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Empat buah kapasitor masing-masing kapasitasnya C, dirangkai seperti gambar di bawah ini. Rangkaian yang memiliki kapasitas 0,6 C adalah ....

Pertanyaan

Empat buah kapasitor masing-masing kapasitasnya C, dirangkai seperti gambar di bawah ini. Rangkaian yang memiliki kapasitas 0,6 C adalah ....space 

A. Aulia

Master Teacher

Jawaban terverifikasi

Jawaban

jawaban yang tepat adalah D.

Pembahasan

Diketahui : C subscript 1 equals C subscript 2 equals C subscript 3 equals C subscript 4 equals straight C

Ditanya : rangkaian yang Ctot = 0,6 C

Jawab :

Untuk mengetahui rangkaian mana yang memiliki Ctot = 0,6 C, maka harus dihitung Ctot dari setiap rangkaiannya.

Rangkaian pilihan A : 

table attributes columnalign right center left columnspacing 0px end attributes row cell 1 over C subscript t o t end subscript end cell equals cell 1 over C subscript 1 plus 1 over C subscript 2 plus 1 over C subscript 3 plus 1 over C subscript 4 end cell row cell 1 over C subscript t o t end subscript end cell equals cell 1 over straight C plus 1 over straight C plus 1 over straight C plus 1 over straight C end cell row cell 1 over C subscript t o t end subscript end cell equals cell 4 over straight C end cell row cell C subscript t o t end subscript end cell equals cell 1 fourth straight C end cell row blank equals cell 0 comma 25 space straight C end cell end table

Rangkaian pilihan B :

table attributes columnalign right center left columnspacing 0px end attributes row cell 1 over C subscript t o t end subscript end cell equals cell 1 over C subscript 1 plus fraction numerator 1 over denominator C subscript 2 plus C subscript 3 end fraction plus 1 over C subscript 4 end cell row cell 1 over C subscript t o t end subscript end cell equals cell 1 over straight C plus fraction numerator 1 over denominator straight C plus straight C end fraction plus 1 over straight C end cell row cell 1 over C subscript t o t end subscript end cell equals cell 1 over straight C plus fraction numerator 1 over denominator 2 straight C end fraction plus 1 over straight C end cell row cell 1 over C subscript t o t end subscript end cell equals cell fraction numerator 2 plus 1 plus 2 over denominator 2 straight C end fraction end cell row cell 1 over C subscript t o t end subscript end cell equals cell fraction numerator 5 over denominator 2 straight C end fraction end cell row cell C subscript t o t end subscript end cell equals cell 2 over 5 straight C end cell row blank equals cell 0 comma 4 space straight C end cell end table

Rangkaian pilihan C :

table attributes columnalign right center left columnspacing 0px end attributes row cell 1 over C subscript s subscript 1 end subscript end cell equals cell 1 over C subscript 1 plus 1 over C subscript 2 end cell row cell 1 over C subscript s subscript 1 end subscript end cell equals cell 1 over straight C plus 1 over straight C end cell row cell 1 over C subscript s subscript 1 end subscript end cell equals cell 2 over straight C end cell row cell C subscript s subscript 1 end subscript end cell equals cell 1 half straight C end cell row blank equals cell 0 comma 5 space straight C end cell end table

table attributes columnalign right center left columnspacing 0px end attributes row cell 1 over C subscript s subscript 2 end subscript end cell equals cell 1 over C subscript 3 plus 1 over C subscript 4 end cell row cell 1 over C subscript s subscript 2 end subscript end cell equals cell 1 over straight C plus 1 over straight C end cell row cell 1 over C subscript s subscript 2 end subscript end cell equals cell 2 over straight C end cell row cell C subscript s subscript 2 end subscript end cell equals cell 1 half straight C end cell row blank equals cell 0 comma 5 space straight C end cell end table

table attributes columnalign right center left columnspacing 0px end attributes row cell C subscript t o t end subscript end cell equals cell C subscript s subscript 1 end subscript plus C subscript s subscript 2 end subscript end cell row blank equals cell 0 comma 5 space straight C plus 0 comma 5 space straight C end cell row blank equals straight C end table

Rangkaian pilihan D :

table attributes columnalign right center left columnspacing 0px end attributes row cell 1 over C subscript s subscript 1 end subscript end cell equals cell 1 over C subscript 1 plus 1 over C subscript 2 end cell row cell 1 over C subscript s subscript 1 end subscript end cell equals cell 1 over straight C plus 1 over straight C end cell row cell 1 over C subscript s subscript 1 end subscript end cell equals cell 2 over straight C end cell row cell C subscript s subscript 1 end subscript end cell equals cell 1 half straight C end cell row blank equals cell 0 comma 5 space straight C end cell end table

table attributes columnalign right center left columnspacing 0px end attributes row cell C subscript P end cell equals cell C subscript S subscript 1 end subscript plus C subscript 3 end cell row blank equals cell 0 comma 5 straight C plus straight C end cell row blank equals cell 1 comma 5 space straight C end cell end table

table attributes columnalign right center left columnspacing 0px end attributes row cell 1 over C subscript t o t end subscript end cell equals cell 1 over C subscript P plus 1 over C subscript 4 end cell row cell 1 over C subscript t o t end subscript end cell equals cell fraction numerator 1 over denominator 1 comma 5 straight C end fraction plus 1 over straight C end cell row cell 1 over C subscript t o t end subscript end cell equals cell fraction numerator 1 plus 1 comma 5 over denominator 1 comma 5 straight C end fraction end cell row cell 1 over C subscript t o t end subscript end cell equals cell fraction numerator 2 comma 5 over denominator 1 comma 5 straight C end fraction end cell row cell C subscript t o t end subscript end cell equals cell 15 over 25 straight C end cell row blank equals cell 3 over 5 straight C end cell row blank equals cell 0 comma 6 space straight C end cell end table

Rangkaian pilihan E :

table attributes columnalign right center left columnspacing 0px end attributes row cell 1 over C subscript S end cell equals cell 1 over C subscript 1 plus 1 over C subscript 2 plus 1 over C subscript 3 end cell row cell 1 over C subscript S end cell equals cell 1 over straight C plus 1 over straight C plus 1 over straight C end cell row cell 1 over C subscript S end cell equals cell 3 over straight C end cell row cell C subscript S end cell equals cell 1 third straight C end cell end table

table attributes columnalign right center left columnspacing 0px end attributes row cell C subscript t o t end subscript end cell equals cell C subscript S plus C subscript 4 end cell row blank equals cell 1 third straight C plus straight C end cell row blank equals cell 4 over 3 straight C end cell end table

Berdasarkan perhitungan yang sudah dilakukan, maka diketahui bahwa rangkaian yang Ctot = 0,6 C adalah rangkaian pada pilihan D.


Jadi, jawaban yang tepat adalah D.

4rb+

4.0 (4 rating)

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