Nama : Nita Susanti

NIM : 08121002035

Tugas Metode Numerik

Penyelesaian:Diketahui :R = 280 ohmt= 0.05 sL = 7.5 H

f (C )=e−RT /2Lcos[√ 1L(C )

−( R2 L )2

( t )]−0 .01f (C )=e−280(0.05 )/(2(7 .5) )cos[√ 1

7 .5(C )−(2802(7 .5) )

2

(0 .05)]−0 .01f (C )=e−0 .933cos [√ 1

7 .5(C )−384 .444 (0 .05 )]−0 .01

f (C )=0 .39cos[√ 17 .5(C )

−18 .666 (0 .05 )]−0 .01f (C )=0 .39cos[√ 1

7 .5(C )−0 .933]−0 .01

0 .01=0 .39cos[√ 17 .5(C )

−0 .933]0 .010 .39

=cos [√ 17 .5(C )

−0 .933]0 .025=cos√ 1

7 .5(C )−cos( 0.933 )

0 .025=cos√ 17 .5(C )

−0 .999

0 .025+0 .999=cos√ 17 .5(C )

1 .024=cos√ 17 .5(C )

(1.024 )2=cos 17 .5C

1 .048576=cos1cos7 .5C

1 .048576= 10 .99C

0 .99C= 11 .048576

C= 11.048576(0 .99 )

C= 11.03809024

C=0.96 ≈1

Maka didapatkan nilai C adalah 1