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Chapter 26B - Capacitor Circuits A PowerPoint Presentation by Paul E. Tippens, Professor of Physics Southern Polytechnic State University © 2007 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. 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 + + - Series Circuits Capacitors or other devices connected along a single path are said to be connected in series. See circuit below: + + C1 - + - + -+ -+ C2 Battery - C3 Series connection of capacitors. “+ to – to + …” Charge inside dots is induced. Charge on Capacitors in Series Since inside charge is only induced, the charge on each capacitor is the same. Q1 + + C1 Q2 - + - + Q3 -+ -+ C2 Battery - C3 Charge is same: series connection of capacitors. Q = Q1 = Q2 =Q3 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. V1 + + C1 •A V2 - + - + V3 -+ -+ C2 Battery - C3 • B Total voltage V Series connection Sum of voltages V = V1 + V2 + V3 Equivalent Capacitance: Series V1 + + V2 - + - + C1 C2 V3 -+ -+ - Q Q C ; V V C C3 V = V1 + V2 + V3 Q1= Q2 = Q3 Q Q1 Q2 Q3 C C1 C2 C3 1 1 1 1 Ce C1 C2 C3 Equivalent Ce for capacitors in series: n 1 1 Ce i 1 Ci Example 1. Find the equivalent capacitance of the three capacitors connected in series with a 24-V battery. Ce for series: n 1 1 Ce i 1 Ci 1 1 1 1 Ce 2m F 4m F 6m F 1 0.500 0.250 0.167 Ce 1 1 0.917 or Ce Ce 0.917 C1 C2 C3 + - + -+ + - + -+ 2 mF 4 mF 6 mF 24 V Ce = 1.09 mF Example 1 (Cont.): The equivalent circuit can be shown as follows with single Ce. C1 C2 C3 + - + -+ + - + -+ 2 mF 4 mF 6 mF n 1 1 Ce i 1 Ci Ce 1.09 mF 24 V Ce = 1.09 mF 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) Example 1 (Cont.): What is the total charge and the charge on each capacitor? C1 C2 C3 + - + -+ + - + -+ 2 mF 4 mF 6 mF Ce 1.09 mF 24 V 24 V QT = CeV = (1.09 mF)(24 V); For series circuits: QT = Q1 = Q2 = Q3 Ce = 1.09 mF Q C V Q CV QT = 26.2 mC Q1 = Q2 = Q3 = 26.2 mC Example 1 (Cont.): What is the voltage across each capacitor? Q Q C ; V V C Q1 26.2 m C V1 13.1 V C1 2 mF Q2 26.2 m C V2 6.55 V C2 4 mF Q3 26.2 mC V3 4.37 V C3 6 mF C1 C2 C3 + - + -+ + - + -+ 2 mF 4 mF 6 mF 24 V VT = 24 V Note: VT = 13.1 V + 6.55 V + 4.37 V = 24.0 V Short Cut: Two Series Capacitors The equivalent capacitance Ce for two series capacitors is the product divided by the sum. 1 1 1 ; Ce C1 C2 C1C2 Ce C1 C2 Example: (3 m F)(6 m F) Ce 3 m F 6m F C1 C2 + - + + - + 3 mF 6 mF Ce = 2 mF Parallel Circuits Capacitors which are all connected to the same source of potential are said to be connected in parallel. See below: Parallel capacitors: “+ to +; - to -” - - C3 + + - - C2 + + + + C1 - - Voltages: VT = V1 = V2 = V3 Charges: QT = Q1 + Q2 + Q3 Equivalent Capacitance: Parallel Q C ; Q CV V Parallel capacitors in Parallel: - - Ce = C1 + C2 + C3 C3 + + - - C2 + + + + C1 - - Q = Q1 + Q2 + Q3 Equal Voltages: CV = C1V1 + C2V2 + C3V3 Equivalent Ce for capacitors in parallel: n Ce Ci i 1 Example 2. Find the equivalent capacitance of the three capacitors connected in parallel with a 24-V battery. Ce for parallel: n Ce Ci VT = V1 = V2 = V3 Q = Q1 + Q2 + Q3 i 1 24 V Ce = (2 + 4 + 6) mF 2 mF C1 C2 C3 4 mF 6 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) Example 2 (Cont.) Find the total charge QT and charge across each capacitor. Q = Q1 + Q2 + Q3 24 V 2 mF C1 C2 C3 4 mF 6 mF Ce = 12 mF V1 = V2 = V3 = 24 V Q C ; Q CV 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 Example 3. Find the equivalent capacitance of the circuit drawn below. 24 V C1 4 mF 24 V 4 mF C1 C3,6 C2 3 mF C3 6 mF C3,6 (3m F)(6m F) 2m F 3m F 6m F Ce = 4 mF + 2 mF Ce = 6 mF 24 V 2 mF Ce 6 mF Example 3 (Cont.) Find the total charge QT. Ce = 6 mF 24 V C1 4 mF 24 V 4 mF C1 C2 3 mF C3 6 mF C3,6 Q = CV = (6 mF)(24 V) QT = 144 mC 24 V 2 mF Ce 6 mF Example 3 (Cont.) Find the charge Q4 and voltage V4 across the the 4-mF capacitor. V4 = VT = 24 V 24 V 4 mF C1 C2 3 mF C3 6 mF 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) Q3 = Q6 = 48 mC This can also be found from Q = C3,6V3,6 = (2 mF)(24 V) Example 3 (Cont.) Find the voltages across the 3 and 6-mF capacitors. Q3 = Q6 = 48 mC 24 V 4 mF C1 C2 3 mF C3 6 mF 48m C V3 16.0V 3m F 48m C V6 8.00V 6m F Note: V3 + V6 = 16.0 V + 8.00 V = 24 V Use these techniques to find voltage and capacitance across each capacitor in a circuit. Summary: Series Circuits n 1 1 Ce i 1 Ci Q = Q1 = Q2 = Q3 V = V1 + V2 + V3 For two capacitors at a time: C1C2 Ce C1 C2 Summary: Parallel Circuits n Ce Ci i 1 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. CONCLUSION: Chapter 26B Capacitor Circuits