# 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

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

Let me know if you’ve nay query.

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