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In Fig. P24.59, each capacitance C1 , 6.9μFand each capacitance C2is 4.6μF. (a) Compute the equivalent capacitance of the network between points a and b. (b) Compute the charge on each of the three capacitors nearest a and b when Vab=420V. (c) With 420 V across a and b, compute Vcd.

Short Answer

Expert verified

(a) Ct=2.3μF

(b) Q=Q4=9.66×10aC,Q2=6.44×10-4C.

(c) Vcd=46.67V.

Step by step solution

01

parameters.

C1=6.9μFC2=4.6μF

02

(a) Computing the equivalent capacitance of the network.

The 3 capacitorsC1 are in series so the equivalentC1' is given by

1C1'=1C1+1C1=1C1=1C1C1'=C13=6.93=2.3μF

The capacitanceC1',C2 are in parallel so the equivalentC12' is given by

C12'=C1'+C2=2.3+4.6=6.9μF

The 2 capacitorsrole="math" localid="1664269686387" C1',C2 are in series so the equivalentC12' is given by

1C3=1C1+1C1+1C12'==36.9C3=6.93=2.3μF

The capacitorsrole="math" localid="1664269567298" C3,C2 are in parallel so the equivalent C23is given by

C23=C3+C2=2.3+4.6=6.9μF

The 2 capacitors are in series so the equivalentCt is given by

1C3=1C1+1C1+1C23=36.9Ct=6.93=2.3μF

03

(b) Computing the charge of each of the three capacitors.

Looking to fig (4) the charge flows through series capacitors is the same and is given by

Q=VabCt=420×2.3×10-6=9.66×10-4C

So the voltage on each in fig (4) is given by

V1=QC1=9.66×10-46.9×10-6=140V

So the voltage on V23is given by

V23=Vab-2V2=420-2×140=140V

The voltage across the capacitors on parallel is the same so the charge on C2 is given by

Q2=V23C2=140×4.6×10-6=6.44×10-4C.

04

(c) Computing the Vcd .

Looking to fig (2) knowing the voltage across the capacitorC12' , to get that, the charge flow should be known before.

The charge flow inC12' is given by

Q12'=Q-Q2=9.66-6.44×10-4=3.22×10-4.

So the voltage Vcd is given by

Vcd=Q12'C12'3.22x10-46.9x10-6=46.67V

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