Electronics - RLC Circuits and Resonance - Discussion

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.

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

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.

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.

Vr = 1000*10^-3 = 3v.
VL = 30v.
vc = 18v.

V = square root of(3^2+(30-18)^2).
= square root of (9+144).
= square root of (153).
= 12.3691.

Answer = 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.

Can anyone please difference between impedance and resistance. Can we write v = i*z (z=impedance ).

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.

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].

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.