Physics, 16.04.2020 03:00 leelee85503

A series RLC circuit with L = 12 mH, C = 3.5 mu or micro FF, and R = 3.3 ohm is driven by a generator with a maximum emf of 115 V and a variable angular frequency omega. (a) Find the resonant (angular) frequency omega0. rad/s [1.43 points] 0 attempt(s) made (maximum allowed for credit = 10) (b) Find Irms at resonance. A [1.43 points] 0 attempt(s) made (maximum allowed for credit = 10) When the angular frequency omega = 7600 rad/s, (c) Find the capacitive reactance XC in ohms. ohm [1.43 points] 0 attempt(s) made (maximum allowed for credit = 10) Find the inductive reactance XL in ohms. ohm [1.43 points] 0 attempt(s) made (maximum allowed for credit = 10) (d) Find the impedance Z. (Give your answer in ohms.) ohm [1.43 points] 0 attempt(s) made (maximum allowed for credit = 10) Find Irms. A [1.43 points] 0 attempt(s) made (maximum allowed for credit = 10) (e) Find the phase angle (in degrees). degrees

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Answer from: agray062103

In this exercise we have to know about the series circuits and calculate the following parts:

A. f=800Hz

B. Irms=0.028A

C. $X_c=56.8 \Omega$

D. $X_l=60.3 \Omega$

E. $\phi= 1.51$

Given data:

Inductance: $L=(12)*(10^{-3})H$ Capacitance: $C= (3.5)*(10^{-6})F$ Resistance: $R= 3.3 \ \Omega$ Voltage: V=115V

A. Resonance frequency, in RLC circuit resonance occurs when capacitive reactance equals inductive reactance:

$f=\frac{1}{2\pi} \sqrt{LC} \\f=\frac{1}{2(3.142)} \sqrt{12*10^{-3}*3.5*10^{-6}}\\f=\frac{1}{(6.284)*(0.0002)} \\f=1/0.00125\\f=800Hz$

B. Find Irms at resonance:

$Irms=\frac{R}{V}\\Irms=\frac{3.3}{115} \\Irms=0.028A$

C. Find the capacitive reactance (XC) in Ohms:

$X_c=\frac{1}{2\pi fC} \\X_c= \frac{1}{2(3.142)(800)(3.5)(10^{-6})}\\X_c=1/0.0176\\X_c=56.8 \Omega$

D. To find the inductive reactance:

$X_l=2\pi fL\\X_l=(2)*(3.142)*(800)*(12*10^{-3})\\X_l=60.3 \Omega$

E. Find the impedance (Z)

$Z=\sqrt{R^2+(X_I-X_c)^2}\\Z=\sqrt{(3.3)^2+(60.3-56.8)^2} \\Z=\sqrt{(10.89+12.25}\\Z=\sqrt{23.14}\\Z=4.8 \Omega$

For Phase angle:

$Tan(\phi)=X_c/R\\=56.8/3.3\\Tan(\phi)=17.2\\\phi=Tan-1 (17.2)\\\phi= 1.51$

See more about series circuit at link

Answer from: graycelynn123

Explanation:

Given data

Inductance L=12*10^-³H

Capacitance C= 3.5*10^-6F

Resistance R= 3.3 Ohms

Voltage V=115v

Capacitive reactance Xc=?

inductive reactance Xl=?

Impedance Z=?

Phase angle =?

A. Resonance frequency

In RLC circuit resonance occurs when capacitive reactance equals inductive reactance

f=1/2pi √ LC

f=1/2*3.142 √ 12*10^-³*3.5*10^-6

f=1/6.284*0.0002

f=1/0.00125

f=800HZ

B. Find Irms at resonance.

Irms=R/V

Irms=3.3/115

Irms=0.028amp

Find the capacitive reactance XC in Ohms

Xc=1/2pi*f*C

Xc=1/2*3.142*800*3.5*10^-6

Xc=1/0.0176

Xc=56.8 ohms

To find the inductive reactance

Xl=2pifL

Xl=2*3.142*800*12*10^-3

Xl=60.3ohms

d) Find the impedance Z.

Z=√R²+(Xl-Xc)²

Z=√3.3²+(60.3-56.8)²

Z=√10.89+12.25

Z=√23.14

Z=4.8ohms

Phase angle =

Tan phi=Xc/R=56.8/3.3

Tan phi=17.2

Phi=tan-1 17.2

Phi= 1.51°

Answer from: Quest

would be false hope this out

Answer from: Quest

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