Two capacitors C 1 =5.00 µF and C 2 =14.0 µF are connected in series to a9.00 V

Question

Two capacitors C1 =5.00 µF and C2 =14.0 µF are connected in series to a9.00 V battery. (a) Find the equivalent capacitance of thecombination.          µF (b) Find the potential difference across eachcapacitor. C1 = V                                                                                                  C2 = V
(c) Find the charge on eachcapacitor.                             C1 = µC
                                                                                                 C2 = µC Two capacitors C1 =5.00 µF and C2 =14.0 µF are connected in series to a9.00 V battery. (a) Find the equivalent capacitance of thecombination.          µF (b) Find the potential difference across eachcapacitor. C1 = V                                                                                                  C2 = V
(c) Find the charge on eachcapacitor.                             C1 = µC
                                                                                                 C2 = µC

Explanation / Answer

(a) Find the equivalent capacitance of thecombination. (µF)
(C_eff) = 1/( 1/C1 + 1/C2 ) =
     = 1/(1/(5.0 F) + 1/(14.0 F) )
     = 3.684 F

(b) Find the potential difference across eachcapacitor.
C1= Q/C1 = (33.156 C)/(5.0 F) = 6.631 V
C2= Q/C2 = (33.156 C)/(14.0 F) = 2.368 V

(c) Find the charge on eachcapacitor.
     Q = (C_eff)*V = (3.684 F)*(9.0 V) = 33.16 C
C1= 33.16 µC
C2 = 33.16 µC

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