Notes - Fisika - Alternating Current

ƒ=1/T

y=A*sin(ω*t)

E=(E_m)*sin(ω*t)=N⋅B⋅A⋅ω⋅sin(ω*t)

Where:

• ƒ is frequency

• T is period

• A is amplitude

• ω is phase(?)

Oscilloscope measures maximum voltage, voltmeter measures effective voltage.

(V_E)=(V_max)/√(,2)

(I_E)=(I_max)/√(,2)

Effective voltage is basically the amount of DC voltage needed to match the AC voltage.

How do you read an oscilloscope?

The vertical axis uses a scale based on volts per centimeter.

(V_max)=(s_y)⋅A

The horizontal axis uses a scale based on miliseconds per centimeter.

T=(s_x)⋅λ

Simple Examples

Problem 1

(V_max)=80⋅3=240

T=2.5⋅8=20ms=20/1000=1/50 (in seconds)

ƒ=1/1/50=50Hz

(B)

Problem 2

(V_max)=25⋅2=50

(V_E)=50/√(,2)=25√(,2)

T=4⋅1/10=2/5=2/(5⋅1000)=1/2500 (in seconds)

ƒ=2500Hz

(D)

Problem 3

Subproblem 3.1

(V_max)=4⋅10=40

(V_E)=40/√(,2)=20√(,2)

Subproblem 3.2

T=8⋅2=16ms=16/1000=2/125 (in seconds)

ƒ=1/2/125=125/2 (Hz)

Subproblem 3.3

Amplitudonya jadi 4⋅10/20=2 kotak

Subproblem 3.4

Periodenya jadi 8⋅2/5=16/5=3.2 kotak

Rangkaian AC

I=(I_max)*sin(ω*t)

V=(V_max)*sin(ω*t)

Resistors still resist.

V=L⋅d(I)/d(t)=L⋅d()/d(t)⋅(I_max)⋅sin(ω*t)=L⋅(I_max)⋅ω⋅cos(ω*t)

V=(V_max)⋅cos(ω*t)

Inductors resist. ω*L is the resistance (x_L). ω=2*π*ƒ

Recall that I=(I_max)*sin(ω*t). This means that inductors resist and move the phase of the voltage. The voltage starts 90° earlier.

I=(I_max)⋅sin(ω*t)

V=(V_max)⋅sin(ω*t+90°)

If you draw the phasor diagram, the voltage will always be more anticlockwise.

Rangkaian R-L

V=√(,(V_R)^2+(V_L)^2)

(I*Z)^2=(I*R)^2+(I*(x_L))^2

Z=√(,R^2+(x_L)^2)

I=(V_R)/R=(V_L)/(x_L)=V/Z

tan(ϕ)=(V_L)/(V_R)=(x_L)/R

I=(I_max)*sin(ω*t)

V=(V_max)*sin(ω*t+ϕ)

Circuit Examples

Problem 1

Subproblem 1.1

(x_L)=ω*L=50⋅0.1=5Ω

Subproblem 1.2

Z=√(,12^2+5^2)=13Ω

Subproblem 1.3 and 1.4

tan(ϕ)=(x_L)/R=5/12

I=(260*sin(50*t))/13=20*sin(50*t-22.6198649°)

(I_max)=20

(I_E)=20/√(,2)=10√(,2)

Problem 2

Subproblem 2.1

ω=2*π*50=100*π

V=100*sin(100*π*t)

10=100/R,R=10Ω

8=100/(x_L),(x_L)=12.5Ω

(x_L)=40/50⋅12.5=10Ω

Resistance of inductors is affected by the frequency.

Z=√(,10^2+10^2)=10√(,2)

I=V/Z=150/(10√(,2))=15/2√(,2)

Subproblem 2.2

(V_R)=15/2√(,2)⋅10=75√(,2)

(V_L)=15/2√(,2)⋅10=75√(,2)

Problem 3

Oh my god bruh.

R=60/2=30Ω

Z=100/2=50Ω

50^2=30^2+(x_L)^2

(x_L)=40Ω=2*π⋅50⋅L

L=2/(5*π)

Problem 4

Subproblem 4.1

tan(45°)=(x_L)/25

(x_L)=25Ω=ω*L=50⋅L

L=1/2

Subproblem 4.2

Z=√(,25^2+25^2)=25√(,2)

2*sin(50*t-45°)=V/(25√(,2))

V=50√(,2)*sin(50*t-45°)

But don't forget about the phase shift of +45°.

V=50√(,2)*sin(50*t)

Capacitors

q=C⋅V

d(q)=C*d(V)

d(q)/d(t)=C⋅d(V)/d(t)

(I_E)=C⋅d(V)/d(t)

d(V)=1/C⋅I*d(t)

I=(I_max)*sin(ω*t)

(∫_^)(d(V))=V=1/C⋅(∫_^)(I*d(t))=1/C*(∫_^)((I_max)*sin(ω*t)*d(t))=-1/C(I_max)/ω*cos(ω*t)

V=(I_max)/(ω*C)⋅-cos(ω*t)=(V_max)⋅-cos(ω*t)=(V_max)⋅sin(ω*t-90°)

(x_C)=1/(ω*C)

q adalah muatan, Coulomb

C adalah capacitance, Farad

(x_C) adalah reaktansi kapasitif

Rangkaian R-C

it's the same, but with different symbols.

