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Honest Reviews of Tim Han LMA Course Program.pptxtimhan337
Personal development courses are widely available today, with each one promising life-changing outcomes. Tim Han’s Life Mastery Achievers (LMA) Course has drawn a lot of interest. In addition to offering my frank assessment of Success Insider’s LMA Course, this piece examines the course’s effects via a variety of Tim Han LMA course reviews and Success Insider comments.
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Macroeconomics- Movie Location
This will be used as part of your Personal Professional Portfolio once graded.
Objective:
Prepare a presentation or a paper using research, basic comparative analysis, data organization and application of economic information. You will make an informed assessment of an economic climate outside of the United States to accomplish an entertainment industry objective.
1. ARUS DAN TEGANGAN
BOLAK-BALIK (AC)
DEFINISI :
Arus listrik yang mengalir dalam dua arah dan nilainya selalu berubah-ubah sesuai dengan fungsi waktu.
Contoh : AC, kulkas, kompor listrik, jaringan PLN, Generator
FISIKA KELAS XII SMK
Oleh : Sisca Septiyani, S.Pd
2. RANGKAIAN ARUS BOLAK-BALIK ( AC )
DEFINISI
Pengertian arus listrik AC atau alternating current yaitu listrik yang
besar dan arah arusnya yang selalu berubah-ubah atau bolak-balik.
Listrik arus AC akan membentuk gelombang yang biasa dinamakan
dengan gelombang sinusoida. Dan di Indonesia sendiri, arus AC ini
dikelola dan berada di bawah penguasaan PLN.
MANFAAT ARUS AC
1. Digunakan untuk menghidupkan peralatan rumah tangga
(lampu penerangan, kulkas, pesawat TV, mesin cuci,
mixer, kipas angin).
2. Penerangan jalan raya.
3. Menghidupkan mesin-mesin untuk kebutuhan pabrik.
4. Pengamanan jaringan listrik di rumah
3. ARUS DAN TEGANGAN SINUSODIAL
Perbedaan Tegangan AC dan Dc jika kita amati dengan menggunakan alat ukur osiloskop adalah sebagai berikut :
Sumber arus bolak-balik adalah generator AC yang dapat menghasilkan ggl induksi sebesar :
dan
I = Imaks sin 𝝎 𝒕
4. CONTOH SOAL
1. Sebuah sumber tegangan memiliki persamaan V = 50 𝟐 sin 100 t, besarnya tegangan efektif dan dan kecepatan sudut terhadap t adalah…
Diket : Vmaks = Vm = 50 𝟐, sudut fase = 𝝎𝐭 = 𝟏𝟎𝟎, Dit = Vef dan 𝝎𝒕 ?
Jawab : Vef =
𝑽𝒎𝒂𝒌𝒔
𝟐
=
50 𝟐
𝟐
= 50 volt, 𝝎𝒕 = 100 t
2. Suatu arus bolak – balik mempunyai bentuk persamaan I = 12 sin 200 𝝅𝒕, Besar frekuensi arus bolak – balik adalah…
Diket = Im = 12, 𝝎𝒕 = 200 𝝅𝒕, Dit = f ?
Jawab = 𝝎 = 2𝝅𝒇
200 𝝅 = 2𝝅𝒇
f = 200 𝝅 /2𝝅
f = 100 Hz
5. ARUS DAN TEGANGAN EFEKTIF
Pada tegangan AC terdapat tegangan puncak dan tegangan efektif. Tegangan puncak yaitu tegangan maksimal dari listrik AC sedangkan tegangan efektif
yaitu tegangan yang terukur saat diukur dengan voltmeter. Hubungan matematis antara tegangan puncak atau tegangan max dengan tegangan efektif yaitu :
Keterangan :
Vef = tegangan efektif (Volt)
Vmaks = tegangan maksimum (Volt)
Ief = Arus efektif (A)
Imaks = arus maksimum (A)
Ief =
𝑰𝒎𝒂𝒌𝒔
𝟐
atau Imaks = Ief 𝟐
6. CONTOH SOAL
1. Sebuah sumber tegangan memiliki persamaan I = 25 𝟐 sin 200 t, Besarnya kuat arus maksimum dan efektif adalah…
Diket = Imaks = 25 𝟐, 𝝎𝒕 = 𝟐𝟎𝟎 𝒕
Dit = Im dan Ief ?
