Electronics and Communication Engineering - Electronic Devices and Circuits
Exercise : Electronic Devices and Circuits - Section 5
- 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
16.
For a high pass filter attenuation = 0 at f = 0.
Answer: Option
Explanation:
Attenuation is infinite at f = 0 for high pass filter.
17.
Current having waveform in figure flows through 10 Ω resistance. Average power is
Answer: Option
Explanation:
RMS current = 
= 5.773 A, P = I2R = 333.3 W.
18.
S is closed for a long time and steady state is reached. S is opened at t(0-). The voltage marked V is V0 at t = 0+ and Vf at t = ∞. Then value of V0 and Vf are respectively
Answer: Option
Explanation:
For Thevenin Equivalent Resistance, v = 0, I = 0
.
19.
Two voltages are v1= 100 sin (ωt + 15°) and v2 = 60 cosωt, then
Answer: Option
Explanation:
v1 = 100 sin (ωt + 15°) and v2 = 60 cos ωt = 60 sin (ωt + 90°). Hence, v2 is leading v1 by (90 - 15) = 75°.
20.
If A = 16 ∠64°, (A)0.5 is
Answer: Option
Explanation:
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