Bab
8 Sumber: Young Scientist,1994
Nyala lampu pada malam hari, selain berfungsi sebagai penerangan juga menjadi bagian dari keindahan kota.
Listrik Dinamis Hasil yang harus Anda capai: menerapkan konsep kelistrikan dalam berbagai penyelesaian masalah dan berbagai produk teknologi.
Setelah mempelajari bab ini, Anda harus mampu: • • •
memformulasikan besaran-besaran listrik rangkaian tertutup sederhana (satu loop); mengidentifikasi penerapan listrik AC dan DC dalam kehidupan sehari-hari; menggunakan alat ukur listrik.
#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= ;0B0AC307B4@?0A0=6"0=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;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:
;0:10;8: 30?C= ?030 0@CA ;8AB@8: A40@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;0:10;8: 0:0= 38?4;090@8 ;4187 10=F0: 38 :4;0A .
A. Arus Listrik dan Muatan B. Hukum Ohm dan Hambatan C. Rangkaian Seri dan Paralel D. Hukum II Kirchhoff E. Sumber Arus Searah dari Proses Kimiawi F. Tegangan Listrik Searah dan BolakBalik 169
Tes Kompetensi Awal &$&-5..&.1&-#+#2*,0/3&1*342*,*/#.*3,&2+#,#/-#)30#-30#-$&2*,54%#-#.$5,5-#4*)#/
?0:070@0764@0:0@CA;8AB@8:A40@0734=60=0@0764@0:
#4=60?0 D>;B<4B4@ 30;0< AC0BC @0=6:080= 70@CA 4;4:B@>= 4;0A:0=90E010==30 38?0A0=6A420@0?0@0;4; 0@8?>B4=A80;<0=0:4?>B4=A80;<0=0:070@CA;8AB@8: *C;8A:0=;07 2>=B>7 AC<14@ B460=60= ;8AB@8: F0=6 <4=60;8@ =30:4B07C8 *C;8A:0=;0750:B>@50:B>@F0=6<4<4=60@C7814A0@=F0 AC0BC70<10B0=?4=670=B0@
A. Arus Listrik dan Muatan 8 )#& =30 B4;07 <4=A4? :C0B 0@CA 30= B460=60= ;8AB@8:&0301018=8=300:0=:4<10;8<4
1.
e
e I e e
e
I
I e I
I
Gambar 8.1 Arah arus listrik (I) dan arah elektron (e) berlawanan arah. sumber tegangan
031*'120')030;07 *'0 ,+3 2 ,*'120')1#2' .1#* ,%4 )232#02#,230;0< AC0BC ?4=670=B0@ =4;4:B@>= F0=6 1410A 14@64@0: &@>A4A 0;8@0= 0@CA ;8AB@8: <8@8? 34=60= 0;8@0= ?0=0A 30@8 AC0BC14=3014@AC7CB8=668:414=3014@AC7C;4187@4=307;8@0=?0=0A 0:0=14@74=B8A4B4;07:43C0AC7C14=30B4@A41CBA0<0A4B8<10=6B4@<0; 0;0< 0;8@0= ;8AB@8: 9C60 34<8:80= 98:0 :43C0 B8B8: B4;07 <4<8;8:8 B460=60= A0<0 0;8@0= = 14@;0E0=0= 34=60= 0@07 0@CA ;8AB@8: ?4@ 70B8:0=#.$#20;0<A41C07?4=670=B0@A4AC=66C7=F0?4<10E0 = -0;0C?C= 34<8:80= B4;07 38A4?0:0B8 107E0 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 #.$#2 0:0= B4@9038 1430 ?>B4=A80; 0=B0@0 :43C0 C9C=6 ?4=670=B0@ 430 ?>B4=A80; 8=8;07 F0=6 <4=F4101:0=
2. Gambar 8.2 Arus listrik akan mengalir dalam rangkaian tertutup.
Pengertian Arus Listrik
Kuat Arus Listrik
3 2 031*'120')383458=8A8:0=A410608 ,6 ),6 +3 2 ,*'120')6 ,% +#,% *'0" * +13 23.#,%& ,2 01#2' .1 231 23 ,4 )23)420@0<0B4<0B8A 38BC;8A:0= A410608 14@8:CB / K 2
Contoh 8.1 !4B8:07C90=;410BB4@9038:8;0B30=B4@34B4:A80@CA;8AB@8:A414A0@ :8;>0= 8BC=6
170
/ <0:0 2
/2
Mudah dan Aktif Belajar Fisika untuk Kelas X
/ A A 038
3.
sumber tegangan
Mengukur Kuat Arus
lampu
+=BC: <4=6C:C@ :C0B 0@CA ;8AB@8: 30;0< AC0BC @0=6:080= ;8AB@8: 386C=0:0=0
A:0;0F0=638BC=9C: L10B0AC:C@ A:0;0<0:A8
K
Contoh 8.2 8;01>@0B>@8C<A4:>;07180A0=F0386C=0:0=0@<0A8F0=638?4@>;4730@860<10@B4@A41CBF08BCA:0;0 70 F0=638BC=9C: A:0;0<0:A8
03870A8;?4=6C:C@0==F0
0A8;?4=6C:C@0=
amperemeter
Gambar 8.3 Mengukur kuat arus listrik
0,5 A
voltmeter
Contoh 8.3 @058:14@8:CB<4=C=9C::0=:C0B0@CAF0=6<4=60;8@30;0<AC0BC@0=6:080=B4@BCBC? I(A) 4@30A0@:0=6@058:B4@A41CBB4=BC:0=10=F0:=F0 C;C<1 #7#$ 2 I 0@82 A0;47 2 K A0@CA?030A4;0=6E0:BC8=8 14@B0<107A420@0;8=40@$8;08@0B0@0B0=F0 t(s) 2 4 6 030;07A41060814@8:CB 0 2 0 2 I
P
lampu
arus listrik +
voltmeter V
C<;07;472 AK A A@CA?030A4;0=6E0:BC
/ / /
–
baterai
8=8B4B0? C<;07
Q
× P
lampu
Q
arus listrik + –
4.
Mengukur Beda Potensial
+=BC:<4=6C:C@14A0@1430?>B4=A80;0B0CB460=60=380=B0@0C9C=6C9C=6 ?4=670=B0@386C=0:0=D>;B<4B4@F0=638@0=6:08:0=A4?4@B8?030#.$#2
baterai
Gambar 8.4 Cara mengukur beda potensial.
Listrik Dinamis
171
,>;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=
Kata Kunci • • • • •
arus listrik elektron beda potensial amperemeter voltmeter
Tes Kompetensi Subbab
A
&2+#,#/-#)%#-#.$5,5-#4*)#/ 4;0A:0=F0=638A41CB34=60= 0 ;8AB@8:AB0B8A 1 ;8AB@8:38=0<8A @058:14@8:CB<4=C=9C::0=:C0B0@CAF0=6<4=60;8@ 30;0<70<10B0=A4106085C=6A8E0:BC
I(A) 2
)41C070
3
1
2
3
0,2 0,4
0
1
0
6
8
0,6 A
0,6
A
3A
t(s)
*4=BC:0= 10=F0:=F0 =<4<8;8:8= F0=6 <4=60;8@ 30;0< A41C07 :0E0B ?4=670=B0@F0=6380;8@80@CA;8AB@8: A4;0<0A
terminal 0,6 A
1
2
1
terminal negatif
3
0 0
1
2
3 0
–
0,6 A
0,2 0,4
0,6
A
3A
terminal 3 A
B. Hukum Ohm dan Hambatan &030AC1101=30B4;07<4=A4?0@CA30=B460=60= A4@B0 20@0 ?4=6C:C@0==F0 &030 AC1101 8=8 =30 0:0= <4
Aktivitas Fisika 8.1 Hubungan Tegangan dan Arus Listrik Tujuan Mengetahui hubungan antara tegangan dan arus listrik. Alat-Alat Percobaan 1. Dua buah baterai 2. Hambatan tetap 3. hambatan geser (hambatan yang dapat diubah-ubah) 4. Amperemeter DC 5. Voltmeter DC
tegangan sumber
Langkah-Langkah Percobaan 1. Susunlah semua peralatan seperti pada gambar.
172
Mudah dan Aktif Belajar Fisika untuk Kelas X
hambatan geser hambatan tetap
V
A
2.
3. 4.
1.
Ubahlah hambatan geser dengan cara menggeser-geser kontak luncur, bacalah kuat arus I pada amperemeter dan tegangan hambatan tetap pada voltmeter. Tulislah hasil yang Anda peroleh dalam bentuk tabel. Dari tabel yang Anda tulis, buatlah grafik tegangan V terhadap kuat arus I. Dari grafik tersebut, buatlah kesimpulannya.
Hukum Ohm
4@30A0@:0= 4:A?4@8<4= F0=6 38;0:C:0= &02( *.0/ ). K
3830?0B :4A860<30=<4<4=C78 C:C< %7<:>=AB0=38A41CB>7<8:0B0C;8=40@#8A0;:0=30@8AC0BC70A8; ?4@2>100=38?4@>;47=8;08B460=60= 30=:C0B0@CA0B0B4@A41CB30?0B =30;870B?030 #$&-*014;B4@A41CB30?0B<4=670A8;:0=AC0BC6@058: ;8=40@ A4?4@B8 B0
D>;B
Gambar 8.5 Grafik linear V terhadap I
#$&- $8;08*460=60= 30=!C0B@CA?030 0<10B0=">60<%7<8: &(#/(#/ 60-4
5#4253 #.1&2&
)420@0 <0B4<0B8A ?4@=F0B00= C:C< %7< 30?0B 38BC;8A:0=
K
!4B4@0=60= 1430 ?>B4=A80; , 70<10B0= :C0B 0@CA !4<8@8=60=B0= ?0306@058:B4@A41CB<4@C?0:0=14A0@=F070<10B0= F0=6 <4<8;8:8 =8;08 A0<0 30@8 AC0BC ?4@2>100= :0= B4B0?8 30@8 A4B80? ?4@2>100=B830:A4;0;C<4=670A8;:0=6@058::4<8@8=60=F0=6A0<0 C1C=60= 0=B0@0 70<10B0= 30= :4<8@8=60= 6@058: B0= 38=F0B0:0= 34=60= ?4@A0<00= B0=
K
Tantangan untuk Anda Suatu alat pemanas listrik (heater) memakai arus listrik 11 A jika dihubungkan dengan sumber potensial 220 V. Hitunglah hambatan pemanas tersebut.