Jawab : Ief =
𝑰𝒎
𝟐
Im = 25 𝟐 A
Ief =
25 𝟐
𝟐
Ief = 25 A
7. RANGKAIAN ARUS BOLAK - BALIK
1. REAKTANSI INDUKTOR DAN KAPASITOR 2. RESISTOR
Ket : I = kuat arus (A)
R = hambatan (ohm)
Vmaks = Imaks . R
V = I . R
8. CONTOH SOAL
1. Perhatikan gambar RLC seri berikut ! Jawab : XL = ω.L
XL = 100.8 = 800 ohm
XC =
𝟏
ω.𝑪
XC =
𝟏
𝟏𝟎𝟎.𝟓𝑿𝟏𝟎−𝟔
XC =
𝟏
𝟓𝒙 𝟏𝟎𝟐.𝟏𝟎−𝟔
XC =
𝟏
𝟓 𝒙 𝟏𝟎−𝟒 = 1/5 x 𝟏𝟎𝟒
Besar Reaktansi induktif dan Reaktansi kapasitif adalah ? XC = 10000/5 = 2000 ohm
Diket = R =500 Ω Dit : XL dan XC ?
L = 8 H
C = 5 μF = 5 x 𝟏𝟎−𝟔
F
ω = 100 rad/s vm = 50
9. HUBUNGAN IMPEDANSI TEGANGAN DAN ARUS BOLAK-BALIK
1. Impedansi adalah hambatan atau reaksi pada rangkaian arus bolak-balik
Hubungan antara R, L, C dan Z dapat dinyatakan
dalam suatu diagram yang dinamakan diagram pashor.
Hubungan XR, XL. Dan XC di gambarkan dalam
suatu system sumbu koordinat seperti pada gambar:
Z = 𝑹𝟐 + (𝑿𝑳 − 𝑿𝑪)𝟐
Type equation here.
10. CONTOH SOAL
1. Perhatikan gambar RLC seri berikut !
Jawab = XL = 800 ohm
XC = 200
a. Z = √(R2 + (XL - XC)2)
Tentukan besar impedansi dan tegangan efektif ! Z = 𝟓𝟎𝟎𝟐 + (𝟖𝟎𝟎 − 𝟐𝟎𝟎𝟎)𝟐
Diket : R =500 Ω Dit = Z dan Vef ? Z = 𝟐𝟓𝟎𝟎𝟎𝟎 + (−𝟏𝟐𝟎𝟎)𝟐
L = 8 HL Z = 𝟏𝟔𝟗𝟎𝟎𝟎𝟎 = 1300 ohm
C = 5 μF b. Vef = vm/ 𝟐
ω = 100 rad/s Vef = 50/ 𝟐 = 25 𝟐 Volt
11. IMPEDANSI, TEGANGAN DAN ARUS BOLAK-BALIK
1. IMPEDANSI 2. TEGANGAN TOTAL
Z = 𝑹𝟐 + 𝑿𝑳𝟐
Z = 𝑹𝟐 + 𝑿𝑪𝟐
Tegangan
V = 𝑽𝑹𝟐 + 𝑽𝑳𝟐
V = 𝑽𝑹𝟐 + 𝑽𝑪𝟐
Vmaks = Im . Z
V = I . Z
12. CONTOH SOAL
1. Sebuah rangkaian AC dengan tegangan sumber sebesar 125 Volt diberi beban berupa resistor sebesar 375 Ω dan sebuah kapasitor yang
menghasilkan reaktansi kapasitif sebesar 500 Ω. Tentukanlah nilai impedansi serta arus yang dapat mengalir pada rangkaian tersebut!
