# RLC question (complicated)

by Tino
Tags: complicated
 P: 26 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
 P: n/a 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 ω0, Q and R. You have two equations and two unknowns: L and C.
 P: 26 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 ???
P: n/a
RLC question (complicated)

 Quote by Tino 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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