KUNCI JAWABAN FISIKA LATIHAN SOAL BAB 3
KTSP 2006 SERIBU PENA PENERBIT ERLANGGA
JAWABAN ESSAY UJI KOMPETENSI BAB 3:
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3.DIK:
R1 = 20 OHM
R2 = 30 OHM
GLL = 220 V
DIT: PERBANDINGAN DAYA YANG DISERAP
JWB:
I1 = GLL / R1 = 220 / 20 = 11
I2 = GLL / R2 = 220/30 = 7,3
R2 = 30 OHM
GLL = 220 V
DIT: PERBANDINGAN DAYA YANG DISERAP
JWB:
I1 = GLL / R1 = 220 / 20 = 11
I2 = GLL / R2 = 220/30 = 7,3
DAYA YANG DI SERAP R1 DAN R2
P1 = V1 x I1 = 220 x 11 = 2420 W
P2 = V2 x I2 = 220 x 7,3 = 1606 W
PERBANDINGAN P1 DAN P2 = 2420 : 1606 = 1210 : 803
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5.LAMPU 20 W = 4 BUAH x 4 JAM x 20 W = 320 Wh
LAMPU 60 W = 2 BUAH x 4 JAM x 60 W = 480 Wh
TV = 1 BUAH x 4 JAM x 60 W = 240 Wh
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1040 Wh
= 1, 04 Kwh
=1,04 x 30 HARI = 31, 2 Kwh
=31, 2 Kwh x Rp 75
= Rp 2340
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7DIK =
I = 0,5 A
R = 440 OHM
t = 30 MENIT = 1800 SEKON
DIT =
A. TEGANGAN LISTRIK
V = I x R
= 0,5 x 440
V = 220 volt
B. ENERGI LISTRIK
W = V x I x R
= 220 x 0,5 x 1800
W = 198000 JOULE
C. DAYA LISTRIK
P = I x V
P = 0,5 x 220
P = 110 watt
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