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Analysis of a simple series capacitive circuit and problems

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A series circuits consists of two capacitors C1 = 3 uF, C2 =6 uF connected in series with a DC supply of V(supply) = 6V (SEE ATTACHMENT FOR CIRCUIT).

PART A: Determine the charge stored on C1 = 3 uF

PART B: Determine the charge stored on C2 = 6 uF

PART C: What is the voltage at point 1, the +ve side of C2?

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Solution Summary

A series circuit is presented consisting of 2 capacitors in series with a DC voltage supply. The solution shows how to determine the charge stored on each of the series capacitors and hence determine the voltage that exists between the junction of the two capacitors in the circuit. A step by step approach is used showing how each of these parameters may be derived using simple AC circuit theory and linear algebraic techniques. Great emphasis is made in simplifying the approach to make it easily understandable and to be able to be followed in logical steps.

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PART A: Solution

First need to find the voltage across the capacitor

If we call the first capacitor C1 = 3 uF its impedance (real part) is given by

Z1 = 1/wC1

Similarly for capacitor C2 = 6 uF its impedance is given by

Z2 = 1/wC2

By voltage divider rule the voltage across capacitor C1 is given by

V1 = V(supply)*Z1/(Z1 + Z2) = V(supply)*{1/wC1}/{1/wC1 + 1/wC2}

V1 = ...

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