# Electromagnetic Theory Mcqs

Q:

A) 4 | B) 1 |

C) 7 | D) 1/4 |

Answer & Explanation
Answer: B) 1

Explanation: For a lossless antenna, the input power and the radiated power will be same. The efficiency will be 100%. Thus the radiated power will also be 1 milliwatt.

Explanation: For a lossless antenna, the input power and the radiated power will be same. The efficiency will be 100%. Thus the radiated power will also be 1 milliwatt.

7

Q:

A) To increase the sensitivity of measurement | B) To transmit the signal to a far off place |

C) To study amplitude modulation | D) Because crystal detector fails at microwave frequencies |

Answer & Explanation
Answer: D) Because crystal detector fails at microwave frequencies

Explanation: Since crystal detector fails at microwave frequencies, microwave signal amplitude is modulated at 1kHz.

Explanation: Since crystal detector fails at microwave frequencies, microwave signal amplitude is modulated at 1kHz.

5

Q:

A) E = +8y, H = -4z | B) E = -2y, H = -3z |

C) E = +2z, H = +2y | D) E = -3y, H = +4z |

Answer & Explanation
Answer: B) E = -2y, H = -3z

Explanation: To get wave travelling in the positive direction, the E component has to be in negative y direction and the H component has to be in negative z direction. Thus E = -2y and H = -3z is the right option.

Explanation: To get wave travelling in the positive direction, the E component has to be in negative y direction and the H component has to be in negative z direction. Thus E = -2y and H = -3z is the right option.

3

Q:

A) Complex | B) Purely capacitive |

C) Purely resistive | D) Purely inductive |

Answer & Explanation
Answer: C) Purely resistive

Explanation: Given, voltage minimum point is at load. If minimum voltage or maximum voltage occurs at the load for a lossless transmission line then the load impedance is purely resistive.

Explanation: Given, voltage minimum point is at load. If minimum voltage or maximum voltage occurs at the load for a lossless transmission line then the load impedance is purely resistive.

5

Q:

A) Must be a lossless line | B) Must be a distortionless line |

C) May not be a lossless line | D) May be a lossless and may not be a distortionless line |

Answer & Explanation
Answer: D) May be a lossless and may not be a distortionless line

Explanation: When the characteristic line is a pure resistance, the line must be a distortionless line. But it may not be a lossless line.

Explanation: When the characteristic line is a pure resistance, the line must be a distortionless line. But it may not be a lossless line.

4

Q:

A) 1 | B) 0 |

C) Infinity | D) +j |

Answer & Explanation
Answer: C) Infinity

Explanation: The reflection coefficient is given by R = j50 – 50/j50 + 50 = j – 1/j + 1. Thus the reflection coefficient is given by 1. The VSWR = 1 + R/1 – R = 2/0 = infinity.

Explanation: The reflection coefficient is given by R = j50 – 50/j50 + 50 = j – 1/j + 1. Thus the reflection coefficient is given by 1. The VSWR = 1 + R/1 – R = 2/0 = infinity.

5

Q:

A) Impedance | B) Admittance |

C) Reflection coefficient and VSWR | D) Intrinsic impedance |

Answer & Explanation
Answer: D) Intrinsic impedance

Explanation: From the Smith chart, the parameters impedance, admittance, reflection coefficient and VSWR can be computed directly. The intrinsic impedance cannot be calculated.

Explanation: From the Smith chart, the parameters impedance, admittance, reflection coefficient and VSWR can be computed directly. The intrinsic impedance cannot be calculated.

6

Q:

A) Radius as well as operating wavelength are halved | B) Radius as well as operating wavelength are doubled |

C) Radius is halved and operating wavelength is doubled | D) Radius is doubled and operating wavelength is halved |

Answer & Explanation
Answer: D) Radius is doubled and operating wavelength is halved

Explanation: The cut off frequency is given by f = 2πa sin α/wavelength, where a is the radius of core. For a single mode step index fiber, f must lie between 0 and 2.405. Thus to get f below 2.405, the radius must be doubled and wavelength must be halved.

Explanation: The cut off frequency is given by f = 2πa sin α/wavelength, where a is the radius of core. For a single mode step index fiber, f must lie between 0 and 2.405. Thus to get f below 2.405, the radius must be doubled and wavelength must be halved.

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