Question 1.SA.7: Determine the equivalent capacitance in series and parallel ......

Electrical Engineering Fundamentals [1024827]

Determine the equivalent capacitance in series and parallel combination circuit shown below. The capacitance values are: $C_1 = 10 μF, C_2 = 10 μF, C_3 = 20 μF$ , and $C_4 = 20 μF.$

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The capacitors in this circuit that lend themselves to linear combination are $C_3$ and $C_4$ . Therefore, the combined capacitance, $C_{34}$ , would be as follows:

C_{34}=C_{3}+C_{4}=20  μF + 20  μF= 40  μF

Then, by applying Eq.1.24 to this special hybrid capacitor combination case:

C_{EQ}=\frac{C_1C_2C_{34}}{C_1C_2+C_2C_{34}+C_1C_{34}}

C_{EQ}=\frac{(10×10^{-6})(10×10^{-6})(40×10^{-6})}{(10×10^{-6})(10×10^{-6})+(10×10^{-6})(40×10^{-6})+(10×10^{-6})(40×10^{-6})}

C_{EQ}=4.44  μF

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