Halaman
A. Arus Listrikdan MuatanB. Hukum Ohmdan HambatanC. Rangkaian Seridan ParalelD. Hukum IIKirchhoffE. Sumber ArusSearah dariProses KimiawiF. Tegangan ListrikSearah dan Bolak-BalikListrik Dinamis#0;0< 70@8 B830: <4=9038 70<10B0= 1068 <0AF0@0:0B :>B0 C=BC:<4;0:C:0= 0:B8D8B0A=F0 0; 8=8 38A4101:0= ?4=4@0=60= ;0<?C;0<?C3870<?8@A4<C0AC3CB:>B0AC307B4@?0A0=6"0<?C;0<?C8=8<4@C?0:0=2>=B>7?4@0;0B0=30;0<:4783C?0=A470@870@8F0=6<4<0=500B:0=4=4@68;8AB@8: A4B4;07 38C107 <4=9038 4=4@68 2070F0 "0;C 10608<0=0 4=4@68;8AB@8: 8=8 30?0B <4=F0;0:0= ;0<?C =30 0:0= <4<?4;090@8=F0 ?030101 8=830 3C0 94=8A 0@CA ;8AB@8: F08BC 0@CA ;8AB@8: 1>;0:10;8: *2#0, 2',%300#,230=0@CA;8AB@8:A40@07'0#!2300#,2&0300@CA;8AB@8: 1>;0:10;8: <C0B0= ;8AB@8: <4=60;8@ 30;0< 3C0 0@07 1>;0:10;8:30?C= ?030 0@CA ;8AB@8: A40@07 <C0B0= ;8AB@8: 70=F0 <4=60;8@ 30;0<A0BC0@07>=B>7?4@0;0B0=;8AB@8:F0=6<4=66C=0:0=0@CAA40@07F08BC:0;:C;0B>@0#+-2#!-,20-* 90< 30= ;0<?C A4=B4@ &030 101 8=8 0:0=38?4;090@8 0@CA ;8AB@8: A40@07 )430=6:0= 0@CA ;8AB@8: 1>;0:10;8: 0:0=38?4;090@8 ;4187 10=F0: 38 :4;0A . 169•memformulasikan besaran-besaran listrik rangkaian tertutup sederhana (satu loop);•mengidentifikasi penerapan listrik AC dan DC dalam kehidupan sehari-hari;•menggunakan alat ukur listrik.Setelah mempelajari bab ini, Anda harus mampu:menerapkan konsep kelistrikan dalam berbagai penyelesaian masalah dan berbagaiproduk teknologi.Hasil yang harus Anda capai:Nyala lampu pada malam hari, selain berfungsi sebagai peneranganjuga menjadi bagian dari keindahan kota.Bab8Sumber: Young Scientist,1994
170Mudah dan Aktif Belajar Fisika untuk Kelas X/ 2A. Arus Listrik dan Muatan8 )#& =30 B4;07 <4<?4;090@8 :>=A4? :C0B 0@CA 30= B460=60=;8AB@8:&0301018=8=300:0=:4<10;8<4<?4;090@8=F0;4187<4=30;0<?0:07 0@CA ;8AB@8: 8BC 0608<0=0 7C1C=60==F0 34=60= B460=60=;8AB@8: 0608<0=0 =30 <4=64B07C8 14A0@=F0 0@CA ;8AB@8: F0=6 <4=60;8@30;0<AC0BC@0=6:080=B4@BCBC?+=BC:<4=90E01?4@B0=F00=?4@B0=F00=B4@A41CB ?4;090@8 AC1101 8=8 34=60= A0:A0<01.Pengertian Arus Listrik 031*'120')030;07 *'0 ,+3 2 ,*'120')1#2' .1#* ,%4 )232#02#,230;0<AC0BC ?4=670=B0@ <C0B0= F0=6 <4=60;8@ 030;07 4;4:B@>=4;4:B@>=F0=61410A 14@64@0: &@>A4A 0;8@0= 0@CA ;8AB@8: <8@8? 