ, on integrating the Div(J) with respect to t, the charge density will be 21t.
b
See lessExplanation: Using continuity equation, the problem can be solved. Div(J) =
– dρ/dt. Div(J) = 20cos x + cos z. At origin, we get 20cos 0 + cos 0 = 21. To get ρ, on
integrating the Div(J) with respect to t, the charge density will be 21t.
Find the electron density when convection current density is 120 units and the velocity is 5m/s.
c Explanation: The convection current density is given by J = ρe x v. To get ρe, put J = 120 and v = 5. ρe = 120/5 = 24 units.
c
See lessExplanation: The convection current density is given by J = ρe x v. To get ρe, put J =
120 and v = 5. ρe = 120/5 = 24 units.
Calculate the electric field when the conductivity is 20 units, electron density is 2.4 units and the velocity is 10m/s. Assume the conduction and convection current densities are same.
d Explanation: The conduction current density is given by J = σE and the convection current density is J = ρe v. When both are equal, ρe v = σE. To get E, put σ = 20, ρe = 2.4 and v = 10, E = 2.4 x 10/20 = 1.2 units.
d
See lessExplanation: The conduction current density is given by J = σE and the convection
current density is J = ρe v. When both are equal, ρe v = σE. To get E, put σ = 20, ρe =
2.4 and v = 10, E = 2.4 x 10/20 = 1.2 units.
On equating the generic form of current density equation and the point form of Ohm’s law, we can obtain V=IR. State True/False.
a Explanation: The generic current density equation is J = I/A and the point form of Ohm’s law is J = σ E. On equating both and substituting E = V/L, we get V = IL/σ A = IR which is the Ohm’s law.
a
See lessExplanation: The generic current density equation is J = I/A and the point form of Ohm’s
law is J = σ E. On equating both and substituting E = V/L, we get V = IL/σ A = IR which
is the Ohm’s law.
Find the electric field if the surface density at the boundary of air is 10-9 .
c Explanation: It is the conductor-free space boundary. At the boundary, E = ρ/εo. Put ρ = 10-9 and εo = 10-9/36π. We get E = 36π units.
c
See lessExplanation: It is the conductor-free space boundary. At the boundary, E = ρ/εo. Put ρ =
10-9 and εo = 10-9/36π. We get E = 36π units.
Find the energy stored by the capacitor 3F having a potential of 12V across it.
c Explanation: The energy stored in a capacitor is given by, E = 0.5 CV2. E = 0.5 x 3 x 122 = 0.5 x 432 = 216 units.
c
See lessExplanation: The energy stored in a capacitor is given by, E = 0.5 CV2.
E = 0.5 x 3 x 122 = 0.5 x 432 = 216 units.
A material with zero resistivity is said to have
c Explanation: Since resistivity is directly proportional to the resistance, when the resistivity is zero, resistance is also zero. Thus we get zero resistance. The option infinite conductance is also possible ideally, but it is not possible practically. As there is always some loss in the form of hRead more
c
See lessExplanation: Since resistivity is directly proportional to the resistance, when the
resistivity is zero, resistance is also zero. Thus we get zero resistance. The option
infinite conductance is also possible ideally, but it is not possible practically. As there is
always some loss in the form of heat, thus infinite conductance is impossible, but a short
circuit (zero resistance) is practically possible.
Find the dissipation factor when series resistance is 5 ohm and capacitive resistance is 10 unit.
b Explanation: The dissipation factor is nothing but the tangent of loss angle of loss tangent. Tan δ = Series resistance/Capacitive resistance = 5/10 = 0.5.
b
See lessExplanation: The dissipation factor is nothing but the tangent of loss angle of loss
tangent. Tan δ = Series resistance/Capacitive resistance = 5/10 = 0.5.