Test: Continuity Equation - Electrical Engineering (EE) MCQ

# Test: Continuity Equation - Electrical Engineering (EE) MCQ

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## 10 Questions MCQ Test Electromagnetic Fields Theory (EMFT) - Test: Continuity Equation

Test: Continuity Equation for Electrical Engineering (EE) 2024 is part of Electromagnetic Fields Theory (EMFT) preparation. The Test: Continuity Equation questions and answers have been prepared according to the Electrical Engineering (EE) exam syllabus.The Test: Continuity Equation MCQs are made for Electrical Engineering (EE) 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Continuity Equation below.
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Test: Continuity Equation - Question 1

### Find the current when the charge is a time function given by q(t) = 3t + t2 at 2 seconds.

Detailed Solution for Test: Continuity Equation - Question 1

Explanation: The current is defined as the rate of change of charge in a circuit ie, I = dq/dt. On differentiating the charge with respect to time, we get 3 + 2t. At time t = 2s, I = 7A.

Test: Continuity Equation - Question 2

### The continuity equation is a combination of which of the two laws?

Detailed Solution for Test: Continuity Equation - Question 2

Explanation:

The continuity equation in the context of electrical engineering is a way to express the principle of conservation of electric charge. This principle is derived from two of Maxwell's equations: Gauss's law for electricity and the Ampere-Maxwell law (which is Ampere's law including Maxwell's addition of the displacement current).

Gauss's law states that the electric charge creates an electric field and that the electric flux through a closed surface is proportional to the charge enclosed by the surface.

Ampere's law with Maxwell's addition links magnetic fields to the electric currents and changes in electric fields that produce them.

Thus, the continuity equation, which reflects the conservation of charge, is a consequence of these two laws.

I = ∫ J.ds is the integral form of Ohm’s law and Div (J) = dq/dt is the Gauss law analogous to D. Through these two equations, we get Div(J) = -dρ/dt. This is the continuity equation.

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Test: Continuity Equation - Question 3

### Calculate the charge density for the current density given 20sin x i + ycos z j at the origin.

Detailed Solution for Test: Continuity Equation - Question 3

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.

Test: Continuity Equation - Question 4

Compute the conductivity when the current density is 12 units and the electric field is 20 units. Also identify the nature of the material.

Detailed Solution for Test: Continuity Equation - Question 4

Explanation: The current density is the product of conductivity and electric field intensity. J = σE. To get σ, put J = 12 and E = 20. σ = 12/20 = 0.6. Since the conductivity is less than unity, the material is a dielectric.

Test: Continuity Equation - Question 5

Find the electron density when convection current density is 120 units and the velocity is 5m/s.

Detailed Solution for Test: Continuity Equation - Question 5

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.

Test: Continuity Equation - Question 6

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.

Detailed Solution for Test: Continuity Equation - Question 6

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.

Test: Continuity Equation - Question 7

Find the mobility of the electrons when the drift velocity is 23 units and the electric field is 11 units.

Detailed Solution for Test: Continuity Equation - Question 7

Explanation: The mobility is defined as the drift velocity per unit electric field. Thus μe = vd/E = 23/11 = 2.1 units.

Test: Continuity Equation - Question 8

Find the resistance of a cylinder of area 200 units and length 100m with conductivity of 12 units.

Detailed Solution for Test: Continuity Equation - Question 8

Explanation: The resistance is given by R = ρL/A = L/σA. Put L = 100, σ = 12 and A = 200, we get R = 100/(12 x 200) = 1/24 units.

Test: Continuity Equation - Question 9

Calculate the potential when a conductor of length 2m is having an electric field of 12.3units.

Detailed Solution for Test: Continuity Equation - Question 9

Explanation: The electric field is given by E = V/L. To get V, put E = 12.3 and L = 2.Thus we get V = E x L = 12.3 x 2 = 24.6 units.

Test: Continuity Equation - Question 10

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.

Detailed Solution for Test: Continuity Equation - Question 10

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

## Electromagnetic Fields Theory (EMFT)

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## Electromagnetic Fields Theory (EMFT)

11 videos|45 docs|73 tests