34=60= 0;8@0= ?0=0A 30@8AC0BC14=3014@AC7CB8=668:414=3014@AC7C;4187@4=307;8@0=?0=0A0:0=14@74=B8A4B4;07:43C0AC7C14=30B4@A41CBA0<0A4B8<10=6B4@<0;0;0< 0;8@0= ;8AB@8: 9C60 34<8:80= 98:0 :43C0 B8B8: B4;07 <4<8;8:8B460=60= A0<0 0;8@0= <C0B0= 0:0= 14@74=B8@07 ?4@64@0:0= 4;4:B@>= 14@;0E0=0= 34=60= 0@07 0@CA ;8AB@8: ?4@70B8:0=#.$#20;0<A41C07?4=670=B0@A4AC=66C7=F0?4<10E0<C0B0= ;8AB@8: 030;07 4;4:B@>= -0;0C?C= 34<8:80= B4;07 38A4?0:0B8107E0 0@07 0@CA ;8AB@8: 14@;0E0=0= 34=60= 0@07 64@0: 4;4:B@>=#C0B0= ;8AB@8: 70=F0 0:0= <4=60;8@ 30;0< @0=6:080= B4@BCBC? &030@0=6:080= B4@BCBC? A4?4@B8 ?030#.$#2 0:0= B4@9038 1430 ?>B4=A80;0=B0@0 :43C0 C9C=6 ?4=670=B0@ 430 ?>B4=A80; 8=8;07 F0=6 <4=F4101:0=<C0B0= ;8AB@8: <4=60;8@ B4@9038 0@CA ;8AB@8:2.Kuat Arus Listrik3 2 031*'120')383458=8A8:0=A410608 ,6 ),6 +3 2 ,*'120')6 ,%+#,% *'0" * +13 23.#,%& ,2 01#2' .1 231 23 ,4 )23)420@0<0B4<0B8A38BC;8A:0= A410608 14@8:CBK
Gambar 8.2Arus listrik akan mengalir dalamrangkaian tertutup.!4B8:07C90=;410BB4@9038:8;0B30=B4@34B4:A80@CA;8AB@8:A414A0@ :8;>0<?4@4<4=60;8@30;0<E0:BC <8;8A4:>= 8BC=6<C0B0=F0=638?8=307:0=30@80E0=14@<C0B0=;8AB@8::4C<8?030A00B8BC#7#$8:4B07C8 2 A /2<0:0/2Contoh 8.1eeeeIeIeIIIsumber teganganGambar 8.1Arah arus listrik (I)dan arah elektron (e) berlawanan arah. &$&-5..&.1&-#+#2*,0/3&1*342*,*/#.*3,&2+#,#/-#)30#-30#-$&2*,54%#-#.$5,5-#4*)#/Tes Kompetensi Awal
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171Listrik DinamisGambar 8.3Mengukur kuat arus listrik3.Mengukur Kuat Arus+=BC: <4=6C:C@ :C0B 0@CA ;8AB@8: 30;0< AC0BC @0=6:080= ;8AB@8:386C=0:0=0<?4@4<4B4@0B0C0<<4B4@&4=6C:C@0=0@CA;8AB@8:30;0<AC0BC?4=670=B0@ 30?0B 38;0:C:0= 34=60= 20@0 <4=67C1C=6:0= 0;0B C:C@0@CA;8AB@8: 0<?4@4<4B4@ A420@0 A4@8 A4?4@B8 ?030#.$#2
0@0 <4<1020 A:0;0 0<?4@4<4B4@ 030;07 A410608 14@8:CB 0A8; ?4=6C:C@0= A:0;0F0=638BC=9C:L10B0AC:C@A:0;0<0:A8<C<K @058:14@8:CB<4=C=9C::0=:C0B0@CAF0=6<4=60;8@30;0<AC0BC@0=6:080=B4@BCBC?4@30A0@:0=6@058:B4@A41CBB4=BC:0=10=F0:=F0<C0B0=;8AB@8:F0=6<4=60;8@30;0<@0=6:080=A4;0<0A?4@B0<030;0<A0BC0=2>C;C<1#7#$I0@82 A0<?082 A38?4@>;472 K A0@CA?030A4;0=6E0:BC8=814@B0<107A420@0;8=40@$8;08@0B0@0B0=F0030;07A41060814@8:CB8;01>@0B>@8C<A4:>;07180A0=F0386C=0:0=0<?4@4<4B4@C=BC:<4=6C:C@:C0B0@CA&030AC0BC?4=6C:C@0=0@CA;8AB@8:3830?0B30B0A4?4@B838BC=9C::0=?03060<10@14@8:CB*4=BC:0=70A8;?4=6C:C@0=0<?4@4<4B4@B4@A41CBContoh 8.2Contoh 8.30 2 0 2
