Electronics and Communication Engineering - Electronic Devices and Circuits
Exercise : Electronic Devices and Circuits - Section 13
- Electronic Devices and Circuits - Section 14
- Electronic Devices and Circuits - Section 27
- Electronic Devices and Circuits - Section 26
- Electronic Devices and Circuits - Section 25
- Electronic Devices and Circuits - Section 24
- Electronic Devices and Circuits - Section 23
- Electronic Devices and Circuits - Section 22
- Electronic Devices and Circuits - Section 21
- Electronic Devices and Circuits - Section 20
- Electronic Devices and Circuits - Section 19
- Electronic Devices and Circuits - Section 18
- Electronic Devices and Circuits - Section 17
- Electronic Devices and Circuits - Section 16
- Electronic Devices and Circuits - Section 15
- Electronic Devices and Circuits - Section 1
- Electronic Devices and Circuits - Section 13
- Electronic Devices and Circuits - Section 12
- Electronic Devices and Circuits - Section 11
- Electronic Devices and Circuits - Section 10
- Electronic Devices and Circuits - Section 9
- Electronic Devices and Circuits - Section 8
- Electronic Devices and Circuits - Section 7
- Electronic Devices and Circuits - Section 6
- Electronic Devices and Circuits - Section 5
- Electronic Devices and Circuits - Section 4
- Electronic Devices and Circuits - Section 3
- Electronic Devices and Circuits - 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
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.
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