Electronics - RLC Circuits and Resonance - Discussion

Vr = IxR = 0.003 Ampx1000 Ohms = 3 Volts.

Now we have VL = 30V, Vc = 18V and Vr = 3V.

Applied voltage is given by.

Vs = SQRT of [Vr^2 + (VL-Vc) ^2].

= [3^2 + (30-18) ^2] = [9 + (12^2) ] = [9 + 144] = 153.

Therefore Vs = Square root of 153.

Vs = 12.369 Volts or Say 12.37 Volts here is the answer.

By tricky its sure answer will be A or B but why B tell me please.

Firstly the formula is v = IZ.

Xl = vl/I = 30/3*10^-3 = 10,000.

Xc = vc/I = 18/3*10^-3 = 6000.

Z = sqr of (R^2 + (xl-xc)).

Z = sqr of (1000^2 + (10000-6000)).

Z = sqr of (1004000) = 1001.998004.

Now,

V = 3 * 10^-3 * 1001.998004 = 3.005994.

Approximately ans = 3.00 v.

I don't know whether my answer is right or wrong. If I am wrong, please tell me my mistake.

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

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].

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

Thanks @ Singh Chouhan.

Thanks @Govind Singh Chouhan.

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.

Why can't we write kvl for this but I'm not getting answer.