Electronics - RLC Circuits and Resonance - Discussion

VR=3v;
vl=30v;
vc=18v;
vi=(3^2+(30-18)^2)^1/2;

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.

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.

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?

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

vs=squrt(squr(It*r)+squar(Vl-Vc))
=squrt(squr(3*1000*0.001)+squr(30-18))
=squrt(9+144)
=12.37V

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

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.

I Would like to give my solution to this problem which is as follow:-

we know that voltage = current*Impedance < V = I*Z >,

We know that Impedance= [Square Root of {R^2 +Square of ((Inductive resistance)-(Capacitive reactance)).

Mathematically "Impedance","Z" can be written as Z = [{R^2+(XL-XC)^2}]^2,

We know that Voltage = Current*Impedance,

Mathematically "Voltage","V"; V = I*Z.

Now I will come to the solution:Z = [(1000^2+(10000-60000^2]^1/2,

V = (4123.105626)(3*10^-3), V = 12.36931688.

Therefore answer to the solution is V = 12.37, [b].

From phasor diagram it is clear that VL > Vc

V = sqrt[Vr^2+(VL-Vc)^2].

On putting the given value on formula as given unit we get exact answer.