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4.Mengukur Beda Potensial+=BC:<4=6C:C@14A0@1430?>B4=A80;0B0CB460=60=380=B0@0C9C=6C9C=6?4=670=B0@ 386C=0:0= D>;B<4B4@ F0=6 38@0=6:08:0= A4?4@B8 ?030#.$#2 Gambar 8.4Cara mengukur beda potensial.voltmeterPQlampuaruslistrikbaterai+–aruslistriklampuPQvoltmeterbaterai×V–+sumber teganganlampuamperemeter/ A A 038<C0B0=F0=638?8=307:0=030;07 A0B0C 0701000,5 A#7#$ =5>@<0A8F0=638?4@>;4730@860<10@B4@A41CBF08BCA:0;0F0=638BC=9C: A:0;0<0:A8<C<
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172Mudah dan Aktif Belajar Fisika untuk Kelas XTes Kompetensi Subbab A&2+#,#/-#)%#-#.$5,5-#4*)#/ 4;0A:0=F0=638A41CB34=60=0;8AB@8:AB0B8A1;8AB@8:38=0<8A @058:14@8:CB<4=C=9C::0=:C0B0@CAF0=6<4=60;8@30;0<70<10B0=A4106085C=6A8E0:BC*4=BC:0= 10=F0:=F0 <C0B0= ;8AB@8: F0=6 <4=60;8@30;0<70<10B0=A4;0<0A 0;0<AC0BC@0=6:080=;8AB@8:<4=60;8@0@CA;8AB@8:A414A0@
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<30=A:0;0 386C=0:0=C=BC:<4=6C:C@:C0B0@CAF0=6<4=60;8@30;0<AC0BC@0=6:080=;8AB@8:&030A00B?4=6C:C@0=90@C<0<?4@4<4B4@<4=C=9C::0=0=6:0
B4=BC:0=14A0@:C0B0@CA?030?4=6C:C@0=8=8 *4=BC:0=70A8;?4=6C:C@0=14@8:CB01,>;B<4B4@ 38ACAC= ?0@0;4; A49090@ 34=60= AC<14@ ;8AB@8: 0B0C?4@0;0B0=;8AB@8: F0=6 0:0= 38C:C@ 1430 ?>B4=A80;=F0&030D>;B<4B4@B4@30?0B3C01C07:CBC1F08BC:CBC1?>A8B8530=:CBC1=460B85 !CBC1:CBC1 8=8 70@CA 387C1C=6:0= A420@0 14@A4AC080= 34=60=:CBC1:CBC1 ?030 @0=6:080=B. Hukum Ohm dan Hambatan&030AC1101=30B4;07<4<?4;090@8:>=A4?0@CA30=B460=60=A4@B0 20@0 ?4=6C:C@0==F0 &030 AC1101 8=8 =30 0:0= <4<?4;090@87C:C<%7<30=70<10B0=0608<0=07C1C=60=0=B0@0B460=60=0@CA;8AB@8:30=70<10B0= C:C<%7<0:0=<4<1070A7C1C=60=B4@A41CB)414;C< <4<?4;090@8 101 8=8 ;4187 90C7 ;0:C:0= :4680B0= 14@8:CBAktivitas Fisika 8.1I(A)t(s)21368hambatangeserhambatantetaptegangansumberAVTujuanMengetahui hubungan antara tegangandan arus listrik.Alat-Alat Percobaan1.Dua buah baterai2.Hambatan tetap3.hambatan geser (hambatan yangdapat diubah-ubah)4.Amperemeter DC5.Voltmeter DCterminalnegatifterminal 0,6 A01233 A0,6 A–012300,20,40,6Aterminalnegatifterminal 3 A01230,60,40,20A3 A0,6 A–0123Kata Kunci•arus listrik•elektron•beda potensial•amperemeter•voltmeterLangkah-Langkah Percobaan1.Susunlah semua peralatan seperti pada gambar.Hubungan Tegangan dan Arus Listrik
173Listrik Dinamis1.Hukum Ohm4@30A0@:0= 4:A?4@8<4= F0=6 38;0:C:0=&02( *.0/ ).
