```TUGAS PEMODELAN β ELECTRIC CIRCUIT PROBLEM
NIM : 4153111002
Kelas : MATEMATIKA DIK A 2015
Problem
1
An RLC Circuit has π = 180 Ξ©, πΆ = 280 πΉ, πΏ = 20 π»; and an applied Voltage
πΈ(π‘) = 10 sin π‘; Assuming to initial charge on the Capacitor, but an initial current
of 1 π΄ at π‘ = 0, when the Voltage is first applied, find the subsequent charge on the
Capacitor; π(0) = 0, π β² (0) = 0.
Diketahui
1
πΆ = 280 πΉ,
πΈ(π‘) = 10 sin π‘, π(0) = 0,
Ditanya
πΏ = 20 π»,
π β² (0) = πΌ(0) = 1
: π(π‘) = . . . . . ?
Penyelesaian :
Dari sistem RLC, diperoleh :
πΈπ + πΈπΆ + πΈπΏ = πΈπ‘
Sehingga,
πΈπ + πΈπΆ + πΈπΏ = 10 sin π‘
1
ππΌ
ππ
π + πΏ = 10 sin π‘ ;
πΌ=
πΆ
ππ‘
ππ‘
ππ 1
π ππ
π
+ π+πΏ .
= 10 sin π‘
ππ‘ πΆ
ππ‘ ππ‘
ππΌ +
π
ππ 1
π2 π
+ π + πΏ 2 = 10 sin π‘
ππ‘ πΆ
ππ‘
πΏ
π2π
ππ 1
+π
+ π = 10 sin π‘
2
ππ‘
ππ‘ πΆ
Dengan mensubstitusi data pada soal, diperoleh :
20
π2 π
ππ
1
+ 180
+
π = 10 sin π‘
2
1
ππ‘
ππ‘
280
π2 π
ππ
20 2 + 180
+ 280 π = 10 sin π‘
ππ‘
ππ‘
π2π
ππ
1
+9
+ 14 π = sin π‘
2
ππ‘
ππ‘
2
; (×
1
)
20
. . . . . . (β)
Mencari solusi PD Homogen, yaitu ππΏ = 0, sehingga :
(π·2 + 9π· + 14)π = 0
π·2 + 9π· + 14 = 0
π2 + 9π + 14 = 0
(π + 2)(π + 7) = 0
π1 = β2
dan
π2 = β7
Maka diperoleh :
ππΏ = π1 π β7π‘ + π2 π β2π‘
Mencari solusi PD tak homogen, yaitu ππ dengan menggunakan metode koefisien
tak tentu dan aturan modifikasi.
Jika π(π₯) = π sin ππ‘ maka ππ = π0 sin ππ‘ + π1 cos ππ‘, sehingga :
πΈπΉ = ππ π¬π’π§ π + ππ ππ¨π¬ π
ππβ² = π0 cos π‘ β π1 sin π‘ + ππ sin π‘ + π1 cos π‘ ; ππ sin π‘ + π1 cos π‘ = 0, maka :
πΈβ²πΉ = ππ ππ¨π¬ π β ππ π¬π’π§ π
Diferensiasi lagi terhadap π‘ (karena berorde 2), diperoleh :
ππβ²β² = βπ0 sin π‘ β π1 cos π‘ + π0 cos π‘ β π1 sin π‘ ; π0 cos π‘ β π1 sin π‘ = 0, maka :
πΈβ²β²
πΉ = βππ π¬π’π§ π β ππ ππ¨π¬ π
Substitusi ππ , ππβ² , ππβ²β² ke persamaan (β), sehingga :
ππβ²β² + 9ππβ² + 14 π =
1
sin π‘
2
1
(βπ0 sin π‘ β π1 cos π‘) + 9( π0 cos π‘ β π1 sin π‘) + 14(π0 sin π‘ + π1 cos π‘) = sin π‘
2
1
2
(βπ0 β 9π1 + 14π0 ) sin π‘ + (βπ1 + 9π0 + 14π1 ) cos π‘ = sin π‘
1
2
(13π0 β 9π1 ) sin π‘ + (9π0 + 13π1 ) cos π‘ = sin π‘
Diperoleh :
1
9
13π0 β 9π1 = 2
β¦ . (1)
| × 9|
117π0 β 81π1 = 2
9π0 + 13π1 = 0
β¦ . (2)
| × 13|
117π0 + 169π1 = 0 β
9
β250π1 = 2
β500π1 = 9
π1 = β
Substitusi π1 ke persamaan (2), diperoleh :
9π0 = β13 (β
9
)
500
β
π0 =
13
500
Maka diperoleh :
ππ =
13
9
sin π‘ β
cos π‘
500
500
Karena ππ‘ = ππΏ + ππ maka :
π(π‘) = π1 π β7π‘ + π2 π β2π‘ +
13
9
sin π‘ β
cos π‘
500
500
π(0) = 0
π1 π β7(0) + π2 π β2(0) +
π1 + π2 β
9
=0
500
13
9
sin 0 β
cos 0 = 0
500
500
9
500
π1 + π2 =
9
500
β¦ . (3)
β²
π(π‘)
= β7π1 π β7π‘ β 2π2 π β2π‘ +
13
9
cos π‘ +
sin π‘
500
500
β²
π(0)
=1
β7π1 π β7(0) β 2π2 π β2(0) +
13
9
cos 0 +
sin 0 = 1
500
500
13
=1
500
487
β7π1 β 2π2 =
β¦ . (4)
500
β7π1 β 2π2 +
Dari persamaan (3) dan (4), diperoleh :
18
500
487
β7π1 β 2π2 =
500
505
β5π1
=
500
2π1 + 2π2 =
π1 = β
+
101
500
Substitusi π1 ke persamaan (3), diperoleh :
β
101
9
+ π2 =
500
500
π2 =
110
500
Sehingga diperoleh :
π(π‘) = β
101 β7π‘ 110 β2π‘
13
9
π
+
π
+
sin π‘ β
cos π‘
500
500
500
500
```
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