Page 180 - MUDAH DAN AKTIF BELAJAR FISIKA SMA KELAS X
P. 180
/ A sumber tegangan
A
038 <C0B0= F0=6 38?8=307:0= 030;07 A 0B0C
3. 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<?4@4<4B4@ 0B0C 0<<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 #.$#2
0@0 <4<1020 A:0;0 0<?4@4<4B4@ 030;07 A410608 14@8:CB
A:0;0 F0=6 38BC=9C:
0A8; ?4=6C:C@0= L 10B0A C:C@ K
A:0;0 <0:A8<C<
Contoh 8.2
8 ;01>@0B>@8C< A4:>;07 180A0=F0 386C=0:0= 0<?4@4<4B4@ C=BC: <4=6C:C@ :C0B 0@CA amperemeter
&030 AC0BC ?4=6C:C@0= 0@CA ;8AB@8: 3830?0B 30B0 A4?4@B8 38BC=9C::0= ?030 60<10@ 14@8:CB
*4=BC:0= 70A8; ?4=6C:C@0= 0<?4@4<4B4@ B4@A41CB Gambar 8.3
#7#$ Mengukur kuat arus listrik
=5>@<0A8 F0=6 38?4@>;47 30@8 60<10@ B4@A41CB F08BC A:0;0
F0=6 38BC=9C: A:0;0 <0:A8<C< 10B0A C:C@0= 70
0 100
0A8; ?4=6C:C@0= L
0,5 A
038 70A8; ?4=6C:C@0==F0
voltmeter
Contoh 8.3
@058: 14@8:CB <4=C=9C::0= :C0B 0@CA F0=6 <4=60;8@ 30;0< AC0BC @0=6:080= B4@BCBC?
I(A) 4@30A0@:0= 6@058: B4@A41CB B4=BC:0= 10=F0:=F0
<C0B0= ;8AB@8: F0=6 <4=60;8@ 30;0< @0=6:080=
4 A4;0<0 A ?4@B0<0 30;0< A0BC0= 2>C;C<1 P lampu Q
arus
#7#$ listrik
2 I 0@8 2 A0<?08 2 A 38?4@>;47
2 K A 0@CA ?030 A4;0=6 E0:BC 8=8 + –
14@B0<107 A420@0 ;8=40@ $8;08 @0B0 @0B0=F0
t(s)
2 4 6 030;07 A410608 14@8:CB baterai
voltmeter
0 2 0 2 V
C<;07 <C0B0==F0 / 2 A
I 0@8 2 A A0<?08 2 A 38?4@>;47 2 A K A A @CA ?030 A4;0=6 E0:BC
8=8 B4B0? C<;07 <C0B0==F0 / 2 A ×
038 9C<;07 <C0B0= F0=6 <4=60;8@ <4;0;C8 @0=6:080= A4;0<0 A 030;07 P lampu Q
arus
/ / / listrik
+ –
4. Mengukur Beda Potensial
baterai
+=BC: <4=6C:C@ 14A0@ 1430 ?>B4=A80; 0B0C B460=60= 38 0=B0@0 C9C=6 C9C=6 Gambar 8.4
?4=670=B0@ 386C=0:0= D>;B<4B4@ F0=6 38@0=6:08:0= A4?4@B8 ?030 #.$#2 Cara mengukur beda potensial.
Listrik Dinamis 171