tiga lampu dengan hambatan masing masing 2 ohm ,3 ohm ,dan 6 ohm di pasang secara paralel lalu dihubungkan dengan tegangan 18 V jika lampu tersebut menyala sela

diket. 3 buah lampu dg R masing "
R1= 2 ohm
R2=3 ohm
R 3 = 6 ohm
V = 18 v
t = 8 jam x60 x 60 =28800 s
dit. W , W1, W2, &W3?
jawab :
1/Rp = 1/R1+1/R2+1/R3
= 1/2+1/3+1/6
=3/6+2/6+1/6
Rp =6/6 =1 ohm
I = V /R =18 /1 = 18 A
a.) W = I x V x t
= 18 x 18 x 28800
=9.331.200 joule
= 9.331,2 Kilo joule
b.) I1 =V / R1 =18/2 = 9 A
W1= I1 x V x t
= 9 x 18 x 28800
=4.665.600 joule
=4.665,6 kilo joule

I2 = V /R2 = 18/ 3 =6 A
W2=I2 x V xt
= 6 x 18 x 28800
= 3.110.400 j
= 3.110,4 kj

I3= V / R3=18/6 =3 A
W3 =I3 x V x t
=3 x 18 x 28800
=1.555.200 j
= 1.555,2 kj