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R1 = 10 ohm
R2 = 9 ohm
R3 = 18 ohm
ditanyakan: R pengganti A-B?
jawab:
R2 dan R3 adalah parallel, maka:
1/Rp = 1/R2 + 1/R3
1/Rp = 1/9 + 1/18
1/Rp = 2+1/18
Rp = 18/3 = 6 ohm
R1 seri dengan paralel dari R2 dan R3, maka:
R pengganti A-B:
Rs = R1 + Rp
Rs = 10 + 6 = 16 ohm (jawaban A)
5.
diketahui:
n1 = 5 lampu
P1 = 100 watt
t1 = 5 jam/hari
n2= 1 tv
P2 = 150 watt
t = 6 jam/hari
harga= Rp. 500/kWh
ditanyakan: biaya 30 hari?
Lampu: 5 x 100 watt = 500 watt
TV : 1 x 150 watt = 150 watt
Energi listrik: W = P x t
Lampu: W = 500 x (5 jam) = 2500 Wh
TV: W = 150 x (6jam) = 900 Wh
TOTAL 1 hari = 2500 + 900 = 3400 Wh (watt-hour) = 3,4 kWh
maka 1 bulan: 30 x 3,4 x 500 = 51 rb (jawaban A)
R2 = 9 Ω
R3 = 18 Ω
ditanya :
rangkaian paralel (R2 dan R3)
1/Rp = 1/R2 + 1/R3
= 1/9 + 1/18
= 2/18 + 1/18
= 3/18
= 1/6
Rp = 6 Ω
Rtotal = R1 + Rp
= 10 Ω + 6 Ω
= 16 Ω
4) diketahui : I = 0,75 A
V = 110 volt
ditanya : P
P = w / t
= V . I .t / t
= V . I
= 110 . 0,75
= 82,5 watt
5) w = Σ P. t
= (5 . 100 .5) + (1 . 150 . 6) x 30 / 1000
= ( 2500 + 900) x 3/100
= 3400 x 3/100
= 34 x 3
= 102 Kwh
biaya = 102 x 500
= 51.000