Electromagnetic theory miscellaneous
- If the diameter of a λ/2 dipole antenna is increased from λ/100 to λ/50 then its—
-
View Hint View Answer Discuss in Forum
See from synopsis provided that smaller diameter result in narrower bandwidth and larger diameter gives wider bandwidth.
Correct Option: A
See from synopsis provided that smaller diameter result in narrower bandwidth and larger diameter gives wider bandwidth.
- The frequency range for satellite communication is—
-
View Hint View Answer Discuss in Forum
NA
Correct Option: D
NA
- A-TEM wave incident normally upon a perfect conductor the E and H fields at the boundary will be respectively—
-
View Hint View Answer Discuss in Forum
At boundary the fields are in phase opposition 180º out of phase and whole energy is reflected back for perfect conductor. Hence E and H fields at the boundary will be maximum and minimum respectively.
Correct Option: D
At boundary the fields are in phase opposition 180º out of phase and whole energy is reflected back for perfect conductor. Hence E and H fields at the boundary will be maximum and minimum respectively.
- The magnitude of open circuit and short circuit input impedances of a transmission line are 100 Ω and 25 Ω respectively the characteristic impedance of the line is—
-
View Hint View Answer Discuss in Forum
Characteristic impedance is given in terms of short circuit and open circuit impedance Zc =√ZSCZOC = √25 × 100 = √2500 = 50Ω
Correct Option: B
Characteristic impedance is given in terms of short circuit and open circuit impedance Zc =√ZSCZOC = √25 × 100 = √2500 = 50Ω
- Find Q of the coil at resonance having frequency 1 MHz—
-
View Hint View Answer Discuss in Forum
At resonance XL = XC
jωL = 1 jωC
orL = 1 = 1 ω20C (2∏ × 106)2 × 50 × 10−12 = 1 50 × 4∏2 Q = ω20L = 106 × 2∏ × 1 R 10 50 × 4∏2 = 103 ∏ Correct Option: A
At resonance XL = XC
jωL = 1 jωC
orL = 1 = 1 ω20C (2∏ × 106)2 × 50 × 10−12 = 1 50 × 4∏2 Q = ω20L = 106 × 2∏ × 1 R 10 50 × 4∏2 = 103 ∏