A series RLC circuit has a 200 kHz resonance frequency. What is the resonance frequency if
a.) the resistor value is doubled?
b.) the capacitor value is doubled and, at the same time, the inductor is halved??
2007-12-16 20:22:41 · 6 answers · asked by biscuits 2 in Science & Mathematics ➔ Physics
I believe that in both cases, the frequency remains the same.
a) I don't think the resistance has an effect on the frequency in RLC circuits.
b) The equation for RLC frequency is:
f = (1/2pi)(1/sqrt(LC))
So as L is halved and C is doubled, LC stays the same. Thus, the frequency shouldn't change.
2007-12-16 20:37:11 · answer #1 · answered by Matt Soave 2 · 3⤊ 0⤋
The Resonant frequency of RLC circuit is
f =
1/(2xpi)Square root <( 1/(LC) + R^2/(4xL^2))>
So plug the values in and have a feel
2007-12-16 20:43:27 · answer #2 · answered by kay kay 4 · 0⤊ 1⤋
resonance frequency will not change in dis situation as
a) resistance has no effect on resonance frequency
b) doubled the value of capacitance, and halved the value of inductance will not affect the value LC,
as f'=1/2pi sq root of(1/LC)
SO RESONANT FREQUENCE REMAIN SAME AS BEFORE.
2007-12-16 20:56:15 · answer #3 · answered by ananta prasad c 1 · 0⤊ 1⤋
As long as your circuit is a theoretical one comprised of perfect components then the mathematically correct answers have been given by the previous posters.
If it is made of real components then the resonant frequency can change from 200kHz. Just something to be aware of :)
2007-12-16 21:47:59 · answer #4 · answered by frothuk 4 · 0⤊ 0⤋
The resonant frequency is given by f = (1/2π)√[1/(LC]
a) no effect
b) no effect
2007-12-16 20:39:44 · answer #5 · answered by gp4rts 7 · 1⤊ 0⤋
the impedance it given to you from the formula square root R^2+(XC-XL)^2 which should give you 360.44 ohms therefore Z multiplies by the amp of 233mA is 83.98V Hope this helps
2016-04-09 21:19:50 · answer #6 · answered by ? 4 · 0⤊ 0⤋