Petra Elang Pradana Pradana 1106023096 1106023096 Teknik Perkapalan Physics Homework 4-
PR 4. Arus Listrik, Resistansi, dan Arus Searah Bab 26 no 3, 8, 10, 16, 24, 34, 40, 44. Bab 27 no 2, 10, 20, 30, 40, 53, 63, 65
3)3) = 5050 ; = 3030 / ; = 10100 షర ଵ Muatan pada permukaan: = ௩௪ = ሺଷ/௦) ଷ/௦)ሺ,ହ) = 6,7 x 10-6 C/m2 8)8) i = 1,2 10ିଵ ; = 2,2,5 = 2,2,5 × 10ିଷ ; =8,49×10ଶ଼ ିଷ a. Besarnya laju arus listrik షభబ = ሺସ)ሺଵ.ଶ×ଵ ) = 2,4 × 10-5 A/m2 = = గሺௗ/ଶ) మ ሺగ)ሺଶ.ହ×ଵషయ)మ
b. Kecepatan elektron షఱ/మ ଶ, ସ ×ଵ = = ሺ଼,଼,ସଽସଽ×ଵమఴ/య)ሺଵ,×ଵషభవ) = 1,8 × 10-15 m/s 10) = ሺ3,00 × 10଼) ଶ; = 2,00 ; = 0,900 ; k =3,0 × 108 Arus yang melalui outer section bounded adalah = නଽ/ଵ . = න,ଽሺ ଶ)2 = ሺ3 × 10଼) ଶ. 2 = 1,5 × 10଼ሺ ସ −0,6561 ସ) =2,59×10ିଷ 16) i= 60,0 ; = 0,150Ω/ ; = 89 8960 / ଷ; = 26 2600 / ଷ 16) a)Tembaga: a) )ሺ,ଵହஐ/) = 5,32 × 105A/m2 = ሺ) ଵ,ଵ,ଽ ଽ×ଵషఴஐ b) Densitas massa untuk tembaga: −8 3ቁሺ1,69×10 Ω ) ቀ 8960 / = 1,01 kg/m ሺ /)= 0,150Ω/
Petra Elang Pradana Pradana 1106023096 1106023096 Teknik Perkapalan
c) Alumunium షరஐ/) ሺ ) , ) ሺଵ, ହ ×ଵ = ଶ,ହ௫ଵషఴஐ = 3,27× 105 A/m2 d) Densitas massa alumunium య൯ሺଶ,ହ×ଵషఴஐ) ൫ଶ/ ቀቁ = = 0,495 kg/m ,ଵହஐ/ 24) a. ଵ = 2,50
10ଷ / , ଶ = 4,00 10ଷ / , ଷ = 1,50 10ଷ / , ଷ =2,00 ଵ ଵ = ଷ ଷ ଵ ଵଶ = ଷ ଷଶ 4 ଶଶ = 1,5 ଷଶ 2,5 ଵଶ= 1,5 ଷଶ ଷ= 2 mm; ଵ=1,55mm b. 4 ଶଶ = 1,5 ଷଶ dengan ଷ= 2 mm, maka ଶ= 1,22 mm 34) ଵ = 4,00 ; ଶ = ଶ଼2,2ଶ଼,00ିଷ ; = 2,2,00 ; ∆ = 10,0 ; = 8,49× 10 ି 1 0 1 0 ଵ = 4ሺ2)ሺ1,69×10ି଼) =73,964 / ଶ = ଵ = ሺ8,49×1073,ଶ଼)ሺ9641,6×10ିଵଽ) =5,445 ×10ିଽ / 40) = 100 , = 3,00 40) =ଶ = మ = ଵଽ = 11,1 Ω 44) = 9.0 ; = 7.0 ; 44) 5 = 18000 = = ௧ = ×ଵ଼ ଽ = 1,4 × 10-4 C
:9:0 :9:000
2:00 :00 ; =
Petra Elang Pradana Pradana 1106023096 1106023096 Teknik Perkapalan
BAB 27 2) ଵ = 15 150 ; ଶ = 5050 ; ଵ =3.0Ω; ଶ =2.0Ω; =100 ହ = 20 = ఌభభିఌାమమ = ଵହି ଷାଶ 20 ; = − ଶ − ଵ =100− 50 − ሺ20×3) =−10 10) = 1,1,0 ; ଵ = 2,2,0 ; ଶ = 3,3,0 ; ଵ = ଶ =3,0Ω 10) a. − ଶ + ଶ − ଵ − ଵ − = 0 = ቀఌమିఌ భቁ − ሺ ଵ + ଶ) = ቀଵଷିଶషయቁ − ሺ6) = 990 Ω b. = ଶ = ሺ1,0×10ିଷ )ଶሺ990Ω) = 9,9 10ିସ 20) R1 &R2 dihubungkan seri R =16,0 Ω; Dihubungkan paralel maka R = 3,0 Ω. 1=1+ 1 1 =1+ 1 3,0Ω ଵ ଶ ଵ ଶ R = R1+R2=16,0 Ω ; 1/3,0 Ω= R2+R1/R1R2; 3,0 Ω =ሺR1R2)/R1+R2; R1
p
s
. ଵ = 10,0 ; ଶ =0,500 ଵ ; R1=R2=R3= 4,00 Ω Loop 1 Loop 2 10V - I1R1 - I3R3=0 5V – I2R2 - I3R3=0 10V - I1R1 – ሺI2+I1)R3 =0 5V – I2R2 – ሺI2+I1)R3 =0
