How Do You Calculate RLC Circuit Parameters and Half-Power Frequencies?

  • Thread starter Thread starter Tino
  • Start date Start date
  • Tags Tags
    Rlc
Click For Summary
Calculating RLC circuit parameters involves using the resonant frequency and quality factor to derive inductance and capacitance values. The resonant frequency is given as 1.2 kHz with a Q factor of 50, and the impedance is 75 ohms. The equations Q = XL/R and Q = XC/R help find the reactance values, leading to the formulas for inductance (L) and capacitance (C). To determine the lower and upper half-power frequencies, the relationship Q = ω0/(ω2 - ω1) is used, where ω0 is the resonant frequency. Understanding these equations allows for effective calculation of circuit parameters.
Tino
Messages
25
Reaction score
0
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:
Engineering news on Phys.org
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.
 
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 ?
 
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.
 
I am trying to understand how transferring electric from the powerplant to my house is more effective using high voltage. The suggested explanation that the current is equal to the power supply divided by the voltage, and hence higher voltage leads to lower current and as a result to a lower power loss on the conductives is very confusing me. I know that the current is determined by the voltage and the resistance, and not by a power capability - which defines a limit to the allowable...

Similar threads

  • · Replies 32 ·
2
Replies
32
Views
3K
  • · Replies 40 ·
2
Replies
40
Views
6K
  • · Replies 9 ·
Replies
9
Views
6K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 7 ·
Replies
7
Views
4K
  • · Replies 3 ·
Replies
3
Views
4K
Replies
3
Views
2K
Replies
2
Views
3K
  • · Replies 12 ·
Replies
12
Views
3K
  • · Replies 8 ·
Replies
8
Views
3K