Test: Electric Potential


10 Questions MCQ Test Electromagnetic Theory | Test: Electric Potential


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

Potential difference is the work done in moving a unit positive charge from one point to another in an electric field. State True/False.

Solution:

Answer: a
Explanation: The electric potential is the ratio of work done to the charge. Also it is the work done in moving a unit positive charge from infinity to a point in an electric field.

QUESTION: 2

A point charge 2nC is located at origin. What is the potential at (1,0,0)?

Solution:

Answer: d
Explanation: V = Q/(4πεr), where r = 1m
V = (2 X 10-9)/(4πε x 1) = 18 volts

QUESTION: 3

Six equal point charges Q = 10nC are located at 2,3,4,5,6,7m. Find the potential at origin.

Solution:

Answer: d
Explanation: V = (1/4πεo) ∑Q/r = (10 X 10-9/4πεo)
(0.5 + 0.33 + 0.25 + 0.2 + 0.166 + 0.142) = 143.35 volts.

QUESTION: 4

A point charge 0.4nC is located at (2, 3, 3). Find the potential differences between (2, 3, 3)m and (-2, 3, 3)m due to the charge.

Solution:

Answer: c
Explanation: Vab = (Q/4πεo)(1/rA) + (1/rB), where rA and rB are position vectors rA = 1m and rB = 4m. Thus Vab = 2.7 volts.

QUESTION: 5

Find the potential of V = 60sin θ/r2 at P(3,60,25)

Solution:

Answer: a
Explanation: V = 60sin θ/r2, put r = 3m, θ = 60 and φ = 25, V = 60 sin 60/32 = 5.774 volts.

QUESTION: 6

Given E = 40xyi + 20x2j + 2k. Calculate the potential between two points (1,-1,0) and (2,1,3).

Solution:

Answer: b
Explanation: V = -∫ E.dl = -∫ (40xy dx + 20x2 dy + 2 dz), from (2,1,3) to (1,-1,0), we get Vpq on integrating from Q to P. Vpq = 106 volts.

QUESTION: 7

The potential difference in an open circuit is

Solution:

Answer: c
Explanation: In an open circuit no current exists due to non-existence of loops. Also voltage/potential will be infinity in an open circuit.

QUESTION: 8

The potential taken between two points across a resistor will be

Solution:

Answer: b
Explanation: The resistor will absorb power and dissipate it in the form of heat energy. The potential between two points across a resistor will be negative.

QUESTION: 9

What is the potential difference between 10sinθcosφ/r2 at A(1,30,20) and B(4,90,60)?

Solution:

Answer: c
Explanation: Potential at A, Va = 10sin30cos20/12 = 4.6985 and Potential at B, Vb = 10sin90cos60/42 = 0.3125. Potential difference between A and B is, Vab = 4.6985 – 0.3125 = 4.386 volts.

QUESTION: 10

The voltage at any point in an ac circuit will be

Solution:

Answer: b
Explanation: In any ac circuit, the voltage measured will not be exact maximum. In order to normalise, we assume the instantaneous voltage at any point be 70.7% of the peak value, which is called the root mean square (RMS)voltage.

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