RL Parallel Circuit - Help Needed

Thread Starter

femaleelec

Joined Feb 26, 2010
3
I have the following question, can anyone help....


When placed in series a coil and capacitor resonate at a frequency of 2251hz and when placed in paralle at a frequency of 2228hz. At Parallelresonance is it found that a current of 5mA flows from the supply when the supply voltage is 25v r.m.s. find values of capacitor, inductance and resistance of the coil.

:confused: please help!
 

hgmjr

Joined Jan 28, 2005
9,029
Can you post your efforts at solving the problem? That way we can see how you set up the problem for solution.

hgmjr
 

Kaija

Joined Feb 26, 2010
1
Do you have any answers to check?
I tried to solve the problem according to your text and got these results:
L=0.979861101 H
C=5.10181567 e-9 F
R=37 638.75695 Ohms (the resistance is in series with the inductor in parallel resonance circuit)
L and C seem ok, but can an inductor have such high resistance... ?

Anyway, try to make a system of equations. You have three variables and you can make three equations.
 

Thread Starter

femaleelec

Joined Feb 26, 2010
3
Do you have any answers to check?
I tried to solve the problem according to your text and got these results:
L=0.979861101 H
C=5.10181567 e-9 F
R=37 638.75695 Ohms (the resistance is in series with the inductor in parallel resonance circuit)
L and C seem ok, but can an inductor have such high resistance... ?

Anyway, try to make a system of equations. You have three variables and you can make three equations.

what equations have you used? im totally lost....
 

t_n_k

Joined Mar 6, 2009
5,455
You'll need equations like ...

\(\omega_{0}ser=\frac{1}{\sqrt{LC}}\)

\(\omega_{0}par=\frac{1}{\sqrt{LC}}\sqrt{1-\frac{R^2C}{L}}\)

\(Rp_{resonance}=\frac{R_{L}}{(1-K^{2})}\)

\(where\)

\(K=\frac{\omega_{0}par}{\omega_{0}ser}\)

\(I have \)

\(R_{L}=101.655\Omega\)

\(C=99.2nF\)

\(L=50.42mH\)
 

t_n_k

Joined Mar 6, 2009
5,455
The value of the effective circuit resistance at parallel resonance is the key. From this and the series / parallel resonance frequencies you can find the values of R, L & C.

The effective resistance at parallel resonance is found from the condition that 5mA current flows from the 25V source.

So at parallel resonance Rp=25V/5mA=5kΩ

The value of K is given by K=ω0par/ω0ser=f0par/f0ser=2228/2251=0.98978

So the equivalent series resistance RL of the coil can now be found from this information.

Then you use the value of RL and the known series & parallel equations to find the unknown values of L & C.
 
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