# RL Parallel Circuit - Help Needed

#### 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.

#### 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.

#### 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 ...

$$\omega_{0}ser=\frac{1}{\sqrt{LC}}$$

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

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

$$where$$

$$K=\frac{\omega_{0}par}{\omega_{0}ser}$$

$$I have$$

$$R_{L}=101.655\Omega$$

$$C=99.2nF$$

$$L=50.42mH$$

#### femaleelec

Joined Feb 26, 2010
3
what equations have you used? im totally lost....

#### 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.