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K !4B4@0=60= 1430 ?>B4=A80; ,70<10B0= :C0B 0@CA !4<8@8=60=B0=?0306@058:B4@A41CB<4@C?0:0=14A0@=F070<10B0= F0=6 <4<8;8:8 =8;08 A0<0 30@8 AC0BC ?4@2>100= :0= B4B0?8 30@8 A4B80??4@2>100=B830:A4;0;C<4=670A8;:0=6@058::4<8@8=60=F0=6A0<0 C1C=60=0=B0@0 70<10B0= 30= :4<8@8=60= 6@058: B0= 38=F0B0:0= 34=60=?4@A0<00=Suatu alat pemanas listrik (heater)memakai arus listrik 11 A jikadihubungkan dengan sumberpotensial 220 V. Hitunglahhambatan pemanas tersebut.Tantanganuntuk Anda
&(#/(#/60-45#4253#.1&2&Gambar 8.5Grafik linear V terhadap I D>;B
174Mudah dan Aktif Belajar Fisika untuk Kelas X0153726454321V(volt)I(A)BA60°30°&4@70B8:0=60<10@14@8:CB*4=BC:0=;07=8;08?4@10=38=60=70<10B0=30=70<10B0=#7#$4=60=<4<4@70B8:0=6@058: K38?4@>;47=8;08?4@10=38=60=F08BCB0= B0=B0= HB0= H
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, 03870<10B0=94=8A?4<0=0A;8AB@8:B4@A41CB030;072.Hambatan (Resistansi))4B80?1070=<0B4@80;108:8BC;>60<0B0C?C=1C:0=;>60<<4<8;8:870<10B0= B4@B4=BC 0; 8=8 18A0 =30 0<0B8 107E0 AC0BC 1070= B830:A4;0;C 30?0B <4=670=B0@:0= 0@CA ;8AB@8: A420@0 108: 0?018;0 380;8@8 0@CA;8AB@8:60@30?0B<4<070<8=F0;0:C:0=,4*6*4#3*3*,# 14@8:CBContoh 8.4Contoh 8.5Aktivitas Fisika 8.2Hubungkan Panjang Kawat dan Nilai Hambatan LogamTujuan PercobaanMenyelidiki pengaruh panjang kawat dari jenis kawat terhadap nilai hambatanlogam.Alat-Alat Percobaan1.Amperemeter digital (0 mA – 1,2 mA)2.Sumber tegangan (DC 1,5 volt)3.Kawat nikrom (d = 0,5 mm) dan kawat tembaga berlapis email (d = 0,5 mm;1,0 mm; dan 1,5 mm)4.Kabel5.Penjepit6.Penggaris7.Mikrometer sekrup8.SpidolGrafik di atas menunjukkan kuat arusyang mengalir dalam suatuhambatan R, sebagai fungsi waktu.Banyaknya muatan listrik yangmengalir dalam hambatan tersebutselama 6 sekon pertama adalah ....a.8b.1 0c.1 4d.1 8e.20Ebtanas, 1990PembahasanDiketahui:Pada t = 0 sampai t = 3sq= I tq= 4 × (3)= 12 ColoumbA= 4,5040 cm3Pada t = 0 sampai t = 3sq= I tq= (3) . (1)= 2 Coloumbqtotal= 12 Coloumb + 6 Coloumb + 2 Coloumb = 20 ColoumbJawab: EPembahasan Soal
175Listrik Dinamis 3.Hubungkan kabel negatif sumber tegangan dengan salah satu ujung kawat(anggap ujung ini sebagai titik nol kawat). Kemudian, hubungkan kabelpositif amperemeter ujung kawat lain yang berjarak 25 cm 4.Catatlah kuat arus yang terbaca pada amperemeter, kemudian tuliskanhasilnya pada tabel berikut.No.PanjangTeganganKuat Arus (A)Hambatan ( )Kawat (cm)Sumber ( V)1.25. . .. . .. . .2.50. . .. . .. . .3.75. . .. . .. . .4.100. . .. . .. . .5.125. . .. . .. . . 5.Perhatikan data yang telah Anda tuliskan dalam tabel. Kesimpulan apakahyang Anda peroleh? Bagaimanakah hubungan antara panjang () denganhambatan (R)? 6.Ulangi langkah 2, 3, dan 4 untuk kawat tembaga dengan diameter 0,5 mmdan panjang 50 cm. 7.Hitunglah nilai hambatannya. Bandingkan dengan nilai hambatan untukkawat nikrom dengan panjang 50 cm. 8.Mengapa nilai hambatannya sama atau mengapa nilai hambatannyaberbeda? 9.Ulangi langkah 2, dan 3 untuk kawat tembaga lain dengan nilai diameter 1,0mm dan 1,5 mm, serta panjangnya 50 cm.10.Catatlah hasilnya pada tabel berikut.11.Bagaimanakah hubungan antara luas penampang kawat tembaga denganhambatan (R)?13.21AmperemeterdigitalPenjepit 1Sumber potensialkawatPenjepit 2Langkah-Langkah Percobaan1.Dengan menggunakan penggaris, ukurlah panjang kawat dari salah satuujungnya sepanjang 10 cm. Kemudian, beri tanda dengan spidol. Lakukanhal yang sama untuk setiap 25 cm berikutnya hingga 100 cm. 2.Susun semua peralatan seperti gambar berikut.No.Diameter(mm)Luas Penampang(mm2)Tegangan(V)Kuat Arus(A)Hambatan()1230,51,01,5. . .. . .. . .. . .. . .. . .. . .. . .. . .. . .. . .. . .
