Electronics - RLC Circuits and Resonance - Discussion

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].

Given data:

R = 1000R.
Vc = 18V.
VL = 30V.
I = 3mA.

We know that,

Z=sqrt((R²)+(XL-XC)²)

And XL = VL/I ==> 30/0.003 = 10K.
And XC = VC/I ==> 18/0.003 = 6K.

So know put the value of XL and XC in in formula of Z so we can say that:

Z = sqrt(1000²+(10K-6K)²)
Z = sqrt(1000000+16000000)
Z = sqrt(17000000)
Z = 4123 or
Z = 4.123K.

Now it applied voltage to the circuit is:

V=I*Z
V=0.003*4123
V=12.369 or 12.37V.

So applied voltage to this circuit is 12.37V.

Thanks.

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.

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.

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.

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.

If we do them as if they are easy they will be easy...n if we think them to be difficult they will be difficult.....ALL THE BEST FRNDS...!!!!!!!!!

V=IZ
z=impedence
Z=(R^2+(XL-XC)^2)^1/2
XL=VL/I=10000
XC=VC/I=6000 ----> Z=4123.1056
V=4123.1056*3*10^-3
V=12.369

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.

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?

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 = Sqrt(VR^2 + VX^2).
Volts = Sqrt (3^2 + 12^2).
Volts = Sqrt(153).
Volts = 12.3693.