Electronics and Communication Engineering - Analog Electronics - Discussion
Discussion Forum : Analog Electronics - Section 5 (Q.No. 15)
15.
In the following figure C = 0.02 μF, and Vth is known to be of frequency ω = 107 rad/sec and rd = 2.5 Ω and ZTh = RTh = 10 Ω Then phase angle between id and Vd
Answer: Option
Explanation:
Zd = (rd)11 (-jxd) 
2 - JΩ
θ = tan-1 (-1/2) 26.56 .
Discussion:
3 comments Page 1 of 1.
Rozalind said:
6 years ago
For me, I just solved for Xc then,
2.5 - 5i.
Then got,
5.6 < -63.43,
Then 90 - 63.43 and got that.
2.5 - 5i.
Then got,
5.6 < -63.43,
Then 90 - 63.43 and got that.
Satyendra sehgal said:
5 years ago
Zd = 2.5-j5 (rd=2.5 and xd=5).
0 = Tan-1(-2.5/5) and solved it.
0 = Tan-1(-2.5/5) and solved it.
Shailesh said:
1 decade ago
I can't understand calculation of Zd? Can someone help me?
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