176Mudah dan Aktif Belajar Fisika untuk Kelas XB
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178Mudah dan Aktif Belajar Fisika untuk Kelas X&4@70B8:0=60<10@@0=6:080=14@8:CB8=8*4=BC:0=14A0@=F0(98:0,030;07
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0@860<10@@0=6:080=?030A>0;=><>@B4=BC:0=14A0@=F030=V (volt)I(ampere)0,023Aktivitas Fisika 8.3Hukum I KirchhoffTujuanMemahami Hukum I Kirchhoff.Alat-Alat Percobaan1.Amperemeter DC (0 – 1 A),2.Tiga lampu kecil (masing-masing 1,5 V)3.Sebuah baterai (1,5 V)4.Kabel penghubung secukupnya.Langkah-Langkah Percobaan1.Susunlah peralatan seperti pada gambar, letakkan amperemeter di posisi A12.Amatilah, apakah semua lampu menyala?3.Catat kuat arus yang ditunjukkan amperemeter, kemudian pindahkanamperemeter ke posisi A2, A3, dan A4.4.Catat kuat arus yang ditunjukkan amperemeter pada semua posisi tersebut5.Apakah A1 dan A2 menunjukkan angka yang sama?6.Jumlahkan angka yang ditunjukkan oleh A3 dan A4. Apakah hasilpenjumlahannya sama dengan angka yang ditunjukkan oleh A1 atau A2?7.Apa kesimpulan Anda dari kegiatan ini?Perjanjian cara penggambaranbaterai (sumber potensial DC) padarangkaian adalah sisi yang lebihpanjang menandakan kutubpositifnya.Ingatlah+–A = amperemeterPA3A4A2A150 mA150 mA60 mAABI2I3CI1R1R2R3
179Listrik Dinamis0@8:4680B0=B4@A41CB38?4@>;47:4A8<?C;0=B4=B0=6 C:C< !8@277>553)3+'0!&&-$$ 14@1C=F8(3+* & 031*'120')6 ,%+ 13). " 13 232'2').#0! ,% ,1 + "#,% ,(3+* & 031*'120')6 ,%)#*3 0" 0'2'2')! ,%2#01# 32 C:C<8=8<4@C?0:0=?4@=F0B00=;08=30@87C:C<:4:4:0;0=<C0B0=F0=6<4=F0B0:0= 107E0 9C<;07 <C0B0= F0=6 <4=60;8@ B830: 14@C107&4@70B8:0=#.$#2
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180Mudah dan Aktif Belajar Fisika untuk Kelas X1 ,B>B KK
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430?>B4=A80;A4B80?70<10B0=A0<014A0@Mari Mencari TahuR1R2R3R4GPSRGVJika Anda telah memahami susunanseri hambatan pada rangkaian,tentukan oleh Anda sehinggadiperoleh persamaan (8 – 12),persamaan (8 –13), dan persamaan(8 – 14) dengan cara menurunkandari persamaan (8 – 9) danpersamaan (8 – 10).Tantanganuntuk AndaI1R1R2I2IVI
182Mudah dan Aktif Belajar Fisika untuk Kelas XACAC=0= @0=6:080= A430=6:0= 90@C< 0;D0=><4B4@ <4=F8<?0=6 :4 :8@80B0C:4:0=0= 4<10B0=30;0<:40300=A4B8<10=60:0=38?4@>;4734=60=<4=664A4@64A4@:>=B0:A4?0=90=6:0E0B&030:40300=A4B8<10=690@C< 0;D0=><4B4@ 0:0= <4=C=9C::0= 0=6:0 =>; A478=660 38?4@>;47 0B0C K
030;07 70<10B0= F0=6 74=30: 38C:C@ A430=6:0= 70<10B0=AB0=30@ F0=6 AC307 38:4B07C8 &0=90=6 :0E0B130=2 30?0B B4@1020<4;0;C8 A:0;0 ?0=90=6 ?030 :0E0B B4@A41CB&4@70B8:0=60<10@14@8:CBR = 120GDCAXVS60 cm&0=90=6 2< 0@C<60;D0=><4B4@0:0=A4B8<10=6:4B8:0:>=B0:14@030 2<30@8C9C=6*4=BC:0==8;0870<10B0=5#7#$ 2<K 2< 2<)F0@0B94<10B0=30;0<:40300=A4B8<10=6030;075
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34=60= 70<10B0= 30;0<=F0B>B0000.K !C0B 0@CA F0=6 <4=60;8@ <4;0;C8 @0=6:080= ?030#.$#2 B4@A41CB <4<4=C78 ?4@A0<00=000K +=BC:, 1C07 AC<14@ B460=60= F0=6 38ACAC= A4@8 14@;0:C, ,0K
