Question

A series RLC circuit consists of a 52.0 Ω resistor, a 4.80 mH inductor, and a 330 nF capacitor. It is connected to an oscillator with a peak voltage of 6.00 V . Part G
Determine the impedance at frequency 5000 Hz.
Part H
Determine the peak current at frequency 5000 Hz.
Part I
Determine phase angle at frequency 5000 Hz.

Answer 1

Answer:

(G) 75.11 ohm

(H) 0.08 A

(I) 46.2 degree

Explanation:

R = 52 ohm

L = 4.8 m H = 4.8 x 106-3 H

C = 330 nF = 330 x 10^-9 F

Vo = 6 V

(G)

f = 5000 Hz

Let the impedance is Z.

$X_{L}= 2 \pi fL = 2 \times 3.14\times 5000\times 4.8\times 10^{-3}=150.72 ohm$

$X_{c}= \frac{1}{2 \pi fC}=\frac{1}{2\times 3.14\times 5000\times 330\times 10^{-9}}=96.51 ohm$

$Z=\sqrt{R^{2}+\left ( X_{L}-X_{c} \right )^{2}}$

$Z=\sqrt{52^{2}+\left (150.72-96.51)^{2}}=75.11 ohm$

(H) Let Io be the peak current

$I_{0}=\frac{V_{0}}{Z}=\frac{6}{75.11}=0.0798 A = 0.08 A$

(I) Let Ф be the phase angle

$tan\phi = \frac{X_{L}-X_{C}}{R}$

$tan\phi =\frac{150.72-96.51}}{52}=1.0425$

Ф = 46.2 degree