When five capacitors with equal capacitances are connected in series, the equiva

Question

When five capacitors with equal capacitances are connected in series, the equivalent capacitance of the combination is 4.75 mF. The capacitors are then reconnected so that a parallel combination of two capacitors is connected in series with a parallel combination of three capacitors. Calculate the equivalent capacitance of this combination. Enter your result in the unit of millifarad. I find Ctot=C/5 C=5(4.75x10^-3) C=23.75mF Then I find C12(series)=.0475 Then C345(parallel)=23.75/3=.007917 Then I find Ceq=(.0475)(.007917)/(.0475+.007917)=.006786 I enter my answer =6.786 mF but this is wrong. Anyone know what I am doing wrong?

Explanation / Answer

Let C be the individual capacitance of each capacitor

In series equivalent Capacitance

Ceq =C/5

C=5Ceq =5*4.75

C=23.75mF

so the equivalent capacitance in reconnected circuit

Ceq =(23.75+23.75)in series (23.75+23.75+23.75)

Ceq=47.5 in series|71.25

Ceq =47.5*71.25/(47.5+71.25)

Ceq =28.5mF

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