In the figure below, a 17.0 V battery is connected across capacitors of capacita
ID: 3163516 • Letter: I
Question
In the figure below, a 17.0 V battery is connected across capacitors of capacitances with the following values. C_1 = C_6 = 2.60 mu F C_3 = C_5 = 4.10 mu F C_2 = C_4 = 2.05 mu F What is the equivalent capacitance C_eq? __________ mu F What is the charge stored by C_eq? __________ C What is V_1 of capacitor 1? ______________ C What is q_1 of capacitor 1? ______________ V What is V_2 of capacitor 2? _____________ V What is q_2 of capacitor 2? ___________ C What Is V_3 of capacitor 3? _______________ V What is q_3 of capacitor 3? _______________ C In the figure below, the battery has potential difference V = 9.0 V, C_2 = 3.0 mu F, C_4 middot 5.5 mu F, and all the capacitors are initially uncharged. When switch S is closed, a total charge of 16 mu C passes through point a and a total charge of 6.0 mu C passes through point b.Explanation / Answer
C10 = C3 series C5 = 1 / [ (1/4.1F) + (1/4.1F) ]= 2.05 F
C11 = C10 parallel C2 = 2.05 F + 2.05 F = 4.1 F
C12 = C11 parallel C4 = C11+C4 = 4.1 F + 2.05 F = 6.15 F
C13 = C1 parallel C6 = C1 + C6 = 2.6 F + 2.6 F = 5.2 F
Ceq = C12 series C13
= (6.15 F)*(5.2 F) / (6.15F + 5.2F)
= 2.817 F
Qeq = (Ceq)*(V)
= (2.817E-06Farads)*(17.0 volts)
= 47.899 E-06 Coulombs
The charge on series capacitors is equivalent; the voltage on parallel capacitors is equivalent.
QC13 =QC12 = 47.899 E-06 Coulombs
VC1 = VC6= VC13 = (QC13)/(C13) = (47.899E-06Coulombs)/(5.2E-06 Farads)
= 9.21 volts
QC1 =(VC1)*(C1) = (9.21 volts)*(2.6E-06 Farads)
= 23.95 E-06Coulombs
e) VC2 = VC4= VC10 = VC11 = VC12= (QC12)/(C12) = (47.899E-06 Coulombs)/(6.15E-06Farads)
= 7.78 volts
f) QC2 =(VC2)*(C2) = (7.78 volts)*(2.05E-06 Farads)
= 15.97E-06Coulombs
g) QC10 =(VC10)*(C10) = (7.78 volts)*(2.05 E-06 Farads) = 15.96 E-06 Coulombs
VC3= (QC10)/(C3) = (15.96 E-06 Coulombs)/(4.1 E-06Farads) = 3.9 volts
h) QC3 =(VC3)*(C3)
= (3.93 volts)*(4.10E-06 Farads)
= 15.96 E-06Coulombs