Electronics - RLC Circuits and Resonance - Discussion

Discussion Forum : RLC Circuits and Resonance - General Questions (Q.No. 1)
1.
What is the applied voltage for a series RLC circuit when IT = 3 mA, VL = 30 V, VC = 18 V, and R = 1000 ohms?
3.00 V
12.37 V
34.98 V
48.00 V
Answer: Option
Explanation:
No answer description is available. Let's discuss.
Discussion:
37 comments Page 3 of 4.

Surya said:   1 decade ago
Can anyone please difference between impedance and resistance. Can we write v = i*z (z=impedance ).

Govind Singh Chouhan said:   1 decade ago
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.

Swapnil said:   1 decade ago
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.

Vinit said:   1 decade ago
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.

ANIL SINGH said:   10 years ago
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.

Mahu said:   10 years ago
By tricky its sure answer will be A or B but why B tell me please.

Leelavathi said:   9 years ago
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.

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

Weerasinghege maduka said:   9 years ago
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.

Ch manikantha said:   9 years ago
Vr = i * r =3mA * 1k = 3v
Vs = √(Vr^2 + (Vc - Vl)^2).
Vs = √(3^2 + (30 - 18)^2) = 12.36.


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