(a) Find the equivalent single capacitance ofthe two capacitors in series and re

Question

(a) Find the equivalent single capacitance ofthe two capacitors in series and redraw the diagram (called diagram1) with this equivalent capacitance.
1 µF

(b) In diagram 1 find the equivalent capacitance of the threecapacitors in parallel and redraw the diagram as a single batteryand single capacitor in a loop.
2-µF

(c) Compute the charge on the single equivalent capacitor.
3-µC

(d) Returning to diagram 1, compute the charge on each individualcapacitor.
4.00 µF capacitor
4 µC

0.500 µF capacitor
5 µC

far right capacitor
6 µC

Does the sum agree with the value found in part (b)?
7---Select---noyes

(e) What is the charge on the 22.0µF capacitor and on the 8-µF capacitor?
8 µC

(f) Compute the voltage drop across the 22.0 µF capacitor.
9 V

(g) Compute the voltage drop across the 0.500 µF capacitor.
10 V 4.00 µF capacitor
4 µC

0.500 µF capacitor
5 µC

far right capacitor
6 µC

Does the sum agree with the value found in part (b)?
7---Select---noyes

Explanation / Answer

1. Same law as resistors in parallel. Ctot = 1/ (1/22 + 1/8) = 5.87F 2. Same law as resistors in series, simply add them. Ctot = 5.87 +4 +0.5 = 10.37 F 3. Charge = Q = C*V = 10.37 * 10-6 * 36 =3.73*10-4 coulomb 4,5,6. Use the same formula as (c) above 7. check and report. It should do. 8. same formula as (3) 9. 36V is across both in series. V1 (for C1) = (36/(22+8)) * 8 = (36/30) * 8 = 9.6V V2 (for the 8F) = (36/(22+8)) * 22 = (36/30) * 22 = 26.4V 10. The terminal voltage of 36V is connected across that capacitorso this seems like a silly question.

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