College Physics-ELECTROMAGNETIC OSCILLATIONS & AC?

Question

I have 5 questions in all, here are the links to them:

http://i152.photobucket.com/albums/s167/gt_badone/31-58.jpg

http://i152.photobucket.com/albums/s167/gt_badone/31-59.jpg

http://i152.photobucket.com/albums/s167/gt_badone/31-60.jpg

http://i152.photobucket.com/albums/s167/gt_badone/31-61.jpg

http://i152.photobucket.com/albums/s167/gt_badone/31-98.jpg

I need formulas more than answers, if you are going to give me answers (don't need solution) be sure to provide formulas!

Answer 1

Ok, for the first problem, we don't need to solve the circuit, but we need to remember some theory :

For a AC, circuit : Ir = Im*sen(wt)

Im = Vm / R

P = Im^2*R*sen^2(wt)

That's for the resistor.

For the inductance :

L = Vm / Lw*sen(wt)

and : XL = W.L >>> inductive reactance

VL = Vmsen(wt) = Im.XL*sen(wt)

For the capacite :

ic = dQ / dt = CVmwcos(wt)

ic = wcVmsen(wt) >>> Vc = Vmsen(wt)

XC = 1 / wc >>>>> Capacitance reactance

So, for the whole circuit :

Vm = sqrt((VL - Vc)^2 + Vr^2))

Vr = Im*R
VL = Im*XL
Vc = Im*Xc

Vm = Im*sqrt((XL - Xc)^2 + R^2))

So for the first problem :

E = 40.1*sen(3910)t

Xc = 1 / wc = 1 / 3910*24.3*10^6 = 10.52 = Xc

XL = w.L = 3910*7.47*10^-3 = 29.2 = XL

Vm = E = 40.1

then using :

Vm = Im*sqrt((XL - Xc)^2 + R^2))

40.1 = Im*sqrt((29.2 -10.52)^2 + 23.9^2)

Im = 1.32 Amperes

Now using :

Vr = Im*R >>>> Vr = 1.32*23.9 = 31.5

VL = Im*XL>>>VL = 1.32*29.2 = 38.5

Vc = Im*Xc>>>Vc = 1.32*10.52 = 13.8

I = Im*sen(wt)

I = 1.32*sen(3910t)

for t = 0.526 s

we can find : P = Im^2*R*sen^2(wt)

P = 1.32^2*23.9*sen^2(3910*0.526)

Pr = 39.4 (Watts)

Now, let's find Pc :

I = 1.32*sen(3910t)

Pc = 1/2*C*E^2*sen^2(wt)

Pc = 1/2*24.3*10^-6*(40.1)^2*sen^2(3910*0.522) = 0.018 Watts.

Well, It was kind of long.

Second problem : Hope you understand, again :

Here you have to calculate : VL, Vc and Vr :

First of all, we calculate : Xc and XL

XL = W.L >>> inductive reactance

VL = Vmsen(wt) = Im.XL*sen(wt)

For the capacite :

ic = dQ / dt = cVmwcos(wt)

ic = wcVmsen(wt) >>> Vc = Vmsen(wt)

XC = 1 / wc

Vr = Im*R
VL = Im*XL
Vc = Im*Xc

Given : Frequency = 512 Hz >>> w = 1024*pi

R = 14,5

L = 25.4*10^-3

C = 3.75*10^-6

Vm = 104

V = 104sen(1024*pi*t)

XL = WL = 1024*pi*25.4*10^-3 = 81.7

Xc = 1 / wc = 1 / 1024*pi*3.75*10^-6 = 82.9

using : Vm = Im*sqrt((XL - Xc)^2 + R^2))

104 = Im*sqrt((XL - Xc)^2 + R^2))

Im = 7.14 Amperes

VL = 7.14*81.7 = 583

Vc = 7.14*82.9 = 592

Vr = 7.14*14.5 =103.53

That's it, you have the current, and the voltage in every component, C, L and R.

Third problem.-

This one is very easy, just remember :

N2*V1 = N1*V2

so then : 100*610 = 51*V2

V2 = 1196 volts

Fourth problem .-

Transformers :

N1 = 470 turns

N2 = 9.6 turns

Vp = 120

Using the same formula :

470*V2 = 9.6*120

V2 = 2.45 volts

rms V2 = 2.45sen(wt)

second part : Secondary has a resistor of 20 ohms

remeber : V1*I1 = V2*I2

secondary has a 20 ohm resistor, then : I2 = 2.45sen(wt) / 20

I2 = 0.12sen(wt)

then : 120sen(wt)*I1 = 0.12sen(wt)*2.45sen(wt)

I1 = 2.45*10^-3sen(wt) amperes

Fifth question .- Last problem

It's like the first problems, a series circuit, RLC :

We just have to find the Vc, VL, and Vr, the voltage on every part of the circuit :

XL = W.L

Xc = 1 / wc

w = 120*pi

R = 280

L = 270*10^-3

C = 20.4*10^-6

Vm = 42

V = 42sen(wt)

Xc = 1 / 120*pi*20.4*10^-6 = 130

XL = 120*pi*270*10^-3 = 101.8

Using : Vm = Im*sqrt((XL - Xc)^2 + R^2))

Im = 0.15 amperes

Vc = 0.15*130 = 19.5

VL = 0.15*101.8 =15.27

Vr = 0.15*280 = 42

Hope that might help you

Note : Im = maximum current

Vm = maximum voltage

C= capacitor o capacite