Test: Power & Poynting Vector - Electrical Engineering (EE) MCQ

# Test: Power & Poynting Vector - Electrical Engineering (EE) MCQ

Test Description

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

Test: Power & Poynting Vector for Electrical Engineering (EE) 2024 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) 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Power & Poynting Vector below.
Solutions of Test: Power & Poynting Vector questions in English are available as part of our Electromagnetic Fields Theory (EMFT) for Electrical Engineering (EE) & Test: Power & Poynting Vector solutions in Hindi for Electromagnetic Fields Theory (EMFT) course. Download more important topics, notes, lectures and mock test series for Electrical Engineering (EE) Exam by signing up for free. Attempt Test: Power & Poynting Vector | 12 questions in 10 minutes | Mock test for Electrical Engineering (EE) preparation | Free important questions MCQ to study Electromagnetic Fields Theory (EMFT) for Electrical Engineering (EE) Exam | Download free PDF with solutions
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

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

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.

 1 Crore+ students have signed up on EduRev. Have you?
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

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

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

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

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

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

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

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

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

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

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.

## Electromagnetic Fields Theory (EMFT)

11 videos|45 docs|73 tests
Information about Test: Power & Poynting Vector Page
In this test you can find the Exam questions for Test: Power & Poynting Vector solved & explained in the simplest way possible. Besides giving Questions and answers for Test: Power & Poynting Vector, EduRev gives you an ample number of Online tests for practice

## Electromagnetic Fields Theory (EMFT)

11 videos|45 docs|73 tests