Diketahui : Jawab : Z = 𝑹𝟐 + 𝑿𝑪𝟐
V = 125 Volt Z = 𝟑𝟕𝟓𝟐 + 𝟓𝟎𝟎𝟐
R = 375 Ohm Z = 𝟏𝟒𝟎𝟔𝟐𝟓 + 𝟐𝟓𝟎𝟎𝟎𝟎
XC = 500 Ohm Z = 𝟑𝟗𝟎𝟔𝟐𝟓 = 625 ohm
Ditanya : Z dan I ? b. I = V/Z
I = 125/625 = 0,2 A
13. DAYA PADA RANGKAIAN LISTRIK
Daya sesaat atau daya semu yang mengalir pada rangkaian arus bolak – balik dirumuskan :
Keterangan :
P = Daya ( Watt )
Sementara Daya aktif dirumuskan sebagai : V = Tegangan ( Volt )
I = Kuat arus ( A )
Vef = Tegangan efektif ( Volt )
Ief = Kuat arus efektif ( A )
Cos 𝜽 = faktor daya
Z = Impedansi ( ohm )
P = V . I
P = Vm.I cos 𝜽
Cos 𝜽 =
𝑹
𝒁
14. CONTOH SOAL
1.
Tentukan besar Daya pada gambar rangkaian RLC di atas !
Diketahui: Vm = 100 Volt
V = (100.sin 1000t) volt 𝜔 = 1000 t Jadi P = Vm.I.cos φ = (100)(0,2)(0,8) = 16
watt
R = 400 ohm Langkah 1 :
C = 5 μF = 5 × 10-6 F XL = ω.L = (1.000 rad/s)(0,5 H) = 500 Ω
L = 0,5 H I = Vm/Z = 100/500 = 0,2 A
Ditanya: P = ... ? Faktor daya, φ = 400/500 = 0,8
15. 1. Diket R = 60 ohm, X L = 120 ohm, XC = 40 ohm, V = 200 volt,
Dit : I, dan Z ?
Z = 𝑅2 + (𝑋𝐿 − 𝑋𝐶)2
Z = 602 + (120 − 40)2
Z = 3600 + 6400 = 10000 = 100 volt
I = V/Z
I = 200/100 = 2 A
16. 2. Diket : R = 30 ohm, L = 40𝑚𝐻 = 40𝑥10 − 3 𝐻, C = 50 𝜇𝐹 = 50 𝑥 10−6 F
Dit : f..?
f =
1
2𝜋 𝐿.𝐶
=
1
2𝜋 40𝑥 10−3 .50𝑥10−6
=
103
2𝜋 2
= 1000/2𝜋 2 = 250/ 𝜋 2 Hz
17. 3. V = 200 volt, f = 25 Hz, R = 200 ohm, C = 100/𝜋𝜇F = 100/𝜋 x 10-6 F
Dit : I ?
Jawab : a. 𝝎 = 𝟐𝝅𝒇
= 2 𝜋. 25 = 50 𝜋 rad/s
b. XC =
1
𝜔.𝐶
=
1
50𝜋.
100
𝜋
10−6
=
106
5𝑥103 = 1000/5 = 200 ohm
c. Z = 𝑅2 + 𝑋𝐶2 = 𝟐𝟎𝟎𝟐 + 𝟐𝟎𝟎𝟐 = 200 𝟐 ohm
d. I = V/Z
= 200/ 200 𝟐 = 1/2 𝟐 A
18. 4. R = 500 ohm, L = 8 H, C = 5x10-6 F, 𝝎 = 100 rad/s,
Dit : a. Z
b. Veff
Jawab : a. XL = 𝝎.L d. Veff = Vm/ 𝟐 = 50/ 𝟐 = 25 𝟐
XL = 100.8 = 800 ohm
b. XC =
𝟏
𝝎.𝑪
=
𝟏
𝟏𝟎𝟎.𝟓𝒙𝟏𝟎−𝟔 =
𝟏𝟎𝟔
𝟓𝟎𝟎
= 2000 ohm
c. Z = 𝑹𝟐 + (𝑿𝑳 − 𝑿𝑪)𝟐 = 𝟓𝟎𝟎𝟐 + ( 𝟖𝟎𝟎 − 𝟐𝟎𝟎𝟎)𝟐
Z = 𝟐𝟓𝟎𝟎𝟎𝟎 + 𝟏𝟒𝟒𝟎𝟎𝟎𝟎 = 1300 ohm
20. 7. L = 1/25 𝝅𝟐
H, C = 25 x 10-6 F, Dit : f ?
