Consider the combination of capacitors shown in the diagram, where C 1 = 3.00 ?

Question

Consider the combination of capacitors shown in the diagram, where C1 = 3.00?F , C2 = 11.0?F , C3 = 3.00?F , and C4 = 5.00?F . (Figure 1).

Part A

Find the equivalent capacitance CA of the network of capacitors.

Express your answer in microfarads.

Part B

Two capacitors of capacitance C5 = 6.00?F and C6 = 3.00?F are added to the network, as shown in the diagram.(Figure 2) Find the equivalent capacitance CB of the new network of capacitors.

Express your answer in microfarads.

fig1.

Fig2.

Consider the combination of capacitors shown in the diagram, where C1 = 3.00?F , C2 = 11.0?F , C3 = 3.00?F , and C4 = 5.00?F . (Figure 1). Part A Find the equivalent capacitance CA of the network of capacitors. Express your answer in microfarads. Part B Two capacitors of capacitance C5 = 6.00?F and C6 = 3.00?F are added to the network, as shown in the diagram.(Figure 2) Find the equivalent capacitance CB of the new network of capacitors. Express your answer in microfarads.

Explanation / Answer

remember these about Combination of capacitors

if C1 and C2 are the individual capacitor that re connected in series,

then their resultant capaciatnce 1/Cnet = 1/C1+1/C2 or Cnet   = C1C2/(C1+C2)

if c1 and C2 are connected in parallel, then their effective capacitancen is Cnet = C1+C2

also in sereis combination same amount of charge passes through each of the capacitiors

and in parallel combination, same PD (voltage) exists across each of the capacitor

so here in fig A:

C3, C4 are in parallel = C' = C3+C4 = 3 + 5 = 8 uF

this 8uF is in parallel with C2, so C'' = C2+C' = 11 + 8 = 19 uF

this

19uF is in series with C1,

So Cnet = C1 * C''/(C1 + C'')

Cnet = 19 * 3/(19+3)

Cnet in fig A CA   = 2.6 uF ------------<<<<<ANswer

----------------------------

in fig B:

C3, C5, C4 are in parallel, so C' = C3+C5 +C4

C' = 3 + 6 + 5 = 14 uF

now this 14 uF is in parallel with ( sereis of C2 and C6)

series of C2 and C6 is C2*C6/(C2+C6)

C' = 11* 3/(11+3)

C' = 2.35 uF

so this 2.35 uF as in parallel with 14uF

C'' = 2.35uF +14 uF = 16.35 uF

finally 16.35 uF is in series with C1

Cnet of fig B = CB = 16.35 * 3/(16.35+ 3)

Cnet = 2.534 uF --------------<<<<<<<<<<Answer

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