1 / 58

Electric Circuits and Power

Electric Circuits and Power. Chapter 20. Series and Parallel Circuits. 20.1. Series Circuits. Have only ONE “LOOP” or circuit for the current to travel through. Resistors in Series. When two or more resistors are connected end-to-end, they are said to be in series

sivan
Télécharger la présentation

Electric Circuits and Power

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Electric Circuits and Power Chapter 20

  2. Series and Parallel Circuits 20.1

  3. Series Circuits • Have only ONE “LOOP” or circuit for the current to travel through.

  4. Resistors in Series • When two or more resistors are connected end-to-end, they are said to be in series • The current is the same in all resistors because any charge that flows through one resistor flows through the other • The sum of the voltages across the resistors is equal to the total voltages across the combination – Kirchoff’s Voltage Law, the Conservation of Voltage

  5. Resistors in Series • Potentials add • ΔV = IR1 + IR2 = I (R1+R2) • Consequence of Conservation of Energy • The equivalent resistance has the effect on the circuit as the original combination of resistors

  6. Equivalent Resistance – Series Re = R1 + R2 + R3 + … • The equivalent resistance of a series combination of resistors is the algebraic sum of the individual resistances and is always greater than any of the individual resistors • Batteries and even wires contribute small amounts of resistance but we can ignore this for now…

  7. Equivalent Resistance – Series An Example • Four resistors are replaced with their equivalent resistance

  8. Resistors in Parallel • The voltage across each resistor is the same because each is connected directly across the battery terminals • The current, I, that enters a point must be equal to the total current leaving that point • I = I1 + I2 • The currents are generally not the same • Consequence of Kirchoff’s Second Law, the Conservation of Charge

  9. Equivalent Resistance – Parallel An Example • Equivalent resistance replaces the two original resistances • Household circuits are wired so the electrical devices are connected in parallel • Circuit breakers may be used in series with other circuit elements for safety purposes

  10. Equivalent Resistance – Parallel • Equivalent Resistance • The inverse of the equivalent resistance of two or more resistors connected in parallel is the algebraic sum of the inverses of the individual resistance • The equivalent is always less than the smallest resistor in the group

  11. Example 1 In the depicted circuit, the voltage supplied by the battery is 12V, and the resistors have values of R1 = 10Ω, R2 = 5Ω and R3 = 15Ω. What is the current flowing through each branch?

  12. Example 1 • The current through each resistor can be found with Ohm’s Law • I1 = 1.2A; I2 = 2.4A; I3 = 0.8A • To check this, find the current through the whole circuit by finding the total resistance, then using Ohm’s Law again. • Rtot= 2.7ΩItot= 4.4A • This matches the sum of the individual currents

  13. Example 2 • If, V = 24V; R1 = 2Ω; R2 = 3.3Ω; R3 = 7Ω and R4 = 12.2Ω • Find the current through the circuit and the current through each resistor.

  14. Example 2 • Find the total resistance: • Rtot = o.973Ω • Use this to find the current through the circuit • I = 24.7A • The current through each resistor is just the voltage divided by each individual resistance: • I1 = 12A; I2 = 7.3A; • I3 = 3.4A I4 = 2A

  15. V1 V2 V3 R1 R2 R3 A R4 B Series Circuit • A series of sources separated by resistors is equivalent to a single source having the net voltage and a single resistor having the combined resistance. R = R1 + R2 + R3 + R4 V = V1 - V2 + V3 • The same current passes through every resistor in a given branch, regardless of the presence of sources in that branch, and the resistors are in series even though they are not directly connected to one another..

  16. R1 R2 V Parallel Circuit • The current in each branch of a parallel circuit depends inversely on the total resistance: the larger the resistance, the less current flows through the branch • If we know I but not V

  17. Analysis Of Circuits 20.2

  18. Review of Circuit Rules For SERIES Circuits: • There is ONE path for the current • Current is CONSTANT • Voltage DROPS across each resistance • Resistors add simply • Additional resistances DECREASE current

  19. Review of Circuit Rules For PARALLEL Circuits: • There are MULTIPLE paths for the current • Current may NOT be CONSTANT • Voltage is CONSTANT to each branch • Resistors add RECIPROCALLY • Additional resistance INCREASES current

  20. Review of Circuit Rules Sometimes you will have BOTH series AND parallel resistors in the SAME circuit!! You then need to SIMPLIFY the circuit in your analysis.

