Multisim rms current and voltage RLC Circuit

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SUMMARY

The discussion focuses on calculating the RMS current and voltage in an RLC circuit using Multisim. The circuit consists of a resistor (R = 500 ohm), an inductor (L = 0.40 mH), and a capacitor (C = 100 pF). The resonant frequency is calculated to be 795.77 kHz, with an RMS current of 0.002 A and RMS voltages across the resistor (Vr = 1V), inductor (Vl = 4V), and capacitor (Vc = 3.99998 V). The user seeks assistance in verifying these calculations within the Multisim environment.

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  • Knowledge of calculating resonant frequency using the formula f = 1/(2π√(LC))
  • Ability to measure RMS voltage and current in a circuit
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k31453
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Trying to find rms current at resonance and rms voltage across each elements (resistor, inductor and capacitor)

I got the answers but i want it do in multisim !

i got :

R = 500 ohm
L = 0.40mH
C = 100pF

The answer is

rms current: 0.002A
rms voltage:
Vr = 1V
Vc = 126.52 V
Vl = 126.52 V

The resonant freq. is 25.16 Khz at 1v RMS ac source.


this is my attempt !

Homework Statement


Homework Equations


The Attempt at a Solution

 
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How did you arrive at the resonant frequency? It looks a bit small to me, given the stated component values.
 
gneill said:
How did you arrive at the resonant frequency? It looks a bit small to me, given the stated component values.


sorry this is the answers

A) resonant frequency
ans: f = 1/2*pi*sqrt(LC)
= 1/(2*pi*sqrt(0.0004 * 100 * 10^-12)
= 795.77 Khz

b) the inductive and capacitive reactance at resonance
ans: XL = 2*pi*f*L
= 2*pi* 795.775*10^3*0.0004
= 2000 ohm

Xc = 1/(2*pi*f*C)
= 1/(2*pi* 795.775*10^3 * 100 * 10^-12)
= 1999.9998 ohm

C) the impedance at resonance

ans : impedence of resonance is resistance of circuit : z = R = 500 ohm

D) the rms current at resonance
ans: I = V/R = 1/500 = 0.002A

E) the rms voltage across each element at resonance
Ans: Vr = IR
= 0.002 * 500
= 1V

VL = IXL
= 0.002 * 2000
= 4V

Vc = IXc
= 0.002 * 1999.99
= 3.99998 V
 
Yes, that looks better.
 
gneill said:
Yes, that looks better.

but still i can't prove into multisim !
 
k31453 said:
but still i can't prove into multisim !

I don't have Multisim, so I can't help with the details of using it. Sorry.

But looking at the image you've provided, I would say that there sure are a lot of multimeters inserted, presumably to measure current. Surely one would suffice for a series circuit? Also, it appears that the inductor's top end is open circuited and it is being bypassed by a multipmeter (to measure voltage?). I'd suggest removing all but one multimeter to measure the series current, and make sure that the circuit is unbroken and operational before adding multimeters in parallel with the components to measure their voltages.
 

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