Para el siguiente condensador: C1 = 2 µF C2 = 4 µF V = 20 Voltios.... Question from @karoledenmejiagomez

December 2023 1 1 Report

Para el siguiente condensador: C1 = 2 µF C2 = 4 µF V = 20 Voltios

Respuesta:

Resolución

CE = 40 μF = 40 . 10-6

C1 = 10 μF = 10 . 10-6

+q1 C1 -q1 +q12 C12 -q12



A +q2 C2 –q2 B A B

CE = C1 + C2 ; 40 . 10-6

F = 10 . 10-6

F + C2

C2 = 40 . 10-6

F – 10 . 10-6

F = 30 . 10-6 F

Explicación:

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