!4B4@0=60= 14A0@70<10B0= B0= :4<8@8=60= 6@058: <0:0 >7<14A0@=F0A0<034=60= D>;B?4@0
Listrik Dinamis
173
Pembahasan Soal
)41C07?4<0=0A;8AB@8:3814@8B460=60= ,A478=660<4=60;8@0@CA;8AB@8:A414A0@ 8BC=670<10B0=?4<0=0AB4@A41CB #7#$ 8:4B07C8 ,
5 4 3 2 1 0
Contoh 8.4
1
2
3
4
5
6
7
Grafik di atas menunjukkan kuat arus yang mengalir dalam suatu hambatan R, sebagai fungsi waktu. Banyaknya muatan listrik yang mengalir dalam hambatan tersebut selama 6 sekon pertama adalah .... a. 8 b. 10 c. 14 d. 18 e. 20 Ebtanas, 1990 Pembahasan Diketahui: Pada t = 0 sampai t = 3s q = I t q = 4 × (3) = 12 Coloumb A = 4,5040 cm3 Pada t = 0 sampai t = 3s q = I t q = (3) . (1) = 2 Coloumb qtotal = 12 Coloumb + 6 Coloumb + 2 Coloumb = 20 Coloumb Jawab: E
, 03870<10B0=94=8A?4<0=0A;8AB@8:B4@A41CB030;07
Contoh 8.5 &4@70B8:0=60<10@14@8:CB V(volt) B A
60°
30° I(A)
*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= F08BC B0= B0= B0= HB0= H 038
2.
Hambatan (Resistansi)
)4B80?1070=<0B4@80;108:8BC;>60<0B0C?C=1C:0=;>60<<4<8;8:8 70<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:CB
Aktivitas Fisika 8.2 Hubungkan Panjang Kawat dan Nilai Hambatan Logam Tujuan Percobaan Menyelidiki pengaruh panjang kawat dari jenis kawat terhadap nilai hambatan logam. Alat-Alat Percobaan 1. 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. Kabel 5. Penjepit 6. Penggaris 7. Mikrometer sekrup 8. Spidol
174
Mudah dan Aktif Belajar Fisika untuk Kelas X
Langkah-Langkah Percobaan 1. Dengan menggunakan penggaris, ukurlah panjang kawat dari salah satu ujungnya sepanjang 10 cm. Kemudian, beri tanda dengan spidol. Lakukan hal yang sama untuk setiap 25 cm berikutnya hingga 100 cm. 2. Susun semua peralatan seperti gambar berikut.
kawat
Penjepit 1
Penjepit 2
Amperemeter digital 13.21
3.
4.
Hubungkan kabel negatif sumber tegangan dengan salah satu ujung kawat (anggap ujung ini sebagai titik nol kawat). Kemudian, hubungkan kabel positif amperemeter ujung kawat lain yang berjarak 25 cm Catatlah kuat arus yang terbaca pada amperemeter, kemudian tuliskan hasilnya pada tabel berikut. No.
Panjang Kawat (cm)
1. 2. 3. 4. 5. 5.
6. 7. 8. 9. 10.
11.
25 50 75 100 125
Tegangan Sumber (V) . . . . .
. . . . .
Kuat Arus (A)
. . . . .
. . . . .
. . . . .
. . . . .
Hambatan ( ) . . . . .
. . . . .
. . . . .
Perhatikan data yang telah Anda tuliskan dalam tabel. Kesimpulan apakah yang Anda peroleh? Bagaimanakah hubungan antara panjang ( ) dengan hambatan (R)? Ulangi langkah 2, 3, dan 4 untuk kawat tembaga dengan diameter 0,5 mm dan panjang 50 cm. Hitunglah nilai hambatannya. Bandingkan dengan nilai hambatan untuk kawat nikrom dengan panjang 50 cm. Mengapa nilai hambatannya sama atau mengapa nilai hambatannya berbeda? Ulangi langkah 2, dan 3 untuk kawat tembaga lain dengan nilai diameter 1,0 mm dan 1,5 mm, serta panjangnya 50 cm. Catatlah hasilnya pada tabel berikut.
No. Diameter (mm) 1 2 3
Sumber potensial
0,5 1,0 1,5
Luas Penampang Tegangan (mm2) (V) ... ... ...
... ... ...
Kuat Arus (A)
Hambatan ( )
... ... ...
... ... ...
Bagaimanakah hubungan antara luas penampang kawat tembaga dengan hambatan (R)?
Listrik Dinamis
175
4@30A0@:0=,4*6*4#3*3*,# =3030?0B<4;47:4A8
8:0 :43C0 :4A8;47 ?4@A0<00= 14@8:CB )4;08=14@60=BC=6?030 30= 9C6014@60=BC=6?03094=8A70< 10B0= 4=8A ?4=670=B0@ B4@A41CB 38E0:8;8 >;47 AC0BC 14A0@0= 70<10B0= 94=8A 70<10B0= 94=8A 30?0B 38BC;8A:0= A410608 14@8:CB
K
#$&- $8;08 0<10B0= 4=8A?030 H #)#/
.&4&2
;C<8=8C< 4A8 <0A &4@0: &;0B8=0 *4<1060 *C=6AB4= $8:@>< !0@1>= 4@<0=8C< )8;8:>= !020
L K L K
L K
L K
L K
L K L K
L K K L K KL K
K
K Sumber: Physics, 2000
$8;08 70<10B0= 94=8A 1414@0?0 1070= 3814@8:0= ?030 #$&- 0<10B0= 94=8A AC0BC ?4=670=B0@ 14@60=BC=6 ?030 AC7C ?4=670=B0@ B4@A41CB )420@0 <0B4<0B8A 7C1C=60= 0=B0@0 70<10B0= 94=8A 30= AC7C 38?4@>;47 30@8
t
0
0
B
K
!4B4@0=60=
B 70<10B0=94=8A0:78@ < 0 70<10B0=94=8A458A84= AC7C 70<10B0= ?4@B0<1070= AC7C H %;47 :0@4=0 70<10B0= A410=38=6 34=60= 70<10B0= 94=8A ?4=60@C7 AC7C B4@7030? 70<10B0= 9C60 30?0B 38BC;8A t 0 2 176
Mudah dan Aktif Belajar Fisika untuk Kelas X
2
Kata Kunci K
B !4B4@0=60= 270<10B0=0:78@ 70<10B0=
• • • •
beda potensial hambatan (resistor) hambatan jenis koefisien suhu hambatan
Contoh 8.6 *4=BC:0= 70<10B0= A410B0=6 0;C<8=8C< F0=6 ?0=90=6=F0 2< 30= ;C0A ?4=0
L K < 2< < 2< L K< L , +=BC:<4=4=BC:0=:0E0B?4=670=B0@0;C<8=8C<386C=0:0=?4@A0<00= < K L < L K L K < A 4@30A0@:0= C:C<%7< aluminium
L K , K L
0380@CAF0=6<4=60;8@30;0<?4=670=B0@ 0;C<8=8C<030;07
I
V
Contoh 8.7 )41C07B4@<><4B4@30@8:0E0BBC=6AB4=<4<8;8:870<10B0= ?030AC7C H 30=70<10B0==F0<4=9038 ?030A00BAC7C H*4=BC:0=;07AC7CF0=6 38BC=9C::0=B4@<><4B4@B4@A41CB:4B8:070<10B0=:0E0B=F0 #7#$ 8:4B07C8 H 2 H 2 #4;0;C8?4@A0<00=?4@C1070=70<10B0=14A0@70<10B0=94=8A0:0=38?4@>;47 K L K HK
0 H !4B8:070<10B0=:0E0B=F0 AC7C=F0030;07 K H 0
L K / A478=660
K
H K H H 038AC7CF0=638BC=9C::0=B4@<><4B4@030;07 H
Tes Kompetensi Subbab
B
&2+#,#/-#)%#-#.$5,5-#4*)#/
4@0?0:07 :C0B 0@CA F0=6 <4=60;8@ ?030 A41C07 ?4=670=B0@F0=6<4<8;8:870<10B0= 98:03814@8 B460=60=A414A0@ D>;B
)41C07 :>=4= ;8AB@8: 3814@8 B460=60= D>;B
A478=660<4=60;8@:0=0@CA 4@0?0:07:C0B0@CA
F0=6<4=60;8@?030:>=4=B4@A41CB98:03814@8 B460=60= D>;B Listrik Dinamis
177
0A8; ?4@2>100= 38?4@>;47 V (volt) 6@058:7C1C=60=B460=60= 3 30=:C0B0@CA?030A41C07 @4A8AB>@A4?4@B8?03060<10@ 14@8:CB 8:0 D>;BB4= BC:0= 14A0@ :C0B 0@CA F0=6 I 0,02 (ampere) <4=60;8@ 0@860<10@@0=6:080=?030A>0;=><>@B4=BC:0= 14A0@=F0 30=
&4@70B8:0=60<10@@0=6:080=14@8:CB8=8 50 mA R1
C I3
150 mA A
I1
B I2
R2
60 mA
R3
*4=BC:0=14A0@=F0( 98:0,030;07 D>;B
C. Rangkaian Seri dan Paralel &030 AC1101 8=8 =30 0:0= <4=4=:>=4= ;8AB@8: &4@70B8:0= A8AB4< 8=AB0;0A8 ;8AB@8: 38 @C<07@;8AB@8:A4B@8:0;8AB@8:B4;4D8A8@038>:>=B>7=30A430=6<4=>=B>=B4;4D8A881CA430=6<4=F4B@8:030=0F07 A430=6 <0=38 ;0B0;0B ;8AB@8: F0=6 A430=6 14@>?4@0A8 A00B 8BC F08BC B4;4D8A8 A4B@8:0 ;8AB@8: 30= ?>=B>7 B4@A41CB <4@C?0:0= A0;07 A0BC :4C=BC=60= ?4=66C=00= @0=6:080= ?0@0;4; 30;0< <4@0=6:08 0;0B0;0B ;8AB@8: 0608<0=0 98:0 0;0B 0;0BB4@A41CB38@0=6:08A4@80;0<<4=4=:>=4= ;8AB@8: F0=6 386C=0:0= 0;0<70;8=8=300:0=<4=66C=0:0=@4A8AB>@A410608:>=4=;8AB@8:=F0
1.
