8. 0/2 points | Previous Answers SerPSE9 26.P022.wi. My Notes Ask Your Teacher C

Question

8. 0/2 points | Previous Answers SerPSE9 26.P022.wi. My Notes Ask Your Teacher Consider the following figure. (t C1 C2 C2 (a) Find the equivalent capacitance between points a and b for the group of capacitors connected as shown in the figure above. Take C-3.00 C3-4.00 6.24 C2 13.0 , and Try redrawing the circuit sliding capacitors Ci and C2 on the left and right halves of the upper loop towards the outside so that they reside on the vertical segment of conducting wire. Note that sliding a capacitor along a conducting wire does not affect the circuit as long as you do not pass a junction with another wire. HF (b) What charge is stored on C3 if the potential difference between points a and b is 60.0 V? HC Need Help? Read" Lwonen" Watch It

Explanation / Answer

C1 and C2 are in series, so

C12 = C1 x C2/(C1 + C2)

C12 = 3 x 13/3 + 13 = 2.44 uF

Now we see that the two C12 and C3 are in parallel

C123 = 2C12 + C3

C123 = 2 x 2.44 + 4 = 8.88 uF

Now the 2 C2 are in parallel

C2 = 2C2 = 2 x 13 = 26 uF

C123 and C2 are in series

Ceq = C123 x C2/(C123 + C2)

Ceq = 8.88 x 26/(8.88 + 26) = 6.62 uF

Hence, Ceq = 6.62 uF

b)We know that,

Q = CV = 6.62 x 60 = 397.2 uC

the drop across C2 is:

V2 = Q2/C2 = 397.2/26 = 15.28 V

so the drop across 123 is:

V123 = 60 - 15.28 = 44.72

V3 = 44.72

Q3 = V3 C3 = 44.72 x 4 = 178.88 uC

Hence, Q3 = 178.88 uC

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