A capacitor, C 1 = 10.0 µF is charged to V i = 15.0 V. It is next connected in s

Question

A capacitor, C1 = 10.0 µF is charged to Vi = 15.0 V. It is next connected in series with an uncharged capacitor, C2 = 9.00 µF. The series combination is finally connected across a battery, Vf = 52.0 V as diagrammed in the figure below. Find the new potential differences across the 9.00 µF and 10.0 µF capacitors.

across 9.00 µF = V across 10.0 µF = V A capacitor, C1 = 10.0 muF is charged to ?Vi = 15.0 V. It is next connected in series with an uncharged capacitor, C2 = 9.00 muF. The series combination is finally connected across a battery, ?Vf = 52.0 V as diagrammed in the figure below. Find the new potential differences across the 9.00 muF and 10.0 muF capacitors. across 9.00 muF = V across 10.0 muF = V

Explanation / Answer

Given Capacitors   C 1 = 10.0 F                           = ( 10 F ) ( 10 -6 F / 1 F )                           = 10 *10-6 F                    C 2 = 9.0 F                           = ( 9 F ) ( 10 -6 F / 1 F )                           = 9 *10-6 F initial charge on capacitors is                   Q    = C 1 V i                          = ( 10 *10-6 F ) ( 15 V )                           = 0.15*10-3 C Equavalent capacitance                 C = C 1 C 2 / ( C 1 + C 2 )                      = (10 *10-6 F ) ( 9 *10-6 F ) / ((10 *10-6 F ) + ( 9 *10-6 F )                      = 4.7368 *10-6 F If the capacitors are in series combination caharge on each capacitor is same                 C = Q / V                   Q' = CV                        = ( 4.7368 *10-6 F   ) ( 52 V )                        = 0.246*10-3 C Total charge on each capacitor is                  Q + Q ' = 0.396*10-3 C Voltage across C 1 is V 1 = ( Q+Q' ) / C 1                                          = ( 0.396*10-3 C ) / ( 10 *10-6 F )                                          = 39.6 V                                    V 2 = ( Q+Q ' ) / C 2                                          = ( 0.396*10-3 C ) / ( 9 *10-6 F )                                          = 44 V                           = ( 9 F ) ( 10 -6 F / 1 F )                           = 9 *10-6 F initial charge on capacitors is                   Q    = C 1 V i                          = ( 10 *10-6 F ) ( 15 V )                           = 0.15*10-3 C Equavalent capacitance                 C = C 1 C 2 / ( C 1 + C 2 )                      = (10 *10-6 F ) ( 9 *10-6 F ) / ((10 *10-6 F ) + ( 9 *10-6 F )                      = 4.7368 *10-6 F If the capacitors are in series combination caharge on each capacitor is same                 C = Q / V                   Q' = CV                        = ( 4.7368 *10-6 F   ) ( 52 V )                        = 0.246*10-3 C Total charge on each capacitor is                  Q + Q ' = 0.396*10-3 C Voltage across C 1 is V 1 = ( Q+Q' ) / C 1                                          = ( 0.396*10-3 C ) / ( 10 *10-6 F )                                          = 39.6 V                                    V 2 = ( Q+Q ' ) / C 2                                          = ( 0.396*10-3 C ) / ( 9 *10-6 F )                                          = 44 V                                          = 44 V

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