Hukum I Kirchhoff
)414;C<<455;0:C:0=;07 :4680B0= 14@8:CB
Aktivitas Fisika 8.3
Ingatlah Perjanjian cara penggambaran baterai (sumber potensial DC) pada rangkaian adalah sisi yang lebih panjang menandakan kutub positifnya. +
178
–
Hukum I Kirchhoff Tujuan Memahami Hukum I Kirchhoff.
A3 A2
P A4
Alat-Alat Percobaan 1. Amperemeter DC (0 – 1 A), A1 2. Tiga lampu kecil (masing-masing 1,5 V) A = amperemeter 3. Sebuah baterai (1,5 V) 4. Kabel penghubung secukupnya. Langkah-Langkah Percobaan 1. Susunlah peralatan seperti pada gambar, letakkan amperemeter di posisi A1 2. Amatilah, apakah semua lampu menyala? 3. Catat kuat arus yang ditunjukkan amperemeter, kemudian pindahkan amperemeter ke posisi A2, A3, dan A4. 4. Catat kuat arus yang ditunjukkan amperemeter pada semua posisi tersebut 5. Apakah A1 dan A2 menunjukkan angka yang sama? 6. Jumlahkan angka yang ditunjukkan oleh A 3 dan A 4. Apakah hasil penjumlahannya sama dengan angka yang ditunjukkan oleh A1 atau A2? 7. Apa kesimpulan Anda dari kegiatan ini?
Mudah dan Aktif Belajar Fisika untuk Kelas X
0@8:4680B0=B4@A41CB38?4@>;47:4A855 3)3+ '0!&&-$$ 14@1C=F8 (3+* & 031 *'120') 6 ,% + 13) . " 13 23 2'2') .#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=55 0:0= 14@;0:C <0AC: :4;C0@
I1
I3 I4 O
K
4=60= 34<8:80= ?030 #.$#2 14@;0:C
I2
I5
Gambar 8.6 Arus listrik yang memasuki dan keluar dari titik percabangan O.
Contoh 8.8 &4@70B8:0=60<10@14@8:CB 8A
2A P
4A
Q
4A
6A I
*4=BC:0=0@0730=14A0@:C0B0@CA;8AB@8: #7#$ 4@30A0@:0=60<10@B4@30?0B3C0B8B8:2010=6F08BCB8B8:&30=' I +=BC:B8B8:2010=6&<8A0;:0=0@CA?0302010=6&'<4<8;8:80@07:4;C0@30@8 B8B8:2010=6& 4@30A0@:0= C:C< !8@277>55
I
masuk
I keluar
&' 038&' 14@0@0730@8&:4' I
+=BC:B8B8:2010=6'<8A0;:0=0@07<0AC::4B8B8:2010=6' 4@30A0@:0= C:C< !8@277>55
I masuk I keluar
038K B0=30=460B85K<4=C=9C::0=107E00@071C:0=<0AC:B4B0?8:4;C0@30@8 B8B8:2010=6' Gambar 8.7 Rangkaian seri tiga buah lampu pijar.
2.
Rangkaian Seri Resistor
)41C07 @0=6:080= ;8AB@8: 38A41CB @0=6:080= A4@8 98:0 30;0< @0=6:080= B4@A41CB 70=F0 030 A0BC ;8=B0A0= F0=6 38;0;C8 0@CA ;8AB@8: &030 @0=6:080= A4@8:C0B0@CA;8AB@8:F0=6<4;0;C8A4B80? :>=4=A0<014A0@E0;0C?C= 70<10B0=A4B80?:>=4=14@1430 70<10B0= A4B80? :>=4= 14@1430 #.$#2 <4=C=9C::0= @0=6:08 0=A4@830@8B8601C07;0B0;0=B0@0B8B8: 30= 030;07 +=BC: 70<10B0=70<10B0= F0=6 38ACAC= A4@8 14@;0:C
a
R1
R2
R3
V1
V2
V3
b
Gambar 8.8 Susunan seri hambatan
Listrik Dinamis
179
K
%;47:0@4=0 30= 2-2 A478=660 B>B
K
+=BC: , 1C07 70<10B0= 14@;0:C 1 ,
R1
V1
R2
V2
V
K
&4@A0<00=?4@A0<00= 14@8:CB C=BC: <4=F434@70=0:0= 30= <4
K
K
V2
Gambar 8.9 Rangkaian seri dua buah hambatan.
1
(0=6:080= A4@8 14@5C=6A8 A410608 ?4<1068 B460=60= B>B B>B
3.
Gambar 8.10 Susunan paralel tiga buah lampu pijar R1 I1
a
I
R2 I2
I b
Rangkaian Paralel Resistor
8:0AC0BC@0=6:080=;8AB@8:<4<14@8:0=;418730@8A0BC;8=B0A0=C=BC: 0;8@0= 0@CA ;8AB@8:=F0 @0=6:080= B4@A41CB 38=0<0:0= @0=6:080= ?0@0;4; &030 @0=6:080= ?0@0;4; B460=60= ?030 A4B80? :>=4= A0<0 14A0@ E0;0C?C= 70<10B0= A4B80? :>=4= 14@1430 414@0?0;055 ?030 #.$#2 :C0B 0@CA ;8AB@8: F0=6 <4;0;C8 30= 030;07 30= 30?C= :C0B 0@CA 0=B0@0 B8B8: 30= 030;07 &030 @0=6:080= ?0@0;4; 14@;0:C K %;47:0@4=0
30= .
4=60= 34<8:80=
R3 I3
. Gambar 8.11
Susunan paralel hambatan
K
C=BC: , 1C07 70<10B0= 14@;0:C
... . ,
180
K
Mudah dan Aktif Belajar Fisika untuk Kelas X
K
)4;08= 30?0B 38ACAC= A420@0 A4@8 30= ?0@0;4; :>=4=:>=4= ;8AB@8: 30?0B ?C;0 38ACAC= A420@0 601C=60= A4@8?0@0;4;
Mari Mencari Tahu
Tantangan
&4@70B8:0=60<10@14@8:CB 41C07@0=6:080=B4@BCBC?F0=6B4@38@80B0A3C01C07@4A8AB>@30=A41C07AC<14@ B460=60= I1 I2
I
R1 R2 I
untuk Anda Jika Anda telah memahami susunan seri hambatan pada rangkaian, tentukan oleh Anda sehingga diperoleh persamaan (8 – 12), persamaan (8 –13), dan persamaan (8 – 14) dengan cara menurunkan dari persamaan (8 – 9) dan persamaan (8 – 10).
V
*C60A =30 030;07 <4<1C:B8:0= ?4@A0<00=?4@A0<00= 14@8:CB &4@A0<00= ?4@A0<00= 14@8:CB 14@6C=0 C=BC: <4=F434@70=0:0= 30= <4
8:0B4@30?0B 70<10B0=38ACAC=?0@0;4;14@;0:C
I
2-2
I
I
2-2
+=BC:,1C0770<10B0=F0=638ACAC=?0@0;4;30=A4B80?70<10B0=14A0@=F0 70<10B0=B>B0;=F0030;07
+
4.
,
0<10B0=?0@0;4;14@5C=6A8A410608?4<10680@CA34=60==8;08?4@10=38=60= :C0B0@CA?030A4B80?2010=6030;07 2-2
2-2
2-2
G R1
430?>B4=A80;A4B80?70<10B0=A0<014A0@
P
R
G R3
R4
Jembatan Wheatstone
4<10B0= &# 212-,#<4@C?0:0=A41C07<4B>34F0=6386C=0:0=C=BC: <4=6C:C@70<10B0=F0=614;C<38:4B07C8)4;08=8BC94<10B0=4&# 212-,# 386C=0:0= C=BC: <4=6>@4:A8 :4A0;070= F0=6 30?0B B4@9038 30;0< ?4=6C:C@0= 70<10B0= <4=66C=0:0= C:C< %7< )CAC=0= @0=6:080= 94<10B0= -740BAB>=4 38BC=9C::0= ?030 #.$#2 8:090@C<60;D0=><4B4@<4=C=9C::0=0=6:0=>;A4B8<10=614@ 0@B8 ?030 60;D0=><4B4@ B830: 030 0@CA ;8AB@8: F0=6 <4=60;8@ :810B=F0 ?030 :40300= 8=8 B460=60= 38 A0<0 34=60= B460=60= 38 30= B460=60= 38 A0<0 34=60= 38 A478=660 98:0 14@;0:C L L
R2
S V
Gambar 8.12 Rangkaian jembatan Wheatstone
G
K
&23#.##/ 9 38:4=0; 34=60= ?@8=A8? 94<10B0= -740BAB>=4 4=BC: A434@70=0 A41C07 94<10B0= -740BAB>=4 38BC=9C::0= A4?4@B8 ?030 #.$#2 !4B8:0 A0:;0@ 387C1C=6:0= 0@CA <4=60;8@ <4;0;C8
RX
R
1
2 S
E
Gambar 8.13 Rangkaian sederhana jembatan Wheatstone
Listrik Dinamis
181
Tugas Anda Turunkan oleh Anda persamaan (8 – 19) berdasarkan Gambar 8.13.
