(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

2.00 µF capacitor
5 µC

far right capacitor
6 µC

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

(e) What is the charge on the 25.0µF capacitor and on the 8-µF capacitor?
8 µC

(f) Compute the voltage drop across the 25.0 µF capacitor.
9 V

(g) Compute the voltage drop across the 2.00 µF capacitor.
10 V 4.00 µF capacitor
4 µC

2.00 µF capacitor
5 µC

far right capacitor
6 µC

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

Explanation / Answer

Part a: 25 F is in series with 8 F Equivalent capcitence = 25*8/(25+8) = 6.1 F Part b: Three capacitors 4 F, 2 F, 6.1 F are inparallel Equivelent capacitence = 4+2+6.1 = 12.1 F Part c: Charge Q = C*V = 12.1*10^-6*36 = 435.6 C Part d-4: charge = 4*36 = 144 C Part d-5: charge = 2*36 = 72 C Part d-6: charge = 6.1 *36 = 219.6 C Part d-7: Sum of charges 144+72+219.6 = 435.6 C This agrees with part c. Part e: charge on 25 uF and 8 F will be same as they are inseries. (219.6 C) Part f: Voltage across 25 F = Q/C = 219.6/25 = 8.784 V Part g: Voltage across 2 F = 72/2 = 36 V

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