c.Sumber Tegangan Disusun Paralel 8:0A49C<;07AC<14@B460=60=F0=6<4<8;8:866;A0<012338ACAC= A420@0 ?0@0;4; <0:0 14@;0:CB>B 0<10B0= 30;0<=F0 38@C<CA:0= A410608 14@8:CBK B>B0000.K Gambar 8.14Rangkaian seri tiga sumbertegangan atau baterai.RI+–+–+–11r22r33r
184Mudah dan Aktif Belajar Fisika untuk Kelas XABGambar 8.15Tiga sumber tegangan disusunparalel.&4@70B8:0=#.$#2 !C0B0@CAF0=6<4=60;8@?030@0=6:080=030;072-2000K C=BC:,1C07AC<14@B460=60=34=60=66; 30= 70<10B0= 30;0<0F0=6 38ACAC= ?0@0;4; 14@;0:C0,K *8601C0710B4@0838ACAC=A420@0A4@8A4?4@B860<10@14@8:CB)4B80?10B4@08<4<8;8:866;
,30=70<10B0=30;0< 8:0:4B8601C0710B4@08B4@A41CB387C1C=6:0=34=60=A41C0770<10B0=
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, )+ 4, 4 038:C0B0@CAF0=6<4=60;8@ R = 4,41,5 V ; 0,21,5 V ; 0,21,5 V ; 0,2C01C0710B4@0838ACAC=A420@0?0@0;4;A4?4@B8?03060<10@14@8:CBI1I2I11r22r2Contoh 8.11Contoh 8.12Empat buah hambatan masing-masing besarnya 1 ohmdihubungkan seperti pada gambarberikut.Hitunglah hambatan total R antaratitik A dan titik B.Tantanganuntuk AndaRI22r11r33r
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192Mudah dan Aktif Belajar Fisika untuk Kelas X 20B0C )41C071>;0;0<?C34=60=A?4A858:0A8
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14@;0:C 98:0 B830: B4@9038 ?4@C1070= EC9C3208@ A414;C< B4@9038 ?4=6C0?0=Contoh 8.15K To k o hAlessandro Volta, adalah Fisikawanyang dilahirkan di Como, Italia. Diamenciptakan electrophorus, yaitusuatu alat untuk membangkitkanlistrik statis pada 1775 danmenemukan gas metana pada 1778.Dia ditetapkan sebagai profesoruntuk filsuf ilmu alam di Pavia.Terinspirasi oleh temannya LuigiGalvani, Volta menemukan bahwaarus listrik dibangkitkan ketika dualogam berbeda berada pada jarakyang sangat dekat, danmengembangkan baterai listrikpertama pada 1800. Namanyadiabadikan untuk satuan bedapotensial listrik, volt.Sumber :www.allbiographies.comSumber:www.physics. com.Alessandro Volta(1745 – 1827)
193Listrik Dinamis=30 B4;07 <4<?4;090@8 0@CA ;8AB@8: A40@07 B460=60= ;8AB@8: A40@0730=AC<14@0@CA;8AB@8:A40@07A4@B0@0=6:080=A434@70=0B4@38@80B0A10B4@0878=66038?4@>;47=8;08 30=F0=638A41CB34=60=@0=6:080=0@CAA40@07@CA 30= B460=60= ;8AB@8: 1>;0:10;8: <4<8;8:8 =8;08 F0=6 A4;0;C 14@C107C107 B4@7030? E0:BC A420@0 ?4@8>38: 108: 14A0@ <0C?C= 0@07=F04A0@0= 0@CA 30= B460=60= 1>;0:10;8: 38;0<10=6:0= 34=60=A430=6:0=0@CA30=B460=60=A40@0738;0<10=6:0=34=60=4E0A08=8 70<?8@ A4<C0 ?4@0;0B0= @C<07 B0=660 38>?4@0A8:0= 34=60= 4=4@68;8AB@8: 0@CA 1>;0:10;8: A4?4@B8 B0<?0: ?030#.$#2 =30 B4;07<4=64B07C8 107E0 ?4@14300= <4=30A0@ 0=B0@0 0@CA 1>;0:10;8: 30=0@CAA40@07 030;07 ?>;0@8B0A=F0+=BC:<4=64B07C8?>;0@8B0A0@CA30=B460=60=A40@07F0=6A4;0;CB4B0?30=0@CA1>;0:10;8:F0=6A4;0;C14@C10730?0B386C=0:0=>A8;>A:>?<8A0;=F0( 2&-"# 61!'*-1!-.# #4;0;C8 0;0B 8=8 9C60 380<0B8 =8;08 5@4:C4=A8 30= 14=BC: 64;><10=6 F0=6 3870A8;:0= A430=6:0= C=BC: <4=6C:C@ =8;08 B460=60= 30= :C0B 0@CA ;8AB@8: 30?0B 386C=0:0= D>;B<4B4@ 30= 0<?4@4<4B4@ 2.Mengamati Tegangan Listrik DC dan Tegangan Listrik AC+=BC: <4=6C:C@ B460=60=30= <0:A8<C< + 30= B460=60=?C=20::4?C=20: ..F0=614@0A0;30@890@8=60=;8AB@8:&"$30?0B38;0:C:0=34=60= <4=66C=0:0= >A8;>A:>? A4?4@B8 ?030#.$#2
