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**Chapter 26B - Capacitor Circuits**

A PowerPoint Presentation by Paul E. Tippens, Professor of Physics Southern Polytechnic State University © 2007

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**Objectives: After completing this module, you should be able to:**

Calculate the equivalent capacitance of a number of capacitors connected in series or in parallel. Determine the charge and voltage across any chosen capacitor in a network when given capacitances and the externally applied potential difference.

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**Electrical Circuit Symbols**

Electrical circuits often contain two or more capacitors grouped together and attached to an energy source, such as a battery. The following symbols are often used: Ground Battery - + + Capacitor -

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Series Circuits Capacitors or other devices connected along a single path are said to be connected in series. See circuit below: Series connection of capacitors. “+ to – to + …” + - Battery C1 C2 C3 Charge inside dots is induced.

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**Charge on Capacitors in Series**

Since inside charge is only induced, the charge on each capacitor is the same. Charge is same: series connection of capacitors. Q = Q1 = Q2 =Q3 Battery C1 C2 C3 + - Q1 Q2 Q3

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**Voltage on Capacitors in Series**

Since the potential difference between points A and B is independent of path, the battery voltage V must equal the sum of the voltages across each capacitor. Total voltage V Series connection Sum of voltages V = V1 + V2 + V3 Battery C1 C2 C3 + - V1 V2 V3 • A B

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**Equivalent Capacitance: Series**

Q1= Q2 = Q3 + - C1 C2 C3 V1 V2 V3 V = V1 + V2 + V3 Equivalent Ce for capacitors in series:

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Example 1. Find the equivalent capacitance of the three capacitors connected in series with a 24-V battery. + - 2 mF C1 C2 C3 24 V 4 mF 6 mF Ce for series: Ce = 1.09 mF

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**Example 1 (Cont.): The equivalent circuit can be shown as follows with single Ce.**

+ - 2 mF C1 C2 C3 24 V 4 mF 6 mF Ce = 1.09 mF 1.09 mF Ce 24 V Note that the equivalent capacitance Ce for capacitors in series is always less than the least in the circuit. (1.09 mF < 2 mF)

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**For series circuits: QT = Q1 = Q2 = Q3**

Example 1 (Cont.): What is the total charge and the charge on each capacitor? + - 2 mF C1 C2 C3 24 V 4 mF 6 mF 1.09 mF Ce 24 V Ce = 1.09 mF QT = 26.2 mC QT = CeV = (1.09 mF)(24 V); For series circuits: QT = Q1 = Q2 = Q3 Q1 = Q2 = Q3 = mC

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**Example 1 (Cont.): What is the voltage across each capacitor?**

+ - 2 mF C1 C2 C3 24 V 4 mF 6 mF VT = 24 V Note: VT = 13.1 V V V = 24.0 V

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**Short Cut: Two Series Capacitors**

The equivalent capacitance Ce for two series capacitors is the product divided by the sum. 3 mF 6 mF + - C1 C2 Example: Ce = 2 mF

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**Parallel capacitors: “+ to +; - to -”**

Parallel Circuits Capacitors which are all connected to the same source of potential are said to be connected in parallel. See below: Voltages: VT = V1 = V2 = V3 Parallel capacitors: “+ to +; - to -” C2 C3 C1 + - Charges: QT = Q1 + Q2 + Q3

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**Equivalent Capacitance: Parallel**

Parallel capacitors in Parallel: C2 C3 C1 + - Q = Q1 + Q2 + Q3 Equal Voltages: CV = C1V1 + C2V2 + C3V3 Equivalent Ce for capacitors in parallel: Ce = C1 + C2 + C3

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Example 2. Find the equivalent capacitance of the three capacitors connected in parallel with a 24-V battery. C2 C3 C1 2 mF 4 mF 6 mF 24 V Q = Q1 + Q2 + Q3 VT = V1 = V2 = V3 Ce for parallel: Ce = ( ) mF Ce = 12 mF Note that the equivalent capacitance Ce for capacitors in parallel is always greater than the largest in the circuit. (12 mF > 6 mF)

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**Example 2 (Cont.) Find the total charge QT and charge across each capacitor.**

2 mF 4 mF 6 mF 24 V Q = Q1 + Q2 + Q3 Ce = 12 mF V1 = V2 = V3 = 24 V QT = CeV Q1 = (2 mF)(24 V) = 48 mC QT = (12 mF)(24 V) Q1 = (4 mF)(24 V) = 96 mC QT = 288 mC Q1 = (6 mF)(24 V) = 144 mC

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**Example 3. Find the equivalent capacitance of the circuit drawn below.**

4 mF 3 mF 6 mF 24 V C2 C3 Ce = 4 mF + 2 mF Ce = 6 mF C1 4 mF 2 mF 24 V C3,6 Ce 6 mF 24 V

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**Example 3 (Cont.) Find the total charge QT.**

Ce = 6 mF C1 4 mF 3 mF 6 mF 24 V C2 C3 Q = CV = (6 mF)(24 V) QT = 144 mC C1 4 mF 2 mF 24 V C3,6 Ce 6 mF

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**This can also be found from Q = C3,6V3,6 = (2 mF)(24 V) **

Example 3 (Cont.) Find the charge Q4 and voltage V4 across the the 4-mF capacitor. C1 4 mF 3 mF 6 mF 24 V C2 C3 V4 = VT = 24 V Q4 = (4 mF)(24 V) Q4 = 96 mC The remainder of the charge: (144 mC – 96 mC) is on EACH of the other capacitors. (Series) This can also be found from Q = C3,6V3,6 = (2 mF)(24 V) Q3 = Q6 = 48 mC

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**Example 3 (Cont.) Find the voltages across the 3 and 6-mF capacitors.**

Q3 = Q6 = 48 mC C1 4 mF 3 mF 6 mF 24 V C2 C3 Note: V3 + V6 = 16.0 V V = 24 V Use these techniques to find voltage and capacitance across each capacitor in a circuit.

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**Summary: Series Circuits**

Q = Q1 = Q2 = Q3 V = V1 + V2 + V3 For two capacitors at a time:

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**Summary: Parallel Circuits**

Q = Q1 + Q2 + Q3 V = V1 = V2 =V3 For complex circuits, reduce the circuit in steps using the rules for both series and parallel connections until you are able to solve problem.

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**CONCLUSION: Chapter 26B Capacitor Circuits**

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