From the capacitor array shown, calculate: a) The total capacitance b) The total charge c) The total power energy.(NEED ONLY HANDWRITTEN) F =11ODCoulumbVoltsHEC3 = 7μFC₁ = 3μFVolt=C₂ = 5μFJoulesCoulumbC4 = 9μF C5 = 2μF11Cequivalente C₁110 VHH= ++-C₂ C3++Cequivalente = C₁ + C₂ + C3 + + C₂• Cn151μF = 1X10-6F:Ep=Q²2C

Answered: Image /qna-images/answer/6bca8959-02f5-4503-98f4-86718ed6defc.jpg

c=2 F = Coulumb Volts HE C3 = 7μF C₁ = 3μF Volt= C₂ = 5μF C4 = 9μF C5 = 2μF 1 110 V 1 Cequivalente C₁ HH = + - + C₂ C3 + + /c: Cequivalente = C₁ + C₂ + C3 + + C₂ Joules Coulumb 1μF = 1X10-6F : Ep = Q² 2C

c=2 F = Coulumb Volts HE C3 = 7μF C₁ = 3μF Volt= C₂ = 5μF C4 = 9μF C5 = 2μF 1 110 V 1 Cequivalente C₁ HH = + - + C₂ C3 + + /c: Cequivalente = C₁ + C₂ + C3 + + C₂ Joules Coulumb 1μF = 1X10-6F : Ep = Q² 2C

Introductory Circuit Analysis (13th Edition)

13th Edition

ISBN:9780133923605

Author:Robert L. Boylestad

Publisher:Robert L. Boylestad

Chapter1: Introduction

Section: Chapter Questions

Problem 1P: Visit your local library (at school or home) and describe the extent to which it provides literature...

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Question

From the capacitor array shown, calculate: a) The total capacitance b) The total charge c) The total power energy.(NEED ONLY HANDWRITTEN)

F =
11
OD
Coulumb
Volts
HE
C3 = 7μF
C₁ = 3μF
Volt=
C₂ = 5μF
Joules
Coulumb
C4 = 9μF C5 = 2μF
1
1
Cequivalente C₁
110 V
HH
= +
+
-
C₂ C3
+
+
Cequivalente = C₁ + C₂ + C3 + + C₂
• Cn
15
1μF = 1X10-6F
:
Ep
=
Q²
2C

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