Electronics and Communication Engineering - Networks Analysis and Synthesis - Discussion
Discussion Forum : Networks Analysis and Synthesis - Section 1 (Q.No. 38)
38.
The circuit in figure will act as ideal current source with respect to terminals A and B when frequency is
Answer: Option
Explanation:
When ω = 4 rad/sec ωL and 
are equal.
Discussion:
8 comments Page 1 of 1.
Umair said:
1 decade ago
Is it due to the phenomenon of resonance?
Vinodh said:
1 decade ago
Question related to resonance, concept is that imaginary part due to capacitor and inductor should not appear; clear from explanation.
Shyam said:
1 decade ago
But what is the answer? Is it 4 or 1?
Preeti said:
9 years ago
It is 4, because for ideal current source impedance is infinite. Consider given circuit impedance inductor is parallel to the capacitor and equating result to infinity we will get w=1/ωLC. Solve it we will get 4 rad/sec.
Saranya said:
8 years ago
Convert the volt source into the current source.
So inductor is parallel to the capacitor.
So, WL=1/wc.
Omega square equal to 1/LC. Put the values will get w = 4.
So inductor is parallel to the capacitor.
So, WL=1/wc.
Omega square equal to 1/LC. Put the values will get w = 4.
Pooja said:
7 years ago
Answer is 4rad/sec.
Spidey said:
3 years ago
The right Answer is 4rad/sec.
Prince said:
3 years ago
The Right answer is 4 rad/sec.
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