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Tiga buah kapasitor disusun seperti pada gambar. Diketahui C1= 2μF , C2= 3μF, dan C3=1F. Jika VAB​= 12 V, Muatan pada C1 dan C2 ....

Pertanyaan

Tiga buah kapasitor disusun seperti pada gambar.

Diketahui C 1 equals space 2 mu F space comma space C 2 equals space 3 mu F comma space d a n space C 3 equals 1 F. Jika undefined, Muatan pada C1 dan C2 ....

  1. q 1 space equals space q 2 equals space 10 mu C

  2. q 1 equals q 2 equals 5 mu C

  3. q 1 equals space 6 mu C space d d a n space q 2 equals space 4 mu C

  4. q 1 equals 4 mu C space d a n space q 2 equals space 6 mu C

  5. q 1 equals space 14 mu C space d a n space q 2 equals 16 mu C

A. Aulia

Master Teacher

Jawaban terverifikasi

Pembahasan

Untuk menjawab soal ini kalian harus memahami tentang kapasitansi total kapasitor yang disusun secara seri dan paralel. Pada soal diketahui bahwa

C 1 space equals space 2 mu F  C 2 equals space 3 mu F  C 3 equals space 1 mu F  V subscript A B end subscript equals space 12 space V

C 1 space d a n space C 2 space d i p a s a n g space p a r a r e l comma space m a k a space k a p i s a tan si space p e n g g a n t i space k e d u a n y a space a d a l a h space  C subscript p equals space C 1 space plus space C 2  C subscript P equals space 2 plus space 3 space  C subscript p equals space 5 mu F equals space 5 space x space 10 to the power of negative 6 end exponent F    C subscript p space d a n space C 3 space d i p a s a n g space s e r i comma space m a k a space k a p a s i tan s i space t o t a l space p a d a space r a n g k a i a n space t e r s e b u t space a d a l a h space  1 over C equals 1 over C subscript p space plus space fraction numerator 1 over denominator C 3 end fraction  1 over C equals 1 fifth plus 1 over 1 space equals 6 over 5  C equals 5 over 6 mu F equals 5 over 6 space x space 10 to the power of negative 6 end exponent  M u a tan space t o t a l space p a d a space r a n g k a i a n space y a i t u  Q equals C V  Q equals open parentheses 5 over 6 space x space 10 to the power of negative 6 end exponent close parentheses open parentheses 12 close parentheses  Q equals 10 to the power of negative 5 end exponent  D e n g a n space d e m i k i a n space comma space j u m l a h space m u a tan space p a d a space C 1 space d a n space C 2 space a d a l a h space 10 to the power of negative 5 end exponent. space T e g a n g a n space y a n g space j a t u h space p a d a space k e d u a n y a space y a i t u  V subscript p equals space Q over C subscript p  V subscript p equals fraction numerator 10 to the power of negative 5 end exponent over denominator 5 space x space 10 to the power of negative 6 end exponent end fraction equals 2 space V  B e s a r space m u a tan space p a d a space C 1 space equals space  Q 1 equals space C 1 V subscript p  Q 1 equals open parentheses 2 space space x space 10 to the power of negative 6 end exponent close parentheses open parentheses 2 close parentheses  Q 1 equals space 4 space x space 10 to the power of negative 6 end exponent space C  Q 1 equals 4 mu F  B e s a r space m u a tan space p a d a space C 2 equals  Q 2 equals C 2 V subscript p  Q 2 equals open parentheses 3 space x space 10 to the power of negative 6 end exponent close parentheses open parentheses 2 close parentheses  Q 2 equals space 6 space x space 10 to the power of negative 6 end exponent C  Q 2 equals space 6 mu F

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3.6 (3 rating)

Pertanyaan serupa

Perhatikan susunan kapasitor pada gambar berikut! Hitunglah kapasitas kapasitor pengganti!

139

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