An RLC series circuit has a 1.
An RLC series circuit has a 1.00 kΩ resistor, a 160 mHinductor, and a 25.0 nF capacitor.
(a)
Find the circuit’s impedance (in Ω) at 490 Hz.
12539.57588 Ω
(b)
Find the circuit’s impedance (in Ω) at 7.50 kHz.
6765.310603 Ω
(c)
If the voltage source has Vrms = 408 V, whatis Irms (in mA) at each frequency?
mA (at 490 Hz)
mA (at 7.50 kHz)
(d)
What is the resonant frequency (in kHz) of the circuit?
kHz
(e)
What is Irms (in mA) at resonance?
mA
†
Additional Materials
- Reading
Answer:
Part A.
Impedance of RLC circuit is given by:
Z = sqrt (R^2 + (XL – Xc)^2)
XL = Inductive reactance = w*L = 2*pi*f*L
Xc = Capacitive reactance = 1/(w*C) = 1/(2*pi*f*C)
So,
Z = sqrt (R^2 + (2*pi*f*L – 1/(2*pi*f*C))^2)
Using given values:
Z = sqrt (1000^2 + (2*pi*490*160*10^-3 -1/(2*pi*490*25.0*10^-9))^2)
Z = 12539.57 Ohm
Part B
when f = 7.50 kHz = 7500 Hz
Using given values:
Z = sqrt (1000^2 + (2*pi*7500*160*10^-3 -1/(2*pi*7500*25.0*10^-9))^2)
Z = 6765.31 Ohm
Part C.
Using ohm’s law:
Vrms = Irms*Z
Irms = Vrms/Z
Given that Vrms = 408 V
So, when f = 490 Hz
Irms = 408/12539.57 = 0.032537
Irms = 32.537*10^-3 A = 32.537 mA
when f = 7500 Hz
Irms = 408/6765.31 = 0.060308
Irms = 60.308*10^-3 A = 60.308 mA
Part D.
At resosnance frequency:
XL = Xc
2*pi*f0*L = 1/(2*pi*f0*C)
f0 = 1/(2*pi*sqrt (L*C))
f0 = 1/(2*pi*sqrt (160*10^-3*25.0*10^-9))
f0 = 2516.46 Hz
f0 = resonance frequency = 2.516*10^3 Hz = 2.516kHz
Part E.
Since at resonance XL = Xc, So
Z = sqrt (R^2 + (XL – Xc)^2) = sqrt (R^2 + 0^2) = R
Z = 1000 ohm
So,
Irms = Vrms/Z
Irms = 408/1000 = 408*10^-3 A
Irms = 408 mA
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