Electronics and Communication Engineering - Electronic Devices and Circuits
Exercise : Electronic Devices and Circuits - Section 4
- 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
41.
In active filters inductances are divided because they
Answer: Option
Explanation:
Active filters consist of electronic components and have very small size.
42.
Three resistance of 15 Ω each are connected in delta. The resistance of equivalent star will have a value of
Answer: Option
Explanation:
43.
The circuit in figure(1) is replaced by that in figure(2). It remains same R0 will be
Answer: Option
Explanation:
Net resistance of circuit is 4.571 R and that of circuit is 0.571 R.
Hence R0 be 4R.
44.
In a varistor the v-i graph is linear.
Answer: Option
Explanation:
The v-i graph, in varistor, is non-linear.
45.
In the circuit of figure the current through 5 Ω resistance at t = 0+ is
Answer: Option
Explanation:
Since 3 ohm resistance is short-circuited by capacitor, I = 20/5 A.
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