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Alycia Zemlak

Feb 20, 2025

Determine the equivalent capacitance between A and B for the group of capacitors in the drawing. Let C1 = 12 µF and C2 = 3.0 µF.?

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Start on the right.The 24, 8.0 and C1 are in series. Use the series formula to figure out the equivalent of those three 1/C = 1/24 + 1/8 + 1/C1.That’s in parallel with the 4.0. So it simply adds.That’s in series with C2 and the 5.0. So use the series formula again.... Show More
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24 μF in series with C₁ and 8.0 μFC(eq1) = 1/24 + 1/12 + 1/8.0 = (1+2+3)/24 = 7/24 μF4 μF // C(eq1)C(eq₂) = 4 + (7/24) = 103/24 μF5.0 μF in series with C(eq₂) and 3.0 μFC(A-B) = 1/5.0 + 1/(103/24) + 1/3 = 1/5 + 24/103 + 1/3 = (309+360+515) /1545 = 1184/1545 = 0.77 μF... Show More
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Capacitors in series add indirectly, 1/x + 1/y = 1/z. z is the equivalent capacitance.Capacitors in parallel add directly, x + y = z. z is the equivalent capacitance.Your challenge is to identify which are in series and which are in parallel. If they share a voltage, they are in parallel. If they share a current, then they are in series.C = ((1/5 + 1/C2 + (((1/24 + 1/C1 + 1/8)^(-1)) + 4)^(-1)))^(-1)Now, can you jump through those hoops?... Show More
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