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?
Discussion:
37 comments Page 3 of 4.
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
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
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
=squrt(squr(3*1000*0.001)+squr(30-18))
=squrt(9+144)
=12.37V
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 :-).
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?
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?
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.
R= 1000 ohms, so voltage across "R" is 1000*3mA = 3volt.
So applied voltage = 30V + 18V + 3V = 51V.
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.
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.
Yugandhar said:
1 decade ago
VR=3v;
vl=30v;
vc=18v;
vi=(3^2+(30-18)^2)^1/2;
vl=30v;
vc=18v;
vi=(3^2+(30-18)^2)^1/2;
Manikanta said:
1 decade ago
Why can't we write kvl for this but I'm not getting answer.
Stanley said:
1 decade ago
Lets train ourselves to use SI units always, by so doing you will not mess. Thanks.
Ankur said:
1 decade ago
We have relation
Vm= (VR^2+(VL-VC)^2)^1/2
VR= I*R=3*10^-3 * 1000 = 3 volts
VL= 30 volts
VC= 18 volts
therefore, Vm= 12.37
Vm= (VR^2+(VL-VC)^2)^1/2
VR= I*R=3*10^-3 * 1000 = 3 volts
VL= 30 volts
VC= 18 volts
therefore, Vm= 12.37
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