Electronics - RLC Circuits and Resonance - Discussion

Discussion Forum : RLC Circuits and Resonance - General Questions (Q.No. 1)
1.
What is the applied voltage for a series RLC circuit when IT = 3 mA, VL = 30 V, VC = 18 V, and R = 1000 ohms?
3.00 V
12.37 V
34.98 V
48.00 V
Answer: Option
Explanation:
No answer description is available. Let's discuss.
Discussion:
37 comments Page 4 of 4.

Sukumar said:   1 decade ago
According to Lalit Garg
Givendata:-
Vl=30;Vc=18;It=0.03A;
Xl=Vl/It--->30/0.03=1000
Xc=Vc/It--->18/0.03=600
Z=(R^2+(Xc-Xl)^2)^1/2
then z=1077.03
Vsource=Z*It=32.31V
but this is not valid with options.

Biru said:   1 decade ago
Given voltage across inductor = 30v and capacitor=18v.

R= 1000 ohms, so voltage across "R" is 1000*3mA = 3volt.

So applied voltage = 30V + 18V + 3V = 51V.

Saikrishna said:   1 decade ago
Given voltage across inductor = 30v and capacitor=18v.

R= 1000 ohms, so voltage across "R" is 1000*3mA = 3volt.

So applied voltage = 30V + 18V + 3V = 51V.

Because as the answer says 12.37volts. If source is such less how can the drop of 30v occurs at inductor?

Badar said:   1 decade ago
Because votlage across inductor and capacitor are opposite in phase about 180 degree. They cancel the effect of each other :-).

Nilesh said:   1 decade ago
vs=squrt(squr(It*r)+squar(Vl-Vc))
=squrt(squr(3*1000*0.001)+squr(30-18))
=squrt(9+144)
=12.37V

ANSHUMAN said:   1 decade ago
Vl=30;Vc=18;It=0.03A;
Xl=Vl/It--->30/0.03=1000
Xc=Vc/It--->18/0.03=600
Z=(R^2+(Xc-Xl)^2)^1/2
then z=1077.03
Vsource=Z*It=32.31V

Prakash said:   1 decade ago
We have to find impedance because we have been asked to find out AC voltage source. Impedance is the resistance to AC voltage, thats why we take out impedance (Z) with the help of reactance of both L and C.

Rest all the calculation have been explained by others.


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