Consider the combination of capacitors in the figure below. (Let C 1 = 44.0 F an

Question

Consider the combination of capacitors in the figure below. (Let C1 = 44.0 F and C2 = 2.10 F.)

(a) Find the equivalent single capacitance of the two capacitors in series.
6.76 F

(b) In diagram 1 find the equivalent capacitance of the three capacitors in parallel.
12.86 F



(c) Compute the charge on the single equivalent capacitor.
462.96C

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

Does the sum agree with the value found in part (c)?

Yes

(e) What is the charge on the 44.0-F capacitor and on the 8.00-F capacitor?
______C

(f) Compute the voltage drop across the 44.0-F capacitor.
_______V

(g) Compute the voltage drop across the 8.00-F capacitor.

______V

(I NEED HELP WITH PARTS E, F & G) PLEASE WRITE YOUR SOLTUION LEGBILY)

4.00-F capacitor    144 C 2.10-F capacitor    75.6 C 44.0-F and 8.00-F equivalent capacitor    243.36 C 86.0 V 4.00 AuF 8.00 AuF

Explanation / Answer

E.

Charge distribution in series circuit remains same, So

Charge on 44 uF capacitor = Charge on 8 uF capacitor = 243.36 uC

F.

Q = C*V

V = Q/C

V = 243.36*10^-6/(44*10^-6)

V = 5.53 V

G.

V = Q/C

V = 243.36*10^-6/(8*10^-6)

V = 30.42 V

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