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194Mudah dan Aktif Belajar Fisika untuk Kelas X4A0@=F00@CA30=B460=60=1>;0:10;8:30?0B380<0B834=60=0D><4B4@@CA30=B460=60=F0=638BC=9C::0=0;0B8=8<4@C?0:0=70@60454:B85=F0&>;0@8B0A ;8AB@8: F0=6 3870A8;:0= 30@8 90@8=60= &"$ 14@C?0 6@058: A8=CA8>30; 34=60= @4:C4=A8 G@CA 454:B85 30?0B 38@C<CA:0= A410608 14@8:CB#$<0:AK 34=60= 20@0 F0=6 A0<0 C=BC: B460=60= 454:B85 #$ 0:0= 38?4@>;47 ?4@A0<00= #$#$ #$<0:A K #$&-
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2<%;47:0@4=0A:0;0D4@B8:0; ,2<<0:0=F014@C?060@8A;C@CAA430=6:0=;8AB@8:1>;0:10;8:<4<8;8:8B460=60=F0=614@C107C107 A4B80? A00B F08BC 14@14=BC: A8=CA>830; 414@0?0 ?4@14300=?@8=A8? ;8AB@8: A40@07 30= 1>;0:10;8: 30?0B =30 ;870B ?030 #$&- Contoh 8.16I "8AB@8: <C30738B@0=A<8A8:0=30@8?4<10=6:8B:4@C<07@C<07I @CA30=B460=60= <4<8;8:8B460=60=<0:A8<C<B460=60=?C=20::4?C=20:B460=60=A4A00BB460=60=@0B0@0B030==8;08454:B85I "8AB@8:AC;8BC=BC:<4<4=C78:41CBC70=?0A>:0=30;0<9C<;0714A0@I @CA30=B460=60= 70=F0<4<8;8:8=8;08454:B85=F0I "0F0@>A8;>A:>?<4=C=9C::0=60<10@B460=60=1>;0:10;8:0-#,#-*, #) 5.$&2253 #/(,#*#/I "0F0@>A8;>A:>?<4=C=9C::0=60<10@B460=60=A40@07TVpp1 cm1 cmpenahankabel netralInformasiCara menghubungkan kabelpemanas air dapat dilakukansebagai berikut. Ketiga lubang padasteker, yaitu dua ujung terminalpemanas masing-masingdihubungkan ke kutub positif dannegatif, dan kabel lain berwarna birumenuju netral.untuk Andakabelberaruskabel ditanahkansekringInformation for YouTo connect wires of water heater cando as below. Three hole on the plug,that is two edge heater terminalsconnect to positive and negativepoles and other blue wires towardneutral.
195Listrik Dinamis ..
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2Contoh 8.17)41C07@0=6:080=0@CA1>;0:10;8:<4<8;8:870@60B460=60=A4106085C=6A8E0:BCF08BC A8= 2 8BC=6;070B460=60=<0:A8<C< + )14?4@8>341B460=60=?C=20::4?C=20: ..55@4:C4=A8$2B460=60=454:B85 #$6B460=60=A4B4;072
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<0:0 + ,1 .. + ,,2 #$ + , ,Contoh 8.18Persamaan tegangan bolak-baliksuatu rangkaian listrikmemenuhi persamaanV = 314 sin 50 V.Tentukan tegangan rata-ratayang dihasilkan sumber tersebut.Tantanganuntuk Anda
196Mudah dan Aktif Belajar Fisika untuk Kelas X3.Pemasangan Listrik di Rumah Tangga&030C<C<=F00@CA;8AB@8:F0=638A0;C@:0=:4@C<07@C<0714@0A0;30@8 90@8=60= &"$ 34=60= <4=66C=0:0= 0@CA 30= B460=60= 1>;0:10;8: *070?0=<0AC:=F00@CA;8AB@8:30@8B80=690@8=60=:4@C<07030;07A410608 14@8:CB@CA;8AB@8:<0AC::0;8?4@B0<0<4;0;C8
