Answer to Question #154008 in Electric Circuits for Rizwan

Question #154008

Find the equivalent capacitance.

C1=5.0 µF, C2=3.5 µF, C3=8.0 µF, C4=1.5 µF, C5=0.75µF, C6=15 µF


1
Expert's answer
2021-01-06T13:57:11-0500

There are 2 types of placements of capacitors: in series and parallel.

The formulas for these:

i) In series 1Cs=1C1+1C2+1C3+1C4+1C5+1C6;\frac{1}{C_s}=\frac{1}{C_1}+\frac{1}{C_2}+\frac{1}{C_3}+\frac{1}{C_4}+\frac{1}{C_5} +\frac{1}{C_6};


ii) parallel Cp=C1+C2+C3+C4+C5+C6;C_p=C_1+C_2+C_3+C_4+C_5+C_6;


We examine both cases for this task:

i) 1Cs=15.0+13.5+18.0+11.5+10.75+115\frac{1}{C_s}=\frac{1}{5.0}+\frac{1}{3.5}+\frac{1}{8.0}+\frac{1}{1.5}+\frac{1}{0.75}+\frac{1}{15} = 1Cs=15.0+27+18.0+23+43+115\frac{1}{C_s}=\frac{1}{5.0}+\frac{2}{7}+\frac{1}{8.0}+\frac{2}{3}+\frac{4}{3}+\frac{1}{15} =


=168+240+105+1736840=2249840=2569840;\frac{168+240+105+1736}{840}=\frac{2249}{840}=2\frac{569}{840};


ii) Cp=5.0+3.5+8.0+1.5+0.75+15=33.75C_p=5.0+3.5+8.0+1.5+0.75+15=33.75 µF


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