If necessary use any of the above equations to solve the following problem. Cons

Question

If necessary use any of the above equations to solve the following problem.

Consider the circuit shown in Figure 2 below.

All of the capacitors have the same capacitance, C = 100 ?F, and the emf has a potential difference E = 10 V.

a) Calculation Question: The 5 capacitors in the circuit could be replaced with a single e ective capacitor. What would the capacitance of the effective capacitor be?

b) Calculation Question: How much charge is on the capacitor between points c and d? Please show your work in detail.

c) Calculation Question: What is the potential difference across the capacitor between points b and d? Please show your work in detail.

d) Calculation Question: What is the potential difference between the points a and d? Please show your work in detail.

e) Calculation Question: How much charge is on the capacitor between points b and e? Please show your work in detail.

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Explanation / Answer

capacitor inbetween be and de are in series so equivalent of these two is C' then

1/C' = 1/C + 1/C

C' = 0.5C

now C' and bd are in parallel connection.

so C" = C' + C = 1.5C

now C", ab and cd are in series.

1/Ceq = 1/1.5C + 1/C + 1/C

Ceq = 0.375C = 0.375 x 100uF = 37.5 uF

b) capacitors in series will have same charge and also same as on equivalent .

Q = Ceq V = 37.5 uF x 10 = 375uC

c) charge on C" will also be 375 uC .

so PD across C" = Q /C" = 375uC / (1.5 x 100uF) = 2.5 Volt

C' and bd are in parallel so they will have same PD across them

hence PD across bd = 2.5 V

d) PD across ab = 375uC / (100uF) = 3.75 V

Vad = 3.75 + 2.5 =6.25 Volt

e) PD across C' is 2.5 V.

so charge on C' = 2.5 x 0.5C = 2.5 x 0.5 x 100uF = 125 uC

and C' is equvalent of be and de in series.

in series capacitor have same charge.

charge on capacitor inbetween b and e = 125 uC

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