Two capacitors, C1 =4400 pF and C2=1500pF, areconnected in series to a 15.0 V ba

Question

Two capacitors, C1 =4400 pF and C2=1500pF, areconnected in series to a 15.0 V battery. The capacitors arelater disconnected from the battery and connected directly to eachother, positive plate to positive plate, and negative plate tonegative plate. What then will be the charge on each capacitor? Q1=, Q2= Two capacitors, C1 =4400 pF and C2=1500pF, areconnected in series to a 15.0 V battery. The capacitors arelater disconnected from the battery and connected directly to eachother, positive plate to positive plate, and negative plate tonegative plate. What then will be the charge on each capacitor? Q1=, Q2=

Explanation / Answer

Given C = 4400pF           C ' =1500 pF voltage V = 15 V resulatant capacitance when they are connected in series C " = ( C C ' ) / ( C + C ' )                                                                                            = 1118.64 pF So, charge in capacitor C and C '  is Q = C " V                                                                = 16779.66 pC                                                               = 16779.66 * 10 ^ -12 C Voltage across C is v = Q / C=3.8135 Volt voltage across C ' is v ' = Q / C ' = 11.186 Volt when they are conneted parallel to each other . common voltage V ' = ( Q + Q ) / ( C + C ' )                                 = 5.6888 Volt Charge on capacitor C is  q = C V '                                      = 25030.72 pC charge on capacitor C ' is q ' = C ' V '                                              = 8533.2 pC

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