2Objectives: 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.
3Electrical 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:GroundBattery-++Capacitor-
4Series CircuitsCapacitors or other devices connected along a single path are said to be connected in series. See circuit below:Series connection of capacitors. “+ to – to + …”+-BatteryC1C2C3Charge inside dots is induced.
5Charge 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 =Q3BatteryC1C2C3+-Q1Q2Q3
6Voltage 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 voltagesV = V1 + V2 + V3BatteryC1C2C3+-V1V2V3•AB
7Equivalent Capacitance: Series Q1= Q2 = Q3+-C1C2C3V1V2V3V = V1 + V2 + V3Equivalent Ce for capacitors in series:
8Example 1. Find the equivalent capacitance of the three capacitors connected in series with a 24-V battery.+-2 mFC1C2C324 V4 mF6 mFCe for series:Ce = 1.09 mF
9Example 1 (Cont.): The equivalent circuit can be shown as follows with single Ce. +-2 mFC1C2C324 V4 mF6 mFCe = 1.09 mF1.09 mFCe24 VNote that the equivalent capacitance Ce for capacitors in series is always less than the least in the circuit. (1.09 mF < 2 mF)
10For series circuits: QT = Q1 = Q2 = Q3 Example 1 (Cont.): What is the total charge and the charge on each capacitor?+-2 mFC1C2C324 V4 mF6 mF1.09 mFCe24 VCe = 1.09 mFQT = 26.2 mCQT = CeV = (1.09 mF)(24 V);For series circuits: QT = Q1 = Q2 = Q3Q1 = Q2 = Q3 = mC
11Example 1 (Cont.): What is the voltage across each capacitor? +-2 mFC1C2C324 V4 mF6 mFVT = 24 VNote: VT = 13.1 V V V = 24.0 V
12Short Cut: Two Series Capacitors The equivalent capacitance Ce for two series capacitors is the product divided by the sum.3 mF6 mF+-C1C2Example:Ce = 2 mF
13Parallel capacitors: “+ to +; - to -” Parallel CircuitsCapacitors which are all connected to the same source of potential are said to be connected in parallel. See below:Voltages: VT = V1 = V2 = V3Parallel capacitors: “+ to +; - to -”C2C3C1+-Charges: QT = Q1 + Q2 + Q3
14Equivalent Capacitance: Parallel Parallel capacitors in Parallel:C2C3C1+-Q = Q1 + Q2 + Q3Equal Voltages: CV = C1V1 + C2V2 + C3V3Equivalent Ce for capacitors in parallel:Ce = C1 + C2 + C3
15Example 2. Find the equivalent capacitance of the three capacitors connected in parallel with a 24-V battery.C2C3C12 mF4 mF6 mF24 VQ = Q1 + Q2 + Q3VT = V1 = V2 = V3Ce for parallel:Ce = ( ) mFCe = 12 mFNote that the equivalent capacitance Ce for capacitors in parallel is always greater than the largest in the circuit. (12 mF > 6 mF)
17Example 3. Find the equivalent capacitance of the circuit drawn below. 4 mF3 mF6 mF24 VC2C3Ce = 4 mF + 2 mFCe = 6 mFC14 mF2 mF24 VC3,6Ce6 mF24 V
18Example 3 (Cont.) Find the total charge QT. Ce = 6 mFC14 mF3 mF6 mF24 VC2C3Q = CV = (6 mF)(24 V)QT = 144 mCC14 mF2 mF24 VC3,6Ce6 mF
19This 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.C14 mF3 mF6 mF24 VC2C3V4 = VT = 24 VQ4 = (4 mF)(24 V)Q4 = 96 mCThe 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
20Example 3 (Cont.) Find the voltages across the 3 and 6-mF capacitors. Q3 = Q6 = 48 mCC14 mF3 mF6 mF24 VC2C3Note: V3 + V6 = 16.0 V V = 24 VUse these techniques to find voltage and capacitance across each capacitor in a circuit.
21Summary: Series Circuits Q = Q1 = Q2 = Q3V = V1 + V2 + V3For two capacitors at a time:
22Summary: Parallel Circuits Q = Q1 + Q2 + Q3V = V1 = V2 =V3For complex circuits, reduce the circuit in steps using the rules for both series and parallel connections until you are able to solve problem.