RLC question (complicated)

Tino

I find this question very hard and i could not work out the answers, i have looked on the internet for forumals but they don't seem to work ??
Anyone got any methods to work through the question ???

I found a website that is about RLC series circuits but looking into it, it really confuses me with all the equations:

http://www.tpub.com/neets/book2/4l.htm

An RLC series circuit has a resonant frequency of 1.2kHz and Q factor at resonance of 50. If the impedance of the circuit is 75 ohms calculate the values of:

a. the inductance vaule
b. the capacitance value
c. the bandwidth
d. the lower and upper half-power frequencies
e. the vaules of the circuit impedance at the half power frequencies.

Thanks

Last edited:
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SGT

The resonant frequency of a LC circuit is:
$$\omega_0 = \frac{1}{\sqrt{LC}}$$
And the quality factor is:
$$Q = \omega_0 RC$$
Since it is given a real value for the impedance, we must assume that this value is calculated at the resonant frequency.
Z = R.
So, you know &omega;0, Q and R. You have two equations and two unknowns: L and C.

Tino

done it !!!

ok here is the equation i used

Q = XL/R = XC/R

Q x R = XL 50 x 75 = 3750 ohms

Q x R = XC 50 x 75 = 3750 ohms

then just use XL = 2pie FL
and
XC = 1 / 2pie FC

transpose that !!!!

L = 2pie F / XL

C = 1 / 2pie F XC

But one more question, what equation can i use to get the lower and upper half power frequencies ???

SGT

Tino said:
ok here is the equation i used

Q = XL/R = XC/R

Q x R = XL 50 x 75 = 3750 ohms

Q x R = XC 50 x 75 = 3750 ohms

then just use XL = 2pie FL
and
XC = 1 / 2pie FC

transpose that !!!!

L = 2pie F / XL

C = 1 / 2pie F XC

But one more question, what equation can i use to get the lower and upper half power frequencies ???
You have:
$$Q = \frac {\omega_0}{\omega_2 - \omega_1}$$
Where $$\omega_0$$ is the resonant frequency and $$\omega_1$$, $$\omega_2$$ the half power frequencies.

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