Electronics and Communication Engineering - Electronic Devices and Circuits
Exercise : Electronic Devices and Circuits - Section 8
- 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
46.
A two branch parallel tuned circuits has a coil of resistance R and inductance L in one branch and capacitance C in the second branch. The bandwidth is 200 radians/sec. A load of resistance RL is connected in parallel. The new bandwidth will be
Answer: Option
Explanation:
Loading increases bandwidth of parallel tuned circuii.
47.
The correct relation between joules and calories is
Answer: Option
Explanation:
Calories and joules are related by J, i.e., mechanical equivalent of heat.
48.
The filter in figure, uses ideal op-amp. It is
Answer: Option
Explanation:
Capacitor in shunt gives low pass filter.
49.
For a rectified sinusoidal wave clipped at 0.707 its peak value (figure), the average value is
Answer: Option
Explanation:
50.
A voltage v = 100 sin (100 pt + 45°) V is applied to an impedance 10 + j10 ohm. The current is
Answer: Option
Explanation:
Z = 200∠45°Ω ; 
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