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2< L , 2< ,
..
*460=60= <0:A8<C<=F0 030;07
Tantangan
+ )1 .. , , untuk Anda

038 B460=60= <0:A8<C<=F0 030;07 ,
Persamaan tegangan bolak-balik
1 &4@8>34 30;0< 6@058: 2< ):0;0 7>@8G>=B0; <A 2< <0:0 suatu rangkaian listrik
2< L <A 2< <A A memenuhi persamaan
@4:C4=A8 AC<14@ 030;07 V = 314 sin 50 V.
Tentukan tegangan rata-rata
$ G yang dihasilkan sumber tersebut.

Contoh 8.17
)41C07 @0=6:080= F0=6 387C1C=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?0:07 70@60 <0:A8<C< B460=60= 1>;0:
10;8: AC<14@ *C;8A:0= ?C;0 ?4@A0<00==F0 98:0 5@4:C4=A8=F0 G
#7#$
8:4B07C8
, $ G
#$
0 $8;08 B460=60= <0:A8<C< <0:A 3878BC=6 34=60= <4=66C=0:0= ?4@A0<00=
14@8:CB
<0:A #$

<0:A ,
1 &4@A0<00= B460=60= 38?4@>;47 34=60= <4=66C=0:0= ?4@A0<00= 14@8:CB
V maks sin t
A8= 2

Contoh 8.18

)41C07 @0=6:080= 0@CA 1>;0: 10;8: <4<8;8:8 70@60 B460=60= A410608 5C=6A8 E0:BC
F08BC A8= 2 8BC=6;07
0 B460=60= <0:A8<C< + )1 4 ?4@8>34
1 B460=60= ?C=20: :4 ?C=20: 5 5@4:C4=A8 $
..
2 B460=60= 454:B85 6 B460=60= A4B4;07 2 A
#$
3 5@4:C4=A8 0=6C;4@
#7#$
8:4B07C8
&4@A0<00= B460=60= A410608 5C=6A8 E0:BC
A8= 2
%;47 :0@4=0 B460=60= <4@C?0:0= 5C=6A8 A8=CA>830; B4@7030? E0:BC <0:0 ?4@A0<00=
B460=60= 30?0B 38BC;8A:0=
A8= 2
+
@03 A
0 *460=60= <0:A8<C< , B4@9038 ?030 A00B A8= 2
<
<0:0 ,
+
1 , ,
.. +

2 + , ,
#$

Listrik Dinamis 195