ACAC=0= @0=6:080= A430=6:0= 90@C< 0;D0=><4B4@ <4=F8;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 :0E0B 1 30= 2 30?0B B4@1020 <4;0;C8 A:0;0 ?0=90=6 ?030 :0E0B B4@A41CB
Contoh 8.9 &4@70B8:0=60<10@14@8:CB &0=90=6 2< 0@C<60;D0=><4B4@0:0= A4B8<10=6:4B8:0:>=B0:14@030 2<30@8 C9C=6*4=BC:0==8;0870<10B0=5 #7#$ 2<K 2< 2< )F0@0B94<10B0=30;0<:40300=A4B8<10=6 030;07 5 5 2< 2< 5
X
R = 120 G D
A
C
60 cm V
S
Contoh 8.10 *8601C07@4A8AB>@<0A8=6<0A8=6 30= 38ACAC=A4@830=C9C=6C9C=6=F0 387C1C=6:0=34=60=10B4@08 ,A4?4@B8?03060<10@14@8:CB 4
10
6
A
D B
C 60 V
*4=BC:0= 0 :C0B0@CAF0=6<4=60;8@?030@0=6:080= 1 1430?>B4=A80;0=B0@030= 2 1430?>B4=A80;0=B0@030= 3 1430?>B4=A80;0=B0@030= #7#$ 0 , , 038:C0B0@CAF0=6<4=60;8@?030@0=6:080=030;07 1 , 2 , 3 , 038 , ,30= ,
182
Mudah dan Aktif Belajar Fisika untuk Kelas X
5.
Rangkaian Seri dan Paralel Sumber Tegangan
)414;C<<4<1070AACAC=0=A4@8K?0@0;4;AC<14@B460=60=B4@;4187307C;C 0:0=381070A<4=64=08?4@14300=60F064@0:;8AB@8:34=60=B460=60=;8AB@8: a. Perbedaan Gaya Gerak Listrik dengan Tegangan Jepit 6 %#0 ) *'120') %%* 030;07 #" .-2#,1' * ,2 0 3(3,%3(3,% )323 13+ #0 031 *'120') )#2') 13+ #0 031 *'120') 2#01# 32 2'" ) +#,% *'0) , 031 *'120') #% ,% , (#.'2 030;07 #" .-2#,1' * ,2 0 3(3,%3(3,% 13+ #0 031 *'120'))#2') 13+ #0 031*'120')2#01# 322#0 # ,' 2 3+#,% *'0) , 031*'120') C1C=60= 0=B0@0 66; 30= B460=60= 94?8B 030;07
94?8B
K
b. Sumber Tegangan Disusun Seri +=BC:<4=30?0B:0=AC<14@B460=60=F0=6;418714A0@30@8?030B460=60= A4B80?AC<14@B460=60=1414@0?0AC<14@B460=60=70@CA38ACAC=A420@0A4@8 *86010B4@0838ACAC=A420@0A4@8A4?4@B8?030#.$#2 8:0=30 ?4@70B8:0=:4B86010B4@0838ACAC=14@34@4B38<0=0:CBC1:43C010B4@08 F0=6 14@34:0B0= A4;0;C 14@;0E0=0= B0=30 8:0 A49C<;07 AC<14@ B460=60= 0B0C 10B4@08 38ACAC= A420@0 A4@8 14@;0:C
B>B
1 r1 –
+
I
2 r2
3 r3
+
–
–
+
R
Gambar 8.14 Rangkaian seri tiga sumber tegangan atau baterai.
K
34=60= 70<10B0= 30;0<=F0 0B>B 0 0 0 .
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=
0 0 0
K
+=BC: , 1C07 AC<14@ B460=60= F0=6 38ACAC= A4@8 14@;0:C
, ,0
K
c. Sumber Tegangan Disusun Paralel 8:0A49C<;07AC<14@B460=60=F0=6<4<8;8:866;A0<0 1 2 3 38ACAC= A420@0 ?0@0;4; <0:0 14@;0:C
B>B
K
0<10B0= 30;0<=F0 38@C
. 0B>B 0 0 0
K
Listrik Dinamis
183
I
&4@70B8:0=#.$#2 !C0B0@CAF0=6<4=60;8@?030@0=6:080= 030;07
1 r1
2 r2
3 r3
2-2 0 0 0
K
R
Gambar 8.15 Tiga sumber tegangan disusun paralel.
C=BC:,1C07AC<14@B460=60=34=60=66; 30=70<10B0=30;0<0 F0=6 38ACAC= ?0@0;4; 14@;0:C
Tantangan untuk Anda Empat buah hambatan masingmasing besarnya 1 ohm dihubungkan seperti pada gambar berikut.
0 ,
K
Contoh 8.11 *8601C0710B4@0838ACAC=A420@0A4@8A4?4@B860<10@14@8:CB 1,5 V ; 0,2 1,5 V ; 0,2
1,5 V ; 0,2
A B Hitunglah hambatan total R antara titik A dan titik B.
R = 4,4
)4B80?10B4@08<4<8;8:866; ,30=70<10B0=30;0< 8:0:4B8601C0710B4@08 B4@A41CB387C1C=6:0=34=60=A41C0770<10B0= B4=BC:0=:C0B0@CAF0=6 <4=60;8@<4;0;C870<10B0= #7#$ 8:4B07C8
, 0 , +=BC:<4=4=BC:0=:C0B0@CAF0=6<4=60;8@386C=0:0= &23#.##/9 n nr R
, )+ 4, 4
038:C0B0@CAF0=6<4=60;8@
Contoh 8.12 C01C0710B4@0838ACAC=A420@0?0@0;4;A4?4@B8?03060<10@14@8:CB I1
1 r1
2 r2
I I2
2
184
Mudah dan Aktif Belajar Fisika untuk Kelas X
8:0A4B80?10B4@08<4<8;8:866; ,30=70<10B0=30;0<=F0 :4
, 0 ,
Kata Kunci • • • • • • • • •
rangkaian seri rangkaian paralel Hukum I Kirchhoff titik percabangan jembatan Wheatstone galvanometer gaya gerak listrik tegangan jepit hambatan dalam
,
B>B 0B>B 038:C0B0@CAF0=6<4=60;8@ 0
94?8B K0 ,K L , ,
0
1
038B460=60=94?8BA4B80?10B4@08030;07 ,
Tes Kompetensi Subbab
C
&2+#,#/-#)%#-#.$5,5-#4*)#/
&4@70B8:0=60<10@14@8:CB I1 I2
3
4 B
A 12 V
I3
6
*4=BC:0=14A0@=F0 0 70<10B0=B>B0;0=B0@0B8B8:30= 1 2 3
5A I
C
9
6
6
3A
4A 2A
7
A
B
7A
6A
6
9
P Q
&03060<10@@0=6:080=14@8:CB 5
0 :C0B0@CAF0=6<4=60;8@<4;0;C8;0
*4=BC:0=0@0730=14A0@:C0B0@CA *4=BC:0=14A0@30=0@070@CA30@860<10@14@8:CB 9A
4
D
*4=BC:0= 0 70<10B0=;8AB@8:0=B0@0B8B8:30= 1 70<10B0=;8AB@8:0=B0@0B8B8:30= "8<01C0710B4@08<0A8=6<0A8=634=60=66; ,30= 70<10B0=30;0< 38ACAC=A4@8:4
I P 3,5 A
2,5 A
Listrik Dinamis
185
D. Hukum II Kirchhoff 3)3+'0!&&-$$0B0C38A41CB9C60 230 ,*--.3830A0@:0=?030 C:C< !4:4:0;0= =4@68 =4@68 ?030 AC0BC @0=6:080= B4@BCBC? 030;07 :4:0; 3)3+ '0!&&-$$ <4=F0B0:0= 107E0 (3+* & *( 0 .#03 & , 2#% ,% , 6 ,% +#,%#*'*',%' 13 23 0 ,%) ' , 2#02323. *--. 1 + "#,% , ,-* )420@0 <0B4<0B8A 38BC;8A A410608 14@8:CB
R I
2
arah loop
1
Gambar 8.16 Sebuah rangkaian tertutup
K
&4@70B8:0= #.$#2 0F0 64@0: ;8AB@8: 30@8 AC<14@ B460=60= <4=F4101:0= 0@CA ;8AB@8: <4=60;8@ A4?0=90=6 ;>>? @CA ;8AB@8: 38 30;0< ;>>? <4=30?0B 70<10B0= A478=660 <4=60;0<8 ?4=C@C=0= B460=60= &2 3#.##/ 9 30?0B 38BC;8A A410608 14@8:CB
1.
2 r2
1 r1
b
I
arah loop
c
I
Rangkaian dengan Satu Loop
#.$#2 <4=C=9C::0= @0=6:080= A434@70=0 34=60= A0BC ;>>? &030 @0=6:080=B4@A41CB0@CA;8AB@8:F0=6<4=60;8@030;07A0<0F08BC @0=6:080=B4@A41CB0@CA;8AB@8:F0=6<4=60;8@030;07A0<0F08BC#8A0;:0= =30 <4=60<18; 0@07 ;>>? A40@07 34=60= 0@07 F08BC 7 7!7"7 )4;0= 9CB=F0 :C0B 0@CA 30?0B 3878BC=6 34=60= C:C< !8@277>55 14@8:CB #0:0 ?030 #.$#2 14@;0:C
R a
K
K 0 0 d
Gambar 8.17 Rangkaian dengan satu loop
Contoh 8.13 &4@70B8:0=@0=6:080=B4@BCBC?A4?4@B860<10@14@8:CB
1 36 V
2 16 V R1 = 6
r1 2
r2 0,5 B
A
I
R5 = 3
R2 = 6
I R4 = 6
R3 = 5 C
D
4 12 V
3 20 V
r4 0,7
r3 0,8
8BC=6;07 0 :C0B0@CAF0=6<4=60;8@?030@0=6:080= 1 #7#$ 0 #4=C@CB C:C< !8@277>553830;0<@0=6:080=B4@BCBC?B4@A41CB14@;0:C #8A0;:0=0@07;>>?A40@0734=60=?CB0@0=90@C<90<A478=660?4@A0<00=B4@A41CB <4=9038 K 1 K 2 K 3 4 0 0 0 0 K K K
186
Mudah dan Aktif Belajar Fisika untuk Kelas X
1
2.