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Gambar 8.25Arus listrik disalurkan olehjaringan PLN.)41C07A4B@8:0
- ,0:0=38;4=6:0?834=60=A41C07A4:@8=6 8:0A4:@8=6F0=6B4@A438014@=8;08 30=
14@0?0:07=8;08A4:@8=6F0=60:0=38?8;87#7#$!C0B0@CAF0=638?4@;C:0=A4B@8:0030;07
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A478=660F0=638?8;87030;07A4:@8=614@=8;08Sumber: Young Scientists, 1997Contoh 8.193 @03A4 A5$
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,Gambar 8.26Sekringtutup logamtabung kacatutup logamkawat sekringKata Kunci•tegangan listrik searah•tegangan listrik bolak-balik•energi listrik•daya listrik•osiloskop•transmisi daya listrik
197Listrik Dinamis&4<0A0=60= A49C<;07 ;0<?C 38 @C<07 A4108:=F0 387C1C=6:0= A420@0?0@0;4; A4?4@B8#.$#2 30=#.$#2 060@ A4B80? ;0<?C<4=30?0B B460=60= F0=6 A0<0saklar duaarah60 W15 W100 W5 Asekringsumber teganganPLN 220 VGambar 8.28Contoh diagram kabel listrik di rumahsaklarsaklarkotaksekringmeteranstopkontaksakelarstopkontaklampuGambar 8.27Pemasangan lampu secara paralel.pusatpembangkitlistriktransmisitegangantinggigeneratortransformatortransformatortiang listrikpabrikrumahgardu listrikGambar 8.29Bagan transmisi daya listrik jarak jauhTes Kompetensi Subbab F&2+#,#/-#)%#-#.$5,5-#4*)#/
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4;0A:0=0?0F0=638<0:AC334=60=070@60454:B85:C0B0@CA30=B460=60=1>;0:10;8:170@60<0:A8<C<:C0B0@CA30=B460=60=1>;0:10;8:270@60@0B0@0B0:C0B0@CA30=B460=60=1>;0:10;8:)C<14@B460=60=0@CA1>;0:10;8:14A0@=F0
,)41C07A4B@8:0;8AB@8:34=60=70<10B0= 387C1C=6:0= :4 AC<14@ B460=60= B4@A41CB 8BC=6;07 =8;08454:B85=8;08<0:A8<C<30==8;08@0B0@0B0C=BC:0 B460=60=AC<14@1 0@CAF0=6<4=60;8@0@860<10@6@058:14@8:CBB4=BC:0=;0700@CA<0:A8<C<10@CA454:B8525@4:C4=A8=F0 8:0A4:@8=6F0=6B4@A438014@=8;08
30= ;4=6:0?8;07 B014; 14@8:CB C=BC:<4=4=BC:0==8;08A4:@8=6F0=60:0=386C=0:0="0<?C*4;4D8A8&4=64@8=6@0<1CB&4<0=660=6@>B84@4:;8AB@8: &2#-#4#/#8#" &(#/(#/!
&,2*/(10waktu(s)I(A)F0=6 <4=60;8@ 30;0< AC0BC ?4=670=B0@A4B80? A0BC A0BC0= E0:BCRangkuman
@CA ;8AB@8: 030;07 <C0B0= 30@8 ?>B4=A80;B8=668 :4 ?>B4=A80; @4=307 !C0B 0@CA;8AB@8: 383458=8A8:0= A410608 <0B0= ;8AB@8:Sumber: Jendela Iptek, 1997
198Mudah dan Aktif Belajar Fisika untuk Kelas X94=8A=F038C:C@ 34=60=<?4@4<4B4@3894;0A:0= >;47 C:C<!8@277>5538C:C@ 34=60=,>;B<4B4@%A8;>A:>?3878BC=6 34=60= C:C<%7<B4@9038 ?030:><?>=4==F0(0=6:080=*4@BCBC?@CA"8AB@8:*342*,*/#.*3<4<?4;090@838?4=60@C78 >;47)C<14@*460=60=38?4=60@C78 >;47 0<10B0="C0A)C7C 4=8A38A41CB&0=90=6 0<10B0= 4=8A*460=60=3878BC=6 34=60=3878BC=6 34=60=Peta Konsep #4=C@CB C:C< %7< 14@1C=F8 B460=60=0B0C1430?>B4=A80;0C0BC?4=670=B0@;8AB@8:A410=38=634=60=0@CA;8AB@8:F0=6<4=60;8@<4;0;C8?4=670=B0@B4@A41CB&4@10=38=60==F0A4;0;C:>=AB0=F0=638A41CBA41060870<10B0= 0<10B0= AC0BC @4A8AB>@ 14@60=BC=6 ?030?0=90=6 ;C0A ?4=0<?0=6 30= 70<10B0=94=8A 0<10B0= 94=8A 1070= 14@60=BC=6?030 AC7C
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