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

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

A.

B.

C.

D.

E.

Given data:

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

B. Find Irms at resonance:

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

D. To find the inductive reactance:

E. Find the impedance (Z)

For Phase angle:

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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°

the simple machine has three main features. they are:

fulcrum is the center point or pivot point of the machine. effort is input force/ applied force. load is external force/ output.

the keisha 's machine labels are

x : load

y: fulcrum

z: lever

a is the best answer your welcome