Electronics - RLC Circuits and Resonance - Discussion

How can the KVL is not applicable in this series circuit?

Like there are 3 components in this circuit if we add 3 voltages that will equal the voltage applied according to Kirchoff's voltage law. Hence the 2 voltages of the inductor and capacitor are given(30+18=48v) and we add the resistance voltage after calculating by Ohm's law (3v) to get the total voltage of 51 volts, hence the applied voltage should also be 51 volts so as not to violate KVL.

correct me if I am wrong.

Thanks @Govind Singh Chouhan.

Thanks @ Singh Chouhan.

V^2 = Vr^2+(VL-VC)^2.
V^2 = 3^2+(30-18)^2,
=9+144,
V^2= 153.

V = 12.369.

The voltage drop across the resistor is
E = I * R.
E = 0.003 * 1000,
E = 3 volts.

The reactive volts are;
VX = VL - VC.
VX = 30 - 18.
VX = 12 volts.

The applied volts is;
Volts = √(VR^2 + VX^2),
Volts = √(3^2 + 12^2),
Volts = √(153),
Volts = 12.3693.

V=IR=3*10^-3*1000=3V,
3^2+12^2=151,
= 12.37.

Well said, Thanks, @Govind Singh Chouhan.

Vr = i * r =3mA * 1k = 3v
Vs = √(Vr^2 + (Vc - Vl)^2).
Vs = √(3^2 + (30 - 18)^2) = 12.36.

Create voltare trangle.

v^2 = (VL-VC)^2 + VR^2
So,
VR= I * R
VR= 3mA*1000Ohms
VR=3V
V^2=(XL-XC)^2+VR^2,
=(30-18)^2+3^2,
=12^2 + 9,
=144 + 9,
=> V^2 = 153.
V = 12.36.

VL = 30 <90, VC =1 8<-9 0, VR = 3.
hence V = sqrt(12^2 + 3^2) = 12.37.