Electronics and Communication Engineering - Networks Analysis and Synthesis
Exercise : Networks Analysis and Synthesis - Section 13
- Networks Analysis and Synthesis - Section 14
- Networks Analysis and Synthesis - Section 27
- Networks Analysis and Synthesis - Section 26
- Networks Analysis and Synthesis - Section 25
- Networks Analysis and Synthesis - Section 24
- Networks Analysis and Synthesis - Section 23
- Networks Analysis and Synthesis - Section 22
- Networks Analysis and Synthesis - Section 21
- Networks Analysis and Synthesis - Section 20
- Networks Analysis and Synthesis - Section 19
- Networks Analysis and Synthesis - Section 18
- Networks Analysis and Synthesis - Section 17
- Networks Analysis and Synthesis - Section 16
- Networks Analysis and Synthesis - Section 15
- Networks Analysis and Synthesis - Section 1
- Networks Analysis and Synthesis - Section 13
- Networks Analysis and Synthesis - Section 12
- Networks Analysis and Synthesis - Section 11
- Networks Analysis and Synthesis - Section 10
- Networks Analysis and Synthesis - Section 9
- Networks Analysis and Synthesis - Section 8
- Networks Analysis and Synthesis - Section 7
- Networks Analysis and Synthesis - Section 6
- Networks Analysis and Synthesis - Section 5
- Networks Analysis and Synthesis - Section 4
- Networks Analysis and Synthesis - Section 3
- Networks Analysis and Synthesis - Section 2
31.
In a series RLC circuit the dB level of currents at half power frequencies as compared to current at resonance is
Answer: Option
Explanation:
At half power frequencies, current is - 3 dB.
32.
A series RL circuit is initially relaxed. A step voltage is applied to the circuit. If t is the time constant of the circuit, the voltage across R and L will be same at time t equal to
t ln 2
Answer: Option
Explanation:
i = i0 (1 - e-R/Lt) and VR = VL at a given time t.
33.
In figure, the transmission parameters are A = C = 1, B = 2 and D = 3. Then Zin =
Answer: Option
Explanation:
34.
A sinusoidal voltage is written as v = Im (R2 + XC2)0.5 sin 314 t. Its peak value and frequency are
Answer: Option
Explanation:
v2 = sin (ωt + 90°)
35.
If A = 10 + j then logeA is
Answer: Option
Explanation:
A = 14.14Ð45°, loge A = loge 14.14 x 
= 2.65 + j 0.785.
Quick links
Quantitative Aptitude
Verbal (English)
Reasoning
Programming
Interview
Placement Papers