Petra Elang Pradana Pradana 1106023096 1106023096 Teknik Perkapalan
10V – 4I1 – 4I1 – 4I2 = 0 5V – 4I2 – 4I2 – 4I1 = 0 10V – 8I1 – 4I2 =0 5V – 4I1 – 8I2 =0 Eliminasi: 10V – 8I1 – 4I2 =0 ሺ.1) 10V – 8I1 – 4I2 =0 5V – 4I1 – 8I2 =0 ሺ.2) 10V – 8I1 – 18I2 =0 maka I2= 0A; I1= 1,25A b. I3=I2+I1= 1,25A 40) ଵ = 12,0 12,0 ; r= r= 0,2 0,200 00 Ω; R=2, R=2,00r; 00r; a. arus pada resistan internal dirangkaian paralel ଶఌ ଶሺଵଶ, ଶሺଵଶ,) ) = 2i = = ାଶ ሺ,ଶஐ)ାଶሺ,ସஐ) = 24,0 A b. pada rangkaian seri ଶఌ ଶሺଵଶ, ଶሺଵଶ,) ) = = ଶା ଶሺ,ଶஐ) ஐ)ାሺ,ସஐ) = 30,0 A c. Pada rangkaian seri arus lebih besar d. Jika R=r/2,00= 0,100 Ω maka arusnya dalam paralel ଶఌ ଶሺଵଶ, ଶሺଵଶ,) ) = 2i = ାଶ = ሺ,ଶஐ)ାଶሺ,ଵஐ) = 60,0 A e. Jika R=r/2,00= 0,100 Ω maka arusnya dalam seri ଶఌ ଶሺଵଶ, ଶሺଵଶ,) ) = = ଶା ଶሺ,ଶஐ) ஐ)ାሺ,ଵஐ) = 48,0 A f. Pada kondisi R=r/2,00, arus dalam paralel lebih besar 53) ଵ = 3,0 ; r= 100 Ω; R1=250 Ω; R2=300 Ω; Rv = 5,0 k Ω ଵ ଵ − ሺI − ଵ ) ௩ = 0 = ଵ ௩ଵ + ଵ − ଶ− ଵ ଵ− =0 − ሺ ଶ+ )− ଵ ଵ =0 − ሺ ଵ ௩ଵ + ଵ)ሺ ଶ + ) − ଵ ଵ = 0 − ሺሺ ଶ + )ሺ௩ ଵ + ௩) ଵ) − ଵ ଵ = 0 ଵ = ሺ ଶ + )ሺ ଵ +௩ ௩ ) + ଵ ௩
ଵ ଵ=ሺ ଶ+
௩ ଵ ଵ + ௩) + ଵ ௩
Petra Elang Pradana Pradana 1106023096 1106023096 Teknik Perkapalan
)ሺ 3,0 ሺ5000Ω)ሺ250Ω) = ሺ300Ω+100Ω)ሺ 300Ω+100Ω)ሺ250Ω+5000Ω 250Ω+5000Ω)) +250Ωሺ5000Ω) =1,119 ఌభ Jika dianggap ௩ sebagai terlalu besar, maka ଵ ଵ = మାା భ 3,0 ሺ250Ω) = 300Ω+100Ω+250Ω =1,154 % error = |ሺ1,119−1,154)| ሺ1,119−1,154)|/1,154 =3,033% 63) ଵ = 1,2 ; C=6,5 ; R1= R1=R2=R3= R3=0,73M Ω − ଵ ଵ − ሺ ଵ + ଷ) ଶ = 0 ሺ ଵ + ଷ) ଶ − ଷ ଷ = 0 a. ଵ = ଶଷ ሺ1200) = 32ሺ730000 ) ିଷ = 1,096 10 b. ଶ = ଷ 1200 = 3ሺ730000 ) ିସ = 5,479 10 c. ଵ = ଵ = 5,479 10ିସ d. ଵ = ଶ 1200 = 2ሺ730000 ) ିସ = 8,219 10 e. ଶ = ଵ = 8,219 10ିସ f. ଷ = 0 g. − ଵ ଵ − ଶ ଶ = 0 − ଷ −ଵ − 2 ଶ ଶ = 0 ଷ ଶ= 2 ଶ ଶ− ଷ ଷ− =0
3൰ + = ൬ 2 2 = 2 ሺ1 −
ିଶ௧/ଷ)
ଷሺ ) =
=3
ିଶ௧/ଷ
− 2ଷ = 2 − 6 ିଷଶ௧ ଶሺ ) = ଶ ଶ௧ = 2 − 6 ିଷ t=0, ିయೃమ = 1; ଶ = ଶ − = 1200 =4003 ଶሺ ) = 2
h. t=tak hingga,
మ ିయೃ = 0; ଶ = ଶ
൬ −2 ଷ ൰ − − = 0 2 − ሺ 2) − − = 0
Petra Elang Pradana Pradana 1106023096 1106023096 Teknik Perkapalan
= 1200 2 =600
65) R1=10,0k Ω; R2=15,0k Ω; C=0,400 ; ଵ =20,0 Kondisi mula-mula saklar ditutup sampai seimbang dan dibuka saat t=0. Arus pada R 2 saat t=4,00ms? = ଶ ଵ +ଵ ଶ 20,0 =15000Ω 10000Ω+15000Ω =12,0
=
௧ ି
= 12
షయ ସ×ଵ ିଵହሺ ଵହሺ,ସ×ଵషల)
=6,161 = 6,161 = ሺ15000Ω) =4,107×10ିସ -Finish -FinishFinish -
Petra Elang Pradana Pradana 1106023096 1106023096 Teknik Perkapalan
