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
31.
In the circuit of figure the current through 5 Ω resistance at t = ∞ is
Answer: Option
Explanation:
The 5 ohm resistance is short-circuited by inductance.
32.
In an infinite ladder Ckt as shown above each resistance of rΩ then RAB
∞
1.61 Ω
none
Answer: Option
Explanation:
Let RAB = R
R = r + r || R 
R = 1.61r.
33.
A series resonant circuit has R = 2 Ω, L = 0.1 H and C =10 μF. At resonance applied voltage = 10 ∠0 V. Voltage across inductance is likely to be
Answer: Option
Explanation:
Voltage across inductance = QV and leads the applied voltage.
34.
The admittance parameter Y12 in the 2 port network in the figure
Answer: Option
Explanation:
I1 = y11 V1 + y12 V2 y12 = 
V1 = 0
If V1 = 0 then
y12 = 0.1 mho.
35.
The differential equation for the current i(t) in the Ckt is
Answer: Option
Explanation:
Apply KVL
.
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