Consider the combination of capacitors in the figure below. (Let C1 = 26.0 ?F an

Question

Consider the combination of capacitors in the figure below. (Let C1 = 26.0 ?F and C2 = 3.00 ?F.)

(a) Find the equivalent single capacitance of the two capacitors in series and redraw the diagram (called diagram 1) with this equivalent capacitance.....in?F

(b) In diagram 1 find the equivalent capacitance of the three capacitors in parallel and redraw the diagram as a single battery and single capacitor in a loop....-?F

(c) Compute the charge on the single equivalent capacitor......in -?F

(d) Returning to diagram 1, compute the charge on each individual capacitor.

4.00-?F capacitor......in?C

3.00-?F capacitor......in?C

far right capacitor......in?C

(e) What is the charge on the26.0-?F capacitor and on the 8-?F capacitor?

(f) Compute the voltage drop across the26.0-?F capacitor.

(g) Compute the voltage drop across the3.00-?F capacitor.

Explanation / Answer

4 F+3 F=7 F==>26 F*8 F/(26 F+8 F)6.118 F

7 F+6.118 F=13.118 F=Ceq

a) 6.118 F

b) 13.118 F=Ceq

c) C not F is Q+C*V=13.118 F*36 V=472.2353 C or 472.2 C

d) 4 F*36 V=144 C==>3 F*36 V=108 C==>6.118 F*36 V=220.2 C

e) both 26 F and 8 F have 220.2 C

f) 220.2 C/26 F8.47 V

g) v drop across C2 is 36 V

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