Capacitor problem… in series and parallel?
Using the following circuit diagram: http://session.masteringphysics.com/problemAsset/1...
A) What is the charge on each capacitor in the figure? Please enter your answers numerically, separated by commas.
B) What is the potential difference across each capacitor in the figure? Again, please enter your answers numerically, separated by commas.
I'm feeling quite lost with this problem. Any assistance would be greatly appreciated!
4 Answers
V2 = V*C3/(C2+C3 ) = 9*6/10 = 5.4V ....Q2 = V2*C2 = 5.4*4 = 21.60 μC
V3 = V*C2/(C2+C3) = 4*9/10 = 3.6V ....Q3 = V3*C3 = 3.6*6 = 21.60 μC
V1 = V = 9.0 V....................Q1 = V*C1 = 5*9.0 = 45 μC
Jan 26, 2025
Key ideas are charge-voltage relation Q = C.V
C1 is 5 uF with 9V p.d. across it hence Q1 = 5x10^-6 X 9 = 45x10^-6 Coulombs
The series arrangement is a bit more challenging and we can use a number of different
starting points. Lets begin with the idea that the series capacitors each have the SAME
charge i.e Q2 = Q3 so the voltages V split in inverse relation to their capacitances.
The total capacitance 1/C = 1/C2 + 1/C3 =2.4 uF giving
Q2 =Q3 = 2.4*9x10^-6 = 21.6x10^-6 Coulombs
and finally V2 = 21.6/4 = 5.4 V and V3 = 21.6/6 = 3.6 V.
(verify that 5.4 + 3.6 =9 as required and 5.4/3.6 = 6/4 as noted above!!)
The two in series are equivalent to 6•4/10 = 2.4 µF
C1 has 9v on it, and Q1 = CV = 9•5 = 45 µC charge
The series combo has 9v across the pair, and the charge is 9•2.4 = 21.6 µC, and that charge is on both of the pair.
V2 = Q/C = 21.8 / 4 = 5.4 volts
V3 = 21.8 / 6 = 3.6 volts
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