Jawab : f =
𝟏
𝟐𝝅. 𝒍𝒄
=
𝟏
𝟐𝝅.
𝟏
𝟐𝟓𝝅𝟐.𝟐𝟓𝒙𝟏𝟎−𝟔
=
𝟏
𝟐𝝅.
𝟏𝟎−𝟔
𝝅𝟐
=
𝟏
𝟐𝝅.
𝟏𝟎−𝟑
𝝅
= ½. 1000 = 500
8. R = 60 ohm, XL = 120 OHM, XC = 40 ohm, Vm = 200 volt
Dit : a. Im
b. Veff
Jawab : a. Z = 𝑹𝟐 + (𝑿𝑳 − 𝑿𝑪)𝟐 = 𝟔𝟎𝟐 + (𝟏𝟐𝟎 − 𝟒𝟎)𝟐 = 𝟏𝟎𝟎𝟎𝟎 = 100 volt
b. Im = Vm/Z = 200/100 = 2 A
c. Veff : Vm/ 𝟐 = 200/ 𝟐= 100 𝟐
21. 9. R = 600 ohm, XL = 1000 ohm, XC = 200 ohm, V = 100 volt
Dit : P?
Jawab : Z = 𝑅2 + (𝑋𝐿 − 𝑋𝐶)2
Z = 6002 + (1000 − 200)2
Z = 100000 = 1000
I = V/Z = 100/1000 = 0,1
P = V.I = 100. 0,1 = 10 Watt
22. LATIHAN SOAL
1. Perhatikan gambar rangkaian listrik di bawah ini
Jika tegangan maksimum sumber arus bolak-balik = 200 V, maka besar kuat arus maksimum yang mengalir pada rangkaian adalah....
2. Rangkaian RLC dengan R = 30 ohm, L = 40 mH, dan C = 50 μF dihubungkan pada sumber listrik. Rangkaian ini akan beresonansi pada frekuensi ….
3. Sumber arus bolak-balik memiliki amplitudo tegangan 200 V dan frekuensi sudut 25 Hz mengalir melalui hambatan R = 200 Ω dan kapasitor C =
100/π μF yang disusun seri. Kuat arus yang melalui kapasitor tersebut adalah ….
23. LATIHAN SOAL
4. Perhatikan gambar rangkaian RLC berikut. Tentukan besar Impedansi dan tegangan efektif !
5. Pada rangkaian seri RC, jika tegangan resistif dan tegangan kapasitif masing-masing adalah 50 V dan 120 V, maka tegangan totalnya adalah …..
6. Sumber arus bolak-balik memiliki amplitude tegangan 200 V dan frekuensi sudut 25 Hz mengalir melalui hambatan R = 200 Ω dan kapasitor C = 100π μF yang disusun seri.
Kuat arus yang melalui kapasitor tersebut adalah ….
24. LATIHAN SOAL
7. Suatu rangkaian seri R, L, dan C dihubungkan dengan tegangan bolak-balik. Apabila induktansi 1/25π
2 H dan kapasitas
kapasitor 25 μF, maka resonansi rangkaian terjadi pada frekuensi .....
8. Perhatikan gambar rangkaian listrik berikut. Jika tegangan maksimum sumber arus bolak-balik = 200 V, maka besar kuat arus dan
tegangan efektif yang mengalir pada rangkaian adalah....
25. LATIHAN SOAL
9. Sebuah hambatan 600 , induktor 1000 dan kapasitor 200 disusun seri. Jika susunan ini dihubungkan dengan sumber tegangan AC 100 V, maka daya
rangkaiannya adalah….Watt
10.Susunan seri hambatan 80 dan kapasitor dengan reaktasi kapasitif 60 dihubungkan dengan sumber arus bolak-balik, tegangan efektif 300V. Tegangan
efektif pada kapasitor adalah…
11. Sumber arus bolak-balik memiliki amplitudo tegangan 200 V dan frekuensi sudut 25 Hz mengalir melalui hambatan R = 100 Ω dan kapasitor C =
100/π μF yang disusun seri.Tentukan impedansi ….