Question

An RLC series circuit has a 1.00 kΩ resistor, a 160 mH inductor, 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 V_rms = 408 V, what is I_rms (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 I_rms (in mA) at resonance?

mA

†

Additional Materials

• Reading

Answer 1

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 = 2.516 kHz

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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