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A

Anonymous

Oct 23, 2024

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

A
Anonymous

Nov 13, 2024

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

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!!)

A
Anonymous

Feb 06, 2025

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

A
Anonymous

Nov 29, 2024

connect 2 x a million micro farad capacitors in sequence this may will equivalent a .5 farad capacitor. connect those in Parallel with a a million microfarad capacitor and voile! you've a a million.5 microfarad cap

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