, 038:C0B0@CAF0=6<4=60;8@030;07 =3030?0B<4=678BC=6 C=BC:;8=B0A0=F0=6<4=4
K , +=BC:90;0= K K , 038 ,
Ingatlah Perjanjian tanda ggl dan kuat arus dalam rangkaian tertutup (loop). a. Kuat arus bertanda negatif jika searah dengan arah loop, dan bertanda positif jika berlawanan arah dengan arah loop.
(ggl) bertanda negatif jika b. kutub positifnya lebih dahulu dijumpai daripada kutub negatifnya ketika mengikuti arah loop, dan sebaliknya.
Rangkaian dengan Dua Loop atau Lebih
(0=6:080= F0=6 <4<8;8:8 3C0 ;>>? 0B0C ;4187 38A41CB 9C60 @0=6:080= <09455 3 *4B0?:0= ;>>? 14A4@B0 0@07=F0 ?030 A4B80? @0=6:080= B4@BCBC? 4 *C;8A;07?4@A0<00=?4@A0<00=C=BC:A4B80?;>>?<4=66C=0:0= C:C< !8@277>55 5 8BC=6 14A0@0=14A0@0= F0=6 38B0=F0:0= <4=66C=0:0= ?4@A0<00= ?4@A0<00= ?030 ;0=6:07 4
Contoh 8.14 &4@70B8:0=60<10@@0=6:080=;8AB@8:14@8:CB 1
2,5 B
4V
2V 6
r = 0,5
r = 0,5
A 0,5 *4=BC:0= 0 :C0B0@CAF0=6<4=60;8@30;0<70<10B0= 30= 1 1430?>B4=A80;0=B0@0B8B8:30= #7#$ 8:4B07C8 (0=6:080=?030A>0;30?0B38C107<4=9038A4?4@B860<10@14@8:CB 1
I1
I3
2,5
I2
2V
4V 6 r = 0,5
Loop I
r = 0,5 Loop II
0,5
Listrik Dinamis
187
0
Kata Kunci • • •
Hukum II Kirchhoff rangkaian tertutup (loop) perubahan tegangan
4@30A0@:0= C:C< !8@277>55 0B0C K JJJJ 4@30A0@:0= C:C< !8@277>55C=BC:">>? 38?4@>;47
K JJJJ 4@30A0@:0= C:C<!8@277>55 C=BC:">>? 38?4@>;47
K K K JJJJ 4=60=<4=AC1AB8BCA8:0= &23#.##/ :430;0< &23#.##/ <0:038?4@>;47
30= K 038:C0B0@CAF0=6<4=60;8@30;0<70<10B0= 030;07 F0=6<4=60;8@
30;0<70<10B0= 030;07 30=F0=6<4=60;8@30;0<70<10B0= 030;07 B0=30K<4=C=9C::0=107E00@070@CA14@;0E0=0=0@0734=60= 0@07?4<8A0;0=
Tes Kompetensi Subbab
D
&2+#,#/-#)%#-#.$5,5-#4*)#/
&4@70B8:0=60<10@@0=6:080=;8AB@8:14@8:CB
24 25 V
2
8:0 ,30= , B4=BC:0=
4
R1
16 10
6
E2
*4=BC:0=:C0B0@CAF0=6<4=60;8@30;0<70<10B0= 30=:C0B0@CAB>B0;30@8AC<14@B460=60= &4@70B8:0=60<10@@0=6:080=;8AB@8:14@8:CB B 2
8V r = 1
3
5 8
R2
E1
6
0 0@CAF0=6<4=60;8@?030@0=6:080= 1 B460=60=?030A4B80?@4A8AB>@ 8:0
8:0 30= ,<0:0 B4=BC:0=
2V
E
r = 1
R1
R2
A
*4=BC:0= 0 :C0B0@CAF0=6<4=60;8@30;0<70<10B0= 30= 1 1430?>B4=A80;0=B0@0B8B8:30=
188
Mudah dan Aktif Belajar Fisika untuk Kelas X
0 1 2
B460=60=?030A4B80?@4A8AB>@ 0@CAB>B0; 0@CAF0=6<4=60;8@?030A4B80?@4A8AB>@
E. Sumber Arus Searah dari Proses Kimiawi =30=309C60:<0AF0@0:0B38AC0BC304@07B4@?4=28;F0=6<4=66C=0:0=A4; 5>B>D>;B08: A410608 AC<14@ 0@CA ;8AB@8: C=BC: ?4=4@0=60= )4;47AC<14@B460=60=4@8:CB2>=B>7AC<14@AC<14@0@CA;8AB@8:
)C<14@ 4;4:B@><06=4B8: 30?0B <4=670A8;:0= 0@CA ;8AB@8: :0@4=0 6490;08=3C:A84;4:B@><06=4B8:#8A0;=F038=0<>38?4@;870B:0=?030 #.$#2 )C<14@ ;8AB@8: B4@<>4;4:B@8: 0@CA ;8AB@8: 30?0B 3870A8;:0= 30@8 454: B4@<>4;4:B@8: )C<14@ 5>B>;8AB@8: 14@0A0; 30@8 AC0BC ?@>A4A 8A8:0 F0=6 <4=6C107 4=4@68 2070F0 <4=9038 4=4@68 ;8AB@8:
)C<14@ ?84G>4;4:B@8: 3870A8;:0= 30@8 454: ?84G>4;4:B@8: F08BC A850B 1070=F0=60?018;0<4=4@8<0B4:0=0=30@8;C0@30?0B<4=670A8;:0= 0@CA ;8AB@8:
1.
magnet
kumparan komutator
arus arus
Gambar 8.18 Dinamo atau generator adalah contoh dari sumber elektromagnetik.
Sumber Listrik dari Bahan Kimia
&4=4;478;B>B>B>B:0B0: F0=6AC307<0B8<4=F4=B0:98:038A4=BC734=60=3C0;>60<F0=614@1430 &4@:4<10=60=14@8:CB=F0-&33#/%20!0-4#<4=4
*#+#, .0'+#0 F08BC 4;4<4= F0=6 <4<4@;C:0= ?4@60=B80= 1070= 1070= ?4@40:A8 A4B4;07 <4<1410A:0= A49C<;07 4=4@68 <4;0;C8 @0=6:080= ;C0@=F0 ;4<4= ?@8<4@ 8=8 <4=66C=0:0= 1070= :8<80 F0=6 @40:A8 :8<80=F0 B0: 30?0B 3810;8::0= A478=660 4;4<4= ?@8<4@ 70=F0 30?0B 386C=0:0= A0BC :0;8 ?4<0:080= >=B>7=F0 A4; ,>;B0 30= 4;4<4= :4@8=6 10B4@08 *#+#, 1#)3,"#0 F08BC 4;4<4= F0=6 1070=1070= ?4@40:A8=F0 30?0B 38?4@10@C8:4<10;8A4B4;07B830:14@5C=6A8;068>=B>7=F00:C@ 30= 10B4@08 8A8 C;0=6 a. Elemen Primer ;4<4=8=810=F0:<020<=F04@8:CB8=81414@0?0AC<14@0@CA;8AB@8: F0=6 B4@6>;>=6 A410608 4;4<4= ?@8<4@ -&.&/ !0-4# ;4<4=,>;B038B4;47-&33#/%20 !0-4#!0-4# <4=460< 30= ;0@CB0= 0A0< 0B0C 60@0< 30?0B 386C=0:0= A410608 4;4<4= A434@70=0 :0= B4B0?8 180A0=F0 F0=6 386C=0:0= 80;07 ;4;B0 &@>A4A:8<80F0=6B4@9038?0304;4<4=,>;B030?0B3894;0A:0=34=60= ?4@A0<00= :8<80 14@8:CB )% )% K
aliran elektron kutub
Cu
Zn
Larutan asam sulfat encer
Gambar 8.19 Elemen Volta
Listrik Dinamis
189
)4B80?<>;4:C;0A0<AC;50B3830;0<08@?4207<4=9038 8>=783@>64= F0=614@A8B8530= 8>=)% F0=614@<A4=6 F0=6<4;0@CB:430;0<;0@CB0=0A0<AC;50B14@C?0/= )4B80?0B><F0=6 ;0@CB <4=8=660;:0= 3C0 4;4:B@>= ?030 ;4=4;4:B@>= 8=8;07F0=6<4=60;8@30@8A4=6:4B4<1060C<4;0;C8:0E0B?4=670=B0@ A478=660 B4@9038 0@CA ;8AB@8: -&.&/ &2*/( #45 #4&2#* ;4<4=:4@8=60B0C;418738:4=0;34=60=8AB8;0710BC10B4@08<4@C?0 :0=AC<14@0@CA;8AB@8:F0=6?0;8=610=F0:386C=0:0=#.$#2 <4 =C=9C::0= 60<10@ ACAC=0= 30A0@ A41C07 4;4<4= :4@8=6 !CBC1?>A8B854;4<4=:4@8=6B4@1C0B30@8:0@1>=F0=638:4;8;8=688=B8F0=6 B4@1C0B 30@8 20:A830 <0=60= 30= 0@0=6 F0=6 38<0=8C<:;>@8300:A830<0=60=30=:0@1>= 38 A4:4;8;8=6 10B0=6 :0@1>= 14@B8=30: A410608 34?>;0@8A0B>@ ?4=246070= ?4=6:CBC10=
tutup kuningan batang karbon seng pasta kimia
Gambar 8.20 Susunan dasar sebuah elemen kering.
tutup lubang atas
terminal
lapisan pelat pelat pemisah oksida logam asam sulfat
Gambar 8.21 Bagian-bagian akumulator asam sulfat.
