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Besar energi listrik yang tersimpan dalam kapasito...

Besar energi listrik yang tersimpan dalam kapasitor 5 space mu straight Fadalah ... mu straight J.

  1. 9,0

  2. 12,0

  3. 15,0

  4. 18,0

  5. 22,5

Jawaban:

Diketahui:

Ditanya: W5

Jawab:

Kapasitor 5 μF space dan space 7 μF disususn secara paralel, maka

table attributes columnalign right center left columnspacing 0px end attributes row cell C subscript p end cell equals cell 5 μF plus 7 μF end cell row blank equals cell 12 μF end cell end table

Selanjutnya kapasitor totalnya:

table attributes columnalign right center left columnspacing 0px end attributes row cell 1 over C subscript t o t a l end subscript end cell equals cell fraction numerator 1 over denominator 4 μF end fraction plus fraction numerator 1 over denominator 12 μF end fraction end cell row cell 1 over C subscript t o t a l end subscript end cell equals cell fraction numerator 3 over denominator 12 μF end fraction plus fraction numerator 1 over denominator 12 μF end fraction end cell row cell 1 over C subscript t o t a l end subscript end cell equals cell fraction numerator 4 over denominator 12 μF end fraction end cell row cell C subscript t o t a l end subscript end cell equals cell 3 μF end cell end table

Muatan total:

table attributes columnalign right center left columnspacing 0px end attributes row Q equals cell C V end cell row blank equals cell 3 μF cross times 12 space straight V end cell row blank equals cell 36 space μC end cell end table

Menghitung tegangan pada rangkaian paralel

table attributes columnalign right center left columnspacing 0px end attributes row cell V subscript P end cell equals cell straight Q over straight C subscript straight p end cell row blank equals cell fraction numerator 36 space μC over denominator 12 space μF end fraction end cell row blank equals cell 3 space straight V end cell end table 

Besar energi lisrik:

table attributes columnalign right center left columnspacing 0px end attributes row W equals cell 1 half cross times C subscript 5 mu F end subscript cross times V subscript p squared end cell row blank equals cell 1 half cross times 5 space μF cross times 3 squared end cell row blank equals cell 22 comma 5 space μJ end cell end table 

    
Jadi, jawaban yang tepat adalah E.

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