A series RLC circuit consists of a 60.0 resistor, a 5.10 mH inductor, and a 310

Question

A series RLC circuit consists of a 60.0 resistor, a 5.10 mH inductor, and a 310 nF capacitor. It is connected to an AC source with a peak voltage of 5.70 V and a frequency of 3 kHz.

What is the angular frequency of the AC source?

Determine the reactance of the inductor.

Determine the reactance of the capacitor.

Determine the impedance of the circuit.

Determine the peak current

Determine phase angle.

Find the current in the circuit at t=35 s .

Find the voltage across the resistor at t=35 s.

Find the voltage across the inductor at t=35 s.

Find the voltage across the capacitor at t=35 s .

Explanation / Answer

Here ,

R = 60 Ohm

L = 5.1 mH

C = 510 nF

f = 3 kHz = 3000 Hz

Vp = 5.7 V

angular frequency of the AC source = 2pi * f

angular frequency of the AC source = 2pi * 3000 Rad/s

angular frequency of the AC source = 18846 rad/s

--------------------

reactance of inductor = w * L

reactance of inductor = 18846 * 5.1 *10^-3

reactance of inductor = 96.11 Ohm

------------------------------

reactance of capacitor = 1/(w * C)

reactance of capacitor = 1/(18846 * 310 * 10^-9)

reactance of capacitor = 171.2 Ohm

------------------------------
impedance of the circuit = sqrt(R^2 + (Xl - Xc)^2)

impedance of the circuit = sqrt(60^2 + (96.11 - 171.2)^2)

impedance of the circuit = 96.1 Ohm

---------------------------------

for the peak current

peak current = peak voltage/impedance

peak current = 5.7/(96.1)

peak current = 0.0593 A

the peak current is 0.0593 A

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