b. Elemen Sekunder ;4<4= A4:C=34@ 030;07 AC<14@ 0@CA 30@8 1070= :8<80 F0=6 @40:A8 :8<80=F0 30?0B 3810;8: %;47 :0@4=0=F0 4;4<4= 8=8 30?0B 38?4@1070@C8 A420@0 14@C;0=6C;0=6 )0;07 A0BC 2>=B>7 4;4<4= 8=8 F0=6 ?0;8=6 38:4=0; 38 <0AF0@0:0B 030;07 0:C@ 0B0C 0:8 0;0<1018=80:0=381070A3C01C072>=B>74;4<4=A4:C=34@F08BC 0:C@ B8<10; 0A0< AC;50B 30= 0:C@ =8:4; :03<8C< ,5.5-#402 *.$#-3#. 5-'#4 :C@8=810=F0:38B4B>@<>18;0B0C ?030 <4A8=<4A8= F0=6 ;08= A410608 AC<14@ ;8AB@8: &030 0:C@ 94=8A 8=8 1070= ;0@CB0= 4;4:B@>;8B F0=6 386C=0:0= 030;07 0A0< AC;50B BC;07 A4101=F00:C@94=8A8=838A41CB9C600:C@0A0<AC;50B0680= 10680= 0:C@ 0A0< AC;50B 38BC=9C::0= ?030 #.$#2 &030 30A0@=F0 030 3C0 ?@>A4A ?4=B8=6 30;0< 0:C@ &4@B0<0 ?@>A4A ?4=68A80= 0:C@ 30= :43C0 ?@>A4A ?4=66C=00= 0:C@ &030 ?@>A4A ?4=68A80= 0:C@ A49C<;07 0@CA ;8AB@8: 380;8@:0= ?030 0:C@ A434<8:80= 78=660 14@C107 <4=9038 4=4@68 :8<80 8 30;0< 0:C@;0@CB0=4;4:B@>;8B )% B4@C@08<4=9038 30=)% (40:A8 :8<80 F0=6 B4@9038 ?030 ?@>A4A ?4=68A80= 030;07 A410608 14@8:CB I 8 :0B>34 &1)% 4K &1 )% I
8 0=>34 &1)% )% K % &1% )% 4K
&030?@>A4A?4=68A80=8>= 380;8@:0=:4:0B>3030=8>=AC;50B380;8@:0= :4 0=>30 30?C= ?030 ?@>A4A ?4<0:080= :43C0 4;4:B@>30 387C1C=6:0= A478=660B4@90380;8@0=4;4:B@>=30@84;4:B@>30&1<4;0;C81410=<8A0;=F0 ;030 &1% &030 ?@>A4A ?4<0:080= 38 30;0< 0:C@ 0:0= B4@9038 @40:A8 :8<80 A410608 14@8:CB I >= ?>A8B85 0:0= 14@64@0: <4=C9C &1% A478=660 B4@9038 @40:A8 &1&1% )% &1)% % !4?8=6 &1% 14@C107 <4=9038 B8<10; AC;50B &1)%
190
Mudah dan Aktif Belajar Fisika untuk Kelas X
I
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
Tes Kompetensi Subbab
Kata Kunci • • • • • • •
proses kimiawi elektromagnetik termoelektrik fotolistrik piezoelektrik elemen primer elemen sekunder
E
&2+#,#/-#)%#-#.$5,5-#4*)#/
)41CB:0=AC<14@AC<14@0@CA;8AB@8: ?0 F0=6 38<0:AC3 4;4<4= ?@8<4@ 30= 4;4<4= A4:C=34@)41CB:0=2>=B>72>=B>7=F0 *C;8A:0=@40:0A8:8<80F0=6B4@90383830;0<4;4<4=,>;B0
4;0A:0=F0=638<0:AC3?@>A4A?4=68A80=30=?@>A4A ?4=66C=00=0:C@ *C;8A:0=@40:A8:8<80F0=6B4@903838:0B>3430=38 0=>34?030?@>A4A?4=68A80=0:C@
F. Tegangan Listrik Searah dan Bolak-Balik 1.
Energi Listrik
=30 B4;07 <4=64B07C8 107E0 0@CA ;8AB@8: <4=60;8@ 30@8 ?>B4=A80; B8=668:4?>B4=A80;F0=6;4187@4=307)4;08=8BC4;4:B@>=A410608?4<10E0 B4=A80; F0=6 14A0@=F0 B4=A80; ;8AB@8:=F0 ./ %;47 :0@4=0 8BC 4=4@68 ;8AB@8: 030;07 CA070 C=BC: <4<8=307:0=
/ K/
K
/ 2
Ingatlah I=
q J
maka q = I t
K
030;0714A0@4=4@68;8AB@8: 030;07B460=60=030;07:C0B0@CA ;8AB@8: 30= 2 030;07 E0:BC 8:0 <4=AC1AB8BCA8:0= ?4@A0<00= K <4=9038 2 K 0B0C
2
K
Listrik Dinamis
191
Tokoh Alessandro Volta (1745 – 1827)
!4B4@0=60= 4=4@68 ;8AB@8: 1430 ?>B4=A80; , :C0B0@CA 70<10B0= 2 A4;0=6 E0BC A 30?C=30F0;8AB@8:030;079C<;074=4@68?4@A0BC0=E0:BC0F0 ;8AB@8: 30?0B 3878BC=6 34=60= @C
0B0C 2
K
0B0C
K
Sumber: www.physics. com.
Alessandro Volta, adalah Fisikawan yang dilahirkan di Como, Italia. Dia menciptakan electrophorus, yaitu suatu alat untuk membangkitkan listrik statis pada 1775 dan menemukan gas metana pada 1778. Dia ditetapkan sebagai profesor untuk filsuf ilmu alam di Pavia. Terinspirasi oleh temannya Luigi Galvani, Volta menemukan bahwa arus listrik dibangkitkan ketika dua logam berbeda berada pada jarak yang sangat dekat, dan mengembangkan baterai listrik pertama pada 1800. Namanya diabadikan untuk satuan beda potensial listrik, volt. Sumber : www.allbiographies.com
;0B0;0B?4<0=0AF0=610=F0:389C;8AB@8:30=0'!#!--)#0=4@68 ;8AB@8:F0=63870A8;:0=>;474;4<4=?4<0=0A;8AB@8:B4@A41CBA4;0<02A4:>= A414A0@ 2 !4@ A414A0@+! )420@0<0B4<0B8A?4@C1070=4=4@68;8AB@8: <4=9038 4=4@68 :0;>@ B4@A41CB 30?0B 38BC;8A:0= A410608 14@8:CB 2 +!
K
!4B4@0=60= + <0AA0 08@ :6 ! :0;>@94=8A08@ :6H :4=08:0= AC7C 08@ H 0B0B0= &23#.##/ 9
14@;0:C 98:0 B830: B4@9038 ?4@C1070= EC9C3 208@ A414;C< B4@9038 ?4=6C0?0=
Contoh 8.15 )41C071>;0;0B4=A80;
,30=38=F0;0:0=A4;0<0 <4=8B 8BC=64=4@68;8AB@8:F0=6B4@?0:08;0
, 2 <4=8B A 0<10B0=;0=AB0=A478=660
<0:0 L
, L - , 2 - A L
0384=4@68;8AB@8:F0=6B4@?0:08 L
192
Mudah dan Aktif Belajar Fisika untuk Kelas X
Mari Mencari Tahu =4@68;8AB@8:A0=60B38?4@;C:0=30;0<:4783C?0=<0=CA80=4@68;8AB@8:38@C<07 =3014@0A0;30@8&"$=4@68;8AB@8:8=8380=B0@0=F014@0A0;30@8&4<10=6:8B "8AB@8:*4=0608@&"*&4@=07:07=30<4=34=60@&4<10=6:8B"8AB@8: *4=060$C:;8@*C60A=3020@8;078=5>@<0A8<4=64=08&4<10=6:8B"8AB@8: *4=060$C:;8@8=8*C;8A;078=5>@<0A8B4@A41CB30;0<:4@B0A30=38:C;070B0C304@07 :>@0=8=B4@=4B30=<4380;08=C=BC:<4=20@88=5>@<0A8B4@A41CB
=30 B4;07 <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=F0 4A0@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= 4E0A0 8=8 70?4@0A8:0= 34=60= 4=4@68 ;8AB@8: 0@CA 1>;0:10;8: A4?4@B8 B0;0:10;8: 30= 0@CA A40@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&-"# 6 1!'*-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=6 C:C@ =8;08 B460=60= 30= :C0B 0@CA ;8AB@8: 30?0B 386C=0:0= D>;B<4B4@ 30=0
2.
Mengamati Tegangan Listrik DC dan Tegangan Listrik AC
+=BC:<4=6C:C@B460=60= 30= <0:A8A8;>A:>?A4?4@B8?030#.$#2 C1C=6:0=B4@ <8=0;0<1>; / <0B8 0?0:07 90@C< 100=B4@A41CB34=60=<4=67C1C=6:0=0@CA@0=6:080=:4!& ,,#*B460=60= ?030>A8;>A:>?!4;BA38D<8A0;=F0?030?>A8A8 <0B8 64;><10=6 B460=60= 38 ;0F0@ >A8;>A:>? <0:0 0:0= B4@;870B 107E0 B460=60=14@C107A420@0?4@8>38:A4?4@B8?030#.$#2 #:8@0:8@0 :40B0A30= :410E07=30B4;07<4=64B07C8107E0B460=60=F0=6 ?>;0@8B0A=F0A4?4@B864;><10=6030;07B460=60=;8AB@8:1>;0:10;8: 0608<0=0 98:0 0@CA F0=6 38;4E0B:0= 030;07 0@CA A40@07 8:0 =30 <4=60<0B8 B460=60= A40@07 F0=6 3870A8;:0= @0=6:080= 0@CA A40@07 30= 10B4@08 <4=66C=0:0= >A8;>A:>? A4B4;07 =30 <4=67C1C=6:0= 0@CA :4270==4;90@C<0A8;>A:>?0:0=B4@;870B70=F030;0<A0BC0@07A4?4@B8 ?030 #.$#2 $ )4B4;07<4;0:C:0=?4=60<0B0=30@8?4@2>100=B4@A41CB=3030?0B <4=64B07C8 A49C<;07 ?4@14300= 0=B0@0 B460=60= 30= B460=60= "8AB@8:A40@07<4<8;8:8B460=60=F0=6B4B0?A4B80?A00B30=6@058:B460=60=
Sumber: Femina, 1995
Gambar 8.22 Mixer dan bor listrik dioperasikan dengan energi listrik.
