Electronics and Communication Engineering - Networks Analysis and Synthesis
Exercise : Networks Analysis and Synthesis - Section 9
- Networks Analysis and Synthesis - Section 14
- Networks Analysis and Synthesis - Section 27
- Networks Analysis and Synthesis - Section 26
- Networks Analysis and Synthesis - Section 25
- Networks Analysis and Synthesis - Section 24
- Networks Analysis and Synthesis - Section 23
- Networks Analysis and Synthesis - Section 22
- Networks Analysis and Synthesis - Section 21
- Networks Analysis and Synthesis - Section 20
- Networks Analysis and Synthesis - Section 19
- Networks Analysis and Synthesis - Section 18
- Networks Analysis and Synthesis - Section 17
- Networks Analysis and Synthesis - Section 16
- Networks Analysis and Synthesis - Section 15
- Networks Analysis and Synthesis - Section 1
- Networks Analysis and Synthesis - Section 13
- Networks Analysis and Synthesis - Section 12
- Networks Analysis and Synthesis - Section 11
- Networks Analysis and Synthesis - Section 10
- Networks Analysis and Synthesis - Section 9
- Networks Analysis and Synthesis - Section 8
- Networks Analysis and Synthesis - Section 7
- Networks Analysis and Synthesis - Section 6
- Networks Analysis and Synthesis - Section 5
- Networks Analysis and Synthesis - Section 4
- Networks Analysis and Synthesis - Section 3
- Networks Analysis and Synthesis - Section 2
26.
If V = 4 in the figure, the value of Is is given by
Answer: Option
Explanation:
At node V1
at node V2
2V2 - V1 = 0 V1 = 2V2
V2 = 4 V1 = 8 Is = 
x 8 - 4 = 6A.
27.
A current of 1 A in the coil of an iron cored electromagnet causes B = 0.5T. If current is 2A, B =
Answer: Option
Explanation:
If saturation starts B will be less than it.
28.
Two current sources each of rms value 10 A have frequencies 40 Hz and 50 Hz respectively. They are connected in series, the expression for resultant wave is
Answer: Option
Explanation:
Peak value = 14.14 A; ω1 = 2p x 40, ω2 = 2p x 50
29.
To find the current in a 20 Ω resistance connected in a circuit, Norton's theorem is used. IN = 7.5 A. The current though 20 Ω resistance.
Answer: Option
Explanation:
Norton's equivalent is as shown in figure. Some current will flow through RN. There-fore, current through 20Ω resistance is less than 7.5 A.
30.
While calculating Norton's resistance, all current sources are short circuited.
Answer: Option
Explanation:
All current sources are open circuited.
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