Ex. 1

Subproblem 1.1

250⋅10^(-6)=C⋅250

C=1/1000000=10^(-6)

(x_C)=1/(100⋅250⋅10^(-6))=40Ω

Subproblem 1.2

Z=√(,40^2+30^2)=50

Subproblem 1.3

I=(250*sin(100*t))/50=5*sin(100*t)

∴(I_max)=5A

Subproblem 1.4

tan(θ)=40/30=4/3=53°

I=5*sin(100*t-53°)

Ex. 2

62.5 coulomb per volt

ω=50

V=200*sin(50*t)

Subproblem 2.1

(x_C)=1/(50⋅62.5⋅10^(-6))=320Ω

Z=√(,320^2+240^2)=400Ω

Subproblem 2.2

I=(200*sin(50*t))/400=1/2*sin(50*t)

tan(θ)=320/240=4/3

θ=53°

I=1/2*sin(50*t-53°)

Subproblem 2.3

(V_R)=240⋅1/2=120

(V_C)=320⋅1/2=160

Ex. 3

(V_R)=15

(V_max)=25

Subproblem 3.1

(I_max)=(V_R)/R=15/30=1/2

Subproblem 3.2

25=√(,15^2+(V_C)^2)

(V_C)=20=I*(x_C)

20=1/2*(x_C)

(x_C)=40Ω=1/(ω*C)=1/(100*C)

40*C=1/100

C=1/4000=2.5⋅10^(-4)

Rangkaian R-L-C

In the phasor diagram, (V_R) and I have an angle of 0°. (V_L) has an angle of 90° and (V_C) has an angle of 270°. This means that (V_L) and (V_C) have opposing directions. This is essential to understanding the formula below, making sure to simplify it instead of using vectors directly.

V=√(,(V_R)^2+((V_L)-(V_C))^2)

I⋅Z=√(,(I*R)^2+(I*(x_L)-I*(x_C))^2)

Z=√(,R^2+((x_L)-(x_C))^2)

I=V/Z=(V_L)/(x_L)=(V_R)/R=(V_C)/(x_C)

tan(θ)=((x_L)-(x_C))/R

(x_L)>(x_C)→(V_L)>(V_C) bersifat induktif, V mendahului sebesarθ

(x_C)>(x_L)→(V_C)>(V_L) bersifat kapasitif, V tertinggal sebesar θ

(x_L)=(x_C)→(V_C)=(V_L) terjadi resonansi, maka ƒ^2=1/(4*π^2*L*C)→ƒ=1/(2*π√(,L*C))

Ex. 1

ω=100

(x_C)=1/(100⋅125⋅10^(-6))=80Ω

(x_L)=100⋅1.6=160Ω

Z=√(,80^2+(160-80)^2)=80√(,2) (a)

160*sin(100*t)=I⋅80√(,2)

I=√(,2)*sin(100*t)

tan(θ)=80/80=1

∴θ=45°

It's inductive btw, so use the minus sign.

I=√(,2)*sin(100*t-45°) (b)

(I_E)=√(,2)/√(,2)=1

P=1^2⋅80=80 (c)

ƒ=1/(2*π√(,1.6⋅125⋅10^(-6)))=(25√(,2))/π (d)

Ex. 2

wow

(V_R)=80

80=I⋅160

I=0.5

(V_R*L)=100=√(,(V_R)^2+(V_L)^2)=√(,80^2+(V_L)^2)

(V_L)^2=100^2-80^2

(V_L)=60

(x_L)=(V_L)/I=60/0.5=120Ω

Yippee

100=√(,80^2+((V_L)-(V_C))^2)

100^2=80^2+((V_L)-(V_C))^2

((V_L)-(V_C))^2=60^2

(V_L)-(V_C)=60

(V_C)-(V_L)=60

(V_C)=120

(x_C)=120/0.5=240Ω

Ex. 3

V=100√(,2)*sin(100*t)

ω=100

(x_L)=2⋅100=200Ω

(x_C)=1/(10⋅100⋅10^(-6))=1000Ω

tan(θ)=(200-1000)/600=-4/3

θ=-53° (a)

Sifat rangkaian kapasitif karena (x_L)<(x_C) (b)

Z=√(,600^2+(1000-200)^2)=1000Ω

(V_max)=(I_max)*Z

100√(,2)=(I_max)⋅1000

(I_max)=√(,2)/10

∴I=√(,2)/10*sin(100*t+53°) (c)

(I_E)=√(,2)/10⋅1/√(,2)=1/10

(V_L)=1/10⋅200=20 (d)

(V_C)=1/10⋅1000=100 (d)

(V_R)=1/10⋅600=60 (d)

P=1/(10^2)⋅600=6W (e)