Sumber: Phywe
Gambar 8.23 Osiloskop
(a) Galvanometer
Osiloskop 220 2
220 V
0
R
220 2
(b) Galvanometer Osiloskop 220 V 220 V
R
Gambar 8.24 a. Mengamati arus bolak-balik dengan amperemeter DC dan osiloskop. b. Mengamati arus searah dengan amperemeter DC dan osiloskop.
Listrik Dinamis
193
=F014@C?060@8A;C@CAA430=6:0=;8AB@8:1>;0:10;8:<4<8;8:8B460=60=F0=6 14@C107C107 A4B80? A00B F08BC 14@14=BC: A8=CA>830; 414@0?0 ?4@14300= ?@8=A8?;8AB@8:A40@0730=1>;0:10;8:30?0B=30;870B?030 #$&- #$&-
414@0?0&4@14300=&@8=A8?"8AB@8:)40@0730=>;0:0;8: 5.$&2253 #/(,#*#/ 0-#,#-*,
Informasi untuk Anda Cara menghubungkan kabel pemanas air dapat dilakukan sebagai berikut. Ketiga lubang pada steker, yaitu dua ujung terminal pemanas masing-masing dihubungkan ke kutub positif dan negatif, dan kabel lain berwarna biru menuju netral. kabel ditanahkan
#)
I "0F0@>A8;>A:>?<4=C=9C::0= 60<10@B460=60=1>;0:10;8:
I "0F0@>A8;>A:>?<4=C=9C::0= 60<10@B460=60=A40@07
I "8AB@8:
I "8AB@8:AC;8BC=BC:<4<4=C78 :41CBC70=?0A>:0=30;0<9C<;07 14A0@ I @CA30=B460=60= 70=F0 <4<8;8:8=8;08454:B85=F0
4A0@=F00@CA30=B460=60=1>;0:10;8:30?0B380<0B834=60=0D><4B4@ @CA30=B460=60=F0=638BC=9C::0=0;0B8=8<4@C?0:0=70@60454:B85=F0 &>;0@8B0A ;8AB@8: F0=6 3870A8;:0= 30@8 90@8=60= &"$ 14@C?0 6@058: A8 =CA8>30; 34=60= @4:C4=A8 G @CA 454:B85 30?0B 38@C
<0:A
K
34=60= 20@0 F0=6 A0<0 C=BC: B460=60= 454:B85 #$ 0:0= 38?4@>;47 ?4@ A0<00= #$#$ #$
sekring
penahan
<0:A
K
kabel berarus
Contoh 8.16 kabel netral
Information for You To connect wires of water heater can do as below. Three hole on the plug, that is two edge heater terminals connect to positive and negative poles and other blue wires toward neutral.
194
0@8?4=60<0B0=<4;0;C8>A8;>A:>?AC<1CD4@B8:0;380BC@ ?030 B460=60= ,2< A430=6:0= A4;0=6 E0:BC <4=C=9C::0=B0:<4<8;8:8C:C@0=
2<L 2< *4=BC:0= Vpp 0 B460=60=<0:A8
2<%;47:0@4=0A:0;0D4@B8:0; ,2<<0:0
Mudah dan Aktif Belajar Fisika untuk Kelas X
1 cm 1 cm
T
1
.. 2<L ,2< , *460=60=<0:A8
+ )1 .. ,, 038B460=60=<0:A83430;0<6@058: 2<):0;07>@8G>=B0;
G $
Tantangan untuk Anda Persamaan tegangan bolak-balik suatu rangkaian listrik memenuhi persamaan V = 314 sin 50 V. Tentukan tegangan rata-rata yang dihasilkan sumber tersebut.
Contoh 8.17 )41C07@0=6:080=F0=6387C1C=6:0=34=60=AC<14@;8AB@8:1>;0:10;8:38C:C@34=60= D>;B<4B4@30=<4=C=9C::0=0=6:0 ,4@0?0:0770@60<0:A8;0: 10;8:AC<14@*C;8A:0=?C;0?4@A0<00==F098:05@4:C4=A8=F0 G #7#$ 8:4B07C8 #$ ,$ G 0 $8;08B460=60=<0:A8
<0:A , &4@A0<00=B460=60=38?4@>;4734=60=<4=66C=0:0=?4@A0<00=14@8:CB Vmaks sin t A8= 2
Contoh 8.18 )41C07@0=6:080=0@CA1>;0:10;8:<4<8;8:870@60B460=60=A4106085C=6A8E0:BC F08BC A8= 2 8BC=6;07 0 B460=60=<0:A834 1 B460=60=?C=20::4?C=20: .. 55@4:C4=A8$ 2 B460=60=454:B85 #$ 6B460=60=A4B4;072 A 3 5@4:C4=A80=6C;4@ #7#$ 8:4B07C8 &4@A0<00=B460=60=A4106085C=6A8E0:BC A8= 2 %;47:0@4=0B460=60=<4@C?0:0=5C=6A8A8=CA>830;B4@7030?E0:BC<0:0?4@A0<00= B460=60=30?0B38BC;8A:0= + A8= 2 @03A 0 *460=60=<0:A8
<0:0 + , 1 2
.. + , , #$ + , ,
Listrik Dinamis
195
3 4 5 6
3.
Sumber: Young Scientists, 1997
Gambar 8.25 Arus listrik disalurkan oleh jaringan PLN.
kawat sekring
tutup logam tabung kaca tutup logam
Gambar 8.26 Sekring
@03A A
$ G A8= 2 A8= ,
Pemasangan Listrik di Rumah Tangga
&030C;0:10;8: *070?0=<0AC:=F00@CA;8AB@8:30@8B80=690@8=60=:4@C<07030;07 A410608 14@8:CB @CA;8AB@8:<0AC::0;8?4@B0<0<4;0;C8 ','0!3'2 0# )#0 0B0C ?4<10B0A 30F0 F0=6 14@5C=6A8 <4<10B0A8 30F0 <0:A8B0: A4:@8=6 30= 0:78@=F0 :4 A4
8:0 ?4<10B0A 30F0 14A0@=F0 30= B460=60= ;8AB@8: 38 @C<07 ,30F0<0:A8;47 <4;41878 E0BB 8:0 ?4<0:080= 30F0 ;8AB@8: ;4187 14A0@ 30@8?030 E0BB :CB><0B8A A0:;0@ ?030 <4=9038 '' )4:0@0=62>10=3078BC=698:0?030 B4@20B0B=8;080@CA 14@0?0 B>B0; 30F0 ?4@0;0B0= ;8AB@8: 38 @C<07 060@ <4=20?08 50:B>@ :40<0=0= )0:;0@ ?4<10B0A 30F0 0:0= BC@C= A420@0 >B><0B8A :4B8:0 B4@9038 7C1C=60= A8=6:0B :>@A;4B8=6 ;8AB@8: 30= <4=60:810B:0= 0@CA ;8AB@8: B4@?CBCA )4;0=9CB=F0 0@CA ;8AB@8: 70@CA <4;4E0B8 A4:@8=6 C=BC: <4=9060 :40<0=0= 414@0?0 ?4@0;0B0= ;8AB@8: 387C1C=6:0= A420@0 B4@?8A07 :4 A4:@8=6=F0 <0A8=6<0A8=6 F0=6 B4@30?0B ?030 :>B0: A4:@8=6 CB0<0 )4:@8=6$31#B4@1C0B30@8A4CB0A:0E0BB4<1060B8?8A30=0:0=<4=9038 ?0=0A :4B8:00@CA <4=60;8@ <4;0;C8=F0 )4:@8=6 0:0= B4@10:0@:418; A4?430<>B>@?4A0E0B@038>30=B4;4D8A8)4B80?A4:@8=6<4<8;8:8=8;08:C0B 0@CAF0=6B4;0738B4B0?:0=<8A0;=F0
Contoh 8.19
Kata Kunci • • • • • •
tegangan listrik searah tegangan listrik bolak-balik energi listrik daya listrik osiloskop transmisi daya listrik
196
)41C07A4B@8:0 - ,0:0=38;4=6:0?834=60=A41C07A4:@8=6 8:0A4:@8=6F0=6 B4@A438014@=8;08 30= 14@0?0:07=8;08A4:@8=6F0=60:0=38?8;87 #7#$ !C0B0@CAF0=638?4@;C:0=A4B@8:0030;07 , )4:@8=6F0=6386C=0:0=70@CAA438:8B;418714A0@30@8?030 A478=660F0=638?8;87 030;07A4:@8=614@=8;08
Mudah dan Aktif Belajar Fisika untuk Kelas X
&4<0A0=60= A49C<;07 ;0
saklar
meteran lampu stop kontak
kotak sekring
saklar
sakelar stop kontak
sumber tegangan PLN 220 V 60 W
15 W
100 W
transmisi tegangan tinggi
pusat pembangkit listrik
Gambar 8.27
Gambar 8.28
Pemasangan lampu secara paralel.
