Three capacitors of C1= 2.0 %u03BCF, C2= 6.0 %u03BCF, and C3= 4.5 %u03BCF are co

ID: 2121861 • Letter: T

Question

Three capacitors of C1= 2.0 %u03BCF, C2= 6.0 %u03BCF, and C3= 4.5 %u03BCF are connected. C1 and C2 are connected in series and the result circuit is connected to C3 in parallel. How much capacitance in %u03BCF would a single capacitor need to have to replace the three capacitors?
Three capacitors of C1= 2.0 %u03BCF, C2= 6.0 %u03BCF, and C3= 4.5 %u03BCF are connected. C1 and C2 are connected in series and the result circuit is connected to C3 in parallel. How much capacitance in %u03BCF would a single capacitor need to have to replace the three capacitors?
Three capacitors of C1= 2.0 %u03BCF, C2= 6.0 %u03BCF, and C3= 4.5 %u03BCF are connected. C1 and C2 are connected in series and the result circuit is connected to C3 in parallel. How much capacitance in %u03BCF would a single capacitor need to have to replace the three capacitors?
Three capacitors of C1= 2.0 %u03BCF, C2= 6.0 %u03BCF, and C3= 4.5 %u03BCF are connected. C1 and C2 are connected in series and the result circuit is connected to C3 in parallel. How much capacitance in %u03BCF would a single capacitor need to have to replace the three capacitors?

Explanation / Answer

The net capacitance in Series = 1/ C1 +1/C2 = 1/2 +1/6 = 2/3 Farads

Now that circuit is connected in parallel.

If connected in parallel, net capacitance = C1+C2= 2/3+4.5= 5.166. Farads

Therefore, replacing the three capactior would require a capacitance to be 5.166 Farads.

Explanation : Fomula to be used for net capacitance in series and parallel given by ~

For series : 1/C = 1/C1+ 1/C2+ 1/C3....

For parallel : C = C1+ C2+ C3+ ...

if you want the derivations too, please do ask. please rate. :)

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Three capacitors of C1= 2.0 %u03BCF, C2= 6.0 %u03BCF, and C3= 4.5 %u03BCF are co

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