Test: Power & Poynting Vector


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12 Questions MCQ Test Electromagnetic Fields Theory (EMFT) | Test: Power & Poynting Vector

Test: Power & Poynting Vector for Electrical Engineering (EE) 2023 is part of Electromagnetic Fields Theory (EMFT) preparation. The Test: Power & Poynting Vector questions and answers have been prepared according to the Electrical Engineering (EE) exam syllabus.The Test: Power & Poynting Vector MCQs are made for Electrical Engineering (EE) 2023 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Power & Poynting Vector below.
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Test: Power & Poynting Vector - Question 1

The power of the electromagnetic wave with electric and magnetic field intensities given by 12 and 15 respectively is

Detailed Solution for Test: Power & Poynting Vector - Question 1

Answer: b
Explanation: The Poynting vector gives the power of an EM wave. Thus P = EH/2. On substituting for E = 12 and H = 15, we get P = 12 x 15/2 = 90 units.

Test: Power & Poynting Vector - Question 2

The power of a wave of with voltage of 140V and a characteristic impedance of 50 ohm is

Detailed Solution for Test: Power & Poynting Vector - Question 2

Answer: c
Explanation: The power of a wave is given by P = V2/2Zo, where V is the generator voltage and Zo is the characteristic impedance. on substituting the given data, we get P = 1402/(2×50) = 196 units.

Test: Power & Poynting Vector - Question 3

The power reflected by a wave with incident power of 16 units is(Given that the reflection coefficient is 0.5)

Detailed Solution for Test: Power & Poynting Vector - Question 3

Answer: d
Explanation: The fraction of the reflected to the incident power is given by the reflection coefficient. Thus Pref = R2xPinc. On substituting the given data, we get Pref = 0.52 x 16 = 4 units.

Test: Power & Poynting Vector - Question 4

The power transmitted by a wave with incident power of 16 units is(Given that the reflection coefficient is 0.5)

Detailed Solution for Test: Power & Poynting Vector - Question 4

Answer: a
Explanation: The fraction of the transmitted to the incident power is given by the reflection coefficient. Thus Pref = (1-R2) Pinc. On substituting the given data, we get Pref = (1- 0.52) x 16 = 12 units. In other words, it is the remaining power after reflection.

Test: Power & Poynting Vector - Question 5

The incident and the reflected voltage are given by 15 and 5 respectively. The transmission coefficient is

Detailed Solution for Test: Power & Poynting Vector - Question 5

Answer: b
Explanation: The ratio of the reflected to the incident voltage is the reflection coefficient. It is given by R = 5/15 = 1/3. To get the transmission coefficient, T = 1 – R = 1 – 1/3 = 2/3.

Test: Power & Poynting Vector - Question 6

The current reflection coefficient is given by -0.75. Find the voltage reflection coefficient.

Detailed Solution for Test: Power & Poynting Vector - Question 6

Answer: d
Explanation: The voltage reflection coefficient is the negative of the current reflection coefficient. For a current reflection coefficient of -0.75, the voltage reflection coefficient will be 0.75.

Test: Power & Poynting Vector - Question 7

The attenuation is given by 20 units. Find the power loss in decibels.

Detailed Solution for Test: Power & Poynting Vector - Question 7

Answer: a
Explanation: The attenuation refers to the power loss. Thus the power loss is given by 20 units. The power loss in dB will be 10 log 20 = 13.01 decibel.

Test: Power & Poynting Vector - Question 8

The reflection coefficient is 0.5. Find the return loss.

Detailed Solution for Test: Power & Poynting Vector - Question 8

Answer: c
Explanation: The return loss is given by RL = -20log R, where is the reflection coefficient. It is given as 0.5. Thus the return loss will be RL = -20 log 0.5 = 6.02 decibel.

Test: Power & Poynting Vector - Question 9

The radiation resistance of an antenna having a power of 120 units and antenna current of 5A is

Detailed Solution for Test: Power & Poynting Vector - Question 9

Answer: a
Explanation: The power of an antenna is given by Prad = Ia2 Rrad, where Ia is the antenna current and Rrad is the radiation resistance. On substituting the given data, we get Rrad = Prad/Ia2 = 120/52 = 4.8 ohm.

Test: Power & Poynting Vector - Question 10

The transmission coefficient is given by 0.65. Find the return loss of the wave.

Detailed Solution for Test: Power & Poynting Vector - Question 10

Answer: a
Explanation: The transmission coefficient is the reverse of the reflection coefficient, i.e, T + R = 1. When T = 0.65, we get R = 0.35. Thus the return loss RL = -20log R = -20log 0.35 = 9.11 decibel.

Test: Power & Poynting Vector - Question 11

The return loss is given as 12 decibel. Calculate the reflection coefficient.

Detailed Solution for Test: Power & Poynting Vector - Question 11

Answer: c
Explanation: The return loss is given by RL = -20log R. The reflection coefficient can be calculated as R = 10(-RL/20), by anti logarithm property. For the given return loss RL = 12, we get R = 10(-12/20) = 0.25.

Test: Power & Poynting Vector - Question 12

Find the transmission coefficient of a wave, when the return loss is 6 decibel.

Detailed Solution for Test: Power & Poynting Vector - Question 12

Answer: a
Explanation: The return loss is given by RL = -20log R. The reflection coefficient can be calculated as R = 10(-RL/20), by anti logarithm property. For the given return loss RL = 6, we get R = 10(-6/20) = 0.501. The transmission coefficient will be T = 1 –
R = 1-0.501 = 0.498.

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