Contoh diagram kabel listrik di rumah
pabrik
tiang listrik rumah
generator
Gambar 8.29 Bagan transmisi daya listrik jarak jauh transformator
transformator
gardu listrik Sumber: Jendela Iptek, 1997
Tes Kompetensi Subbab
F
&2+#,#/-#)%#-#.$5,5-#4*)#/
)41C07A4B@8:0;8AB@8:34=60=A?4A858:0A8 - , 38?0A0=6?0301430?>B4=A80;
,30=38=F0;0:0= A4;0<0 <4=8B 8BC=64=4@68;8AB@8:F0=6B4@?0:08 A4B@8:0B4@A41CB &030A41C07?4<0=0A;8AB@8:B4@20=BC<A?4A858:0A8 :- ,*4=BC:0= 0 14A0@=F04=4@68F0=63870A8;:0=A4;0<0 <4=8B 1 :C0B0@CA;8AB@8:F0=6<4=60;8@ &4A0E0B*,@0B0@0B038=F0;0:0=90<B80?70@8=F0 8:0 ?4A0E0B*,387C1C=6:0=B460=60= ,0@CAF0=6 <4=60;8@030;07 8:070@60?4@:-7(? B4=BC:0= 70@60 4=4@68 ;8AB@8: F0=6 386C=0:0= C=BC: <4=F0;0:0=*,A4;0<0A41C;0= 1C;0= 70@8
4;0A:0=0?0F0=638<0:AC334=60= 0 70@60454:B85:C0B0@CA30=B460=60=1>;0:10;8: 1 70@60<0:A8;0:10;8: 2 70@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;08 454:B85=8;08<0:A8
0@860<10@6@058:14@8:CBB4=BC:0=;07 I(A) 10 waktu(s)
0 0@CA<0:A8B8 4@4:;8AB@8:
#8# "
&(#/(#/ !
&,2*/(
Rangkuman
@CA ;8AB@8: 030;07 B4=A80; B8=668 :4 ?>B4=A80; @4=307 !C0B 0@CA ;8AB@8: 383458=8A8:0= A410608 <0B0= ;8AB@8:
F0=6 <4=60;8@ 30;0< AC0BC ?4=670=B0@ A4B80? A0BC A0BC0= E0:BC
Listrik Dinamis
197
#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
C:C< !8@277>55 <4=F0B0:0= 107E0 9C<;07 0@CA ;8AB@8: F0=6 <0AC: ?030 AC0BC B8B8: ?4@2010=60= A0<0 34=60= 9C<;07 0@CA ;8AB@8: F0=6 :4;C0@ 30@8 B8B8: 2010=6 B4@A41CB C:C< !8@277>55 <4=F0B0:0= 107E0 9C<;07 0;9010@ ?4@C1070= B460=60= F0=6 <4=64;8;8=68 AC0BC @0=6:080= B4@BCBC? ;>>? A0<0 34=60= =>;
Peta Konsep %A8;>A:>?
,>;B<4B4@
*342*,*/#.*3 <4
38C:C@ 34=60=
*460=60=
@CA"8AB@8:
3878BC=6 34=60=
38C:C@ 34=60=
B4@9038 ?030
C:C< %7<
3878BC=6 34=60=
(0=6:080= *4@BCBC?
C:C< !8@277>55
3894;0A:0= >;47
:>=4==F0 3878BC=6 34=60=
4=8A
0<10B0=
)C<14@ *460=60=
38?4=60@C78 >;47
94=8A=F0
&0=90=6
"C0A
)C7C
38A41CB
0<10B0= 4=8A
38?4=60@C78 >;47
Refleksi Setelah mempelajari bab ini, tentunya Anda dapat membuat rangkaian sederhana dan menganalisisnya menggunakan Hukum Kirchhoff. Dapatkah Anda mengidentifikasi alat-alat listrik di rumah yang
198
Mudah dan Aktif Belajar Fisika untuk Kelas X
menggunakan listrikAC dan DC? Materi manakah yang masih Anda anggap sulit? Diskusikan materi tersebut dengan teman-teman Anda atau dengan guru Fisika Anda.
Tes Kompetensi Bab 8 *-*)-#)3#-#)3#45+#7#$#/8#/(1#-*/(4&1#4%#/,&2+#,#/-#)1#%#$5,5-#4*)#/
0<10B0=;8AB@8:30;0<AC0BC:0E0B030;07 0< 10B0=8=80:0=<4=9038 0 98:0?4=0
1 98:0?0=90=6=F03860=30:0= 2 98:0AC7C=F03860=30:0=
3 98:090@890@8=F03860=30:0=
4 98:0380<4B4@=F03860=30:0= &030@0=6:080=A4?4@B8 2 60<10@66;10B4@08, 2 2 30=70<10B0=30;0< 3 =F0 !C0B0@CAF0=6<4=60;8@?03070<10B0= 030;07 0 3 1 4 2 &4@70B8:0=60<10@14@8:CB 3 2 4 !C0B0@CAF0=6A430=638C:C@14A0@=F0 1 0 5 030;07 5 A 0 1 A 0 3 1 4 2 )C0BC 0
0
=4EB>= D>;B
2>C;><1 >7< 2>C;><1
D>;B
34B8:
3
9>C;4 34B8:
0C;><1 34B8:
D>;B =4EB>= <4B4@ 2>C;><1
)C0BC?4@2>100=;8AB@8:38;0:C:0=34=60=<4=6C107 C107 70<10B0= ?030 1430 ?>B4=A80; F0=6 B4B0? @058:7C1C=60=:C0B0@CA30=70<10B0=70A8; ?4@2>100=B4@A41CB38BC=9C::0=>;476@058:
0
1
I
4
I
R
I
R
R
I
R
R
3
2
I
)49C<;07:0E0B30@81070=F0=6A0<0<4<8;8:8 380<4B4@A0<0B4B0?8?0=90=6=F0 B830:A0<0 8:0 38;0:C:0=?4=6C:C@0=70<10B0=:0E0BB4@A41CB 30= 70A8;=F0 38BC0=6:0= 30;0< 6@058: 7 0:0= 24=34@C=6A4?4@B8 0
R(ohm)
0
3
R(ohm)
1
R(ohm)
I(A)
4
I(A)
2
R(ohm)
I(A)
I(A) R(ohm)
I(A)
(0=6:080=70<10B0=A4?4@B8 ?03060<10@<4=670A8;:0= 3 70<10B0=B>B0;A414A0@ 0 3 1 4 2 6
3 3
3
3
9
3
"8<01C0770<10B0=38 a 18 @0=6:08 A4?4@B8 60<10@ 10 14@8:CB 0<10B0=?4=660=B80=B0@0030=1030;07 0 3 1 4 2
&4@70B8:0=60<10@14@8:CB 4 2 !C0B0@CAF0=6<4;0;C8 2 3 70<10B0= ?030 6 60<10@B4@A41CB030;07 0 3 1 4 2
b
*4@A4380B860;0
, - 30= AC<14@ B460=60= , 60@ 3870A8;:0==F0;0;0
Listrik Dinamis
199
2 3 4
B860;0
"8<0 1C07 ;0
2 "
&4@70B8:0=@0=6:080=14@8:CB 8:00@CA;8AB@8:F0=6<4=60;8@ baterai 20 <4;0;C8 70<10B0= 30 030;07 :C0B0@CAF0=6 <4;4E0B8 70<10B0= 030;07 0 3 1 4 2 A
&4@70B8:0=@0=6:080=14@8:CB *460=60=0=B0@0B8B8:30= 8V ?030 @0=6:080= B4@A41CB 2 030;07, 0 , 3 , 1 , 4 , 2 ,
3V
2V B
2
)41C070:8<4<8;8:866; ,30=70<10B0=30;0< 8:00:88=8388A834=60=0@CA B460=60=0=B0@0 :43C0B4@<8=0;=F0030;07 0
, 3
, 1 , 4 , 2 ,
#7#$-#)1&24#/8##/$&2*,54*/*%&/(#/4&1#4 *860?>B>=6:0E0BF0=6A0<0B410;30=A0<0?0=90=6 14@BC@CBBC@CBB4@1C0B30@814A8109030=B4<106038 A0<1C=6<4=9038A0BC=B0@0:43C0C9C=6:0E0B3814@8 1430?>B4=A80; , 8:070<10B0=94=8A<0A8=6<0A8=6 030;07 K < L K <30= L K < 78BC=61430?>B4=A80;0=B0@0:43C0C9C=6A4B80?:0E0B &4@70B8:0=60<10@14@8:CB a 2
0 1
200
8BC=6 30= 4@0?0 01
4 6V
8V I3 I2
I1 b
10 V
*860 1C07 :0E0B 38ACAC= ?0@0;4; &4@10=38=60= ?4=0
0<10@14@8:CB8=8<4=C= 9C::0= AC0BC @0=6:08 0@CA A40@07
R = 2 I=2A
4A0@ 70<10B0= 30;0< 0 030;07 0 3 1 4 2
r E = 20 V
&0304;4<4=4;4:B@>:8<80A4;0<0B4@9038@40:A8:8<80 0:0=B4@9038?4@C1070=4=4@68<4=90384=4@68 0 ;8AB@8::8<80 3 ;8AB@8::0;>@ 1 :0;>@<4:0=8: 4 :8<80:0;>@ 2 :8<80;8AB@8:
&4@70B8:0=60<10@14@8:CB R 5 8:0A4E0:BC387C1C=6 G :0=34=60=10B4@08B4@ 6 =F0B060;D0=><4B4@<4 8 =C=9C::0= 0=6:0 =>; <0:0=8;0870<10B0=030;07 12 0 3 1 4 2
)4A4>@0=6<4<0:08A4B@8:0;8AB@8:F0=614@BC;8A:0= - ,)4B@8:0B4@A41CB38?0A0=6?030B460=60= ,A4;0<0 90<=4@68;8AB@8:F0=6B4@?0:08>;47 A4B@8:0;8AB@8:8BC030;07 0 L 3 L
1 L 4 L 2 L )41C071>;0;0
8:070<10B0=:0E0B?4@B0<0 >7<78BC=6 70<10B0=?4=660=B8ACAC=0=:0E0B8=8 &4@70B8:0=@0=6:080=;8AB@8:?03060<14@14@8:CB a
b I1 32 V
8BC=6;07 30=
4 I3
c I2 5 10
15 V
d
C0 1C07 ;0
, -30=
, -38ACAC=A4@8 30= 387C1C=6:0= :4 AC<14@ B460=60=
, *4=BC:0= 0 :C0B0@CAF0=6<4=60;8@30;0<@0=6:080= 1 30F038A8?0A8?030;0
, - 2 30F038A8?0A8?030;0
, -