  21. Example 3 • Consider this circuit. • (a) If possible, simplify it and determine an equivalent resistance between C and G. • (b) What current is provided by the source? • (c) What is the voltage across points G and E? • Given: Nine resistors, R = 1.0 kW each, and V = 12V • Find: Re, I, and V between G and E

  22. Example 3 • Solution To solve this one we'll need the equivalent resistance of the circuit • Procedure — Redraw this circuit to make it look more manageable • Lift up the inside square, with the resistor and source attached, and place it outside E-F-G-H – see diagram on right • Branches A-B-C and A-D-C are in parallel, as are E-F-G and E-H-G, and each has a resistance of 1.0 kW + 1.0 kW = 2.0 kW (resistors add in series)

  23. Example 3 • The equivalent circuit is shown to the right • The resistance of each square E-F-G-H and A-B-C-D reduces to • and R = 1.0 kW.

  24. Example 3 • The three 1.0-kW resistors then are in series with the source, • (a) The equivalent resistance is 3.0 kW. • (b) Since V = IRe,

  25. Example 3 • (c) A current of 4.0 mA leaves the battery and splits at C • Because the two branches C-D-A and C-B-A have the same resistance, the current divides into two equal streams of 2.0 mA each • The voltage drop in going from C to A is given by VAC = IR = (2.0 mA)(1.0 kW + 1.0 kW) = 4.0 V • In going from A to E, there is another drop of VEA= IR = (4.0 mA)(1.0 kW) = 4.0 V. C is 12V above G, A is 8.0 V above G, and E is 4.0V above G.

  26. Example 4 Consider the following circuit: A battery supplying 12 volts leads to a resistor (R1 = 1.3Ω), then splits into three branches. The first branch has R2 = 4.5Ω, the second branch has R3 = Ω, and the third branch contains R4 = 5Ω AND R5 = 2.2Ω in series. Finally, the three branches reunite, and lead to R6 = 7Ω before reconnecting to the battery. • Draw a diagram of this circuit • Find the total resistance • Find the overall current in the circuit.

  27. Problem-Solving Strategy, 1 • Combine all resistors in series • They carry the same current • The potential differences across them are not the same • The resistors add directly to give the equivalent resistance of the series combination: Re = R1 + R2 + …

  28. Problem-Solving Strategy, 2 • Combine all resistors in parallel • The potential differences across them are the same • The currents through them are not the same • The equivalent resistance of a parallel combination is found through reciprocal addition:

  29. Problem-Solving Strategy, 3 • A complicated circuit consisting of several resistors and batteries can often be reduced to a simple circuit with only one resistor • Replace any resistors in series or in parallel using steps 1 or 2. • Sketch the new circuit after these changes have been made • Continue to replace any series or parallel combinations • Continue until one equivalent resistance is found

  30. Problem-Solving Strategy, 4 • If the current in or the potential difference across a resistor in the complicated circuit is to be identified, start with the final circuit found in step 3 and gradually work back through the circuits • Use ΔV = I R and the procedures in steps 1 and 2

  31. Equivalent Resistance – Complex Circuit

  32. Capacitors in Circuits • A circuit is a collection of objects usually containing a source of electrical energy (such as a battery) connected to elements that convert electrical energy to other forms • A circuit diagram can be used to show the path of the real circuit

  33. Capacitors in Parallel • When capacitors are first connected in the circuit, electrons are transferred from the left plates through the battery to the right plate, leaving the left plate positively charged and the right plate negatively charged • The flow of charges ceases when the voltage across the capacitors equals that of the battery • The capacitors reach their maximum charge when the flow of charge ceases

  34. Capacitors in Parallel • The total charge is equal to the sum of the charges on the capacitors • Qtotal = Q1 + Q2 • The potential difference across the capacitors is the same • And each is equal to the voltage of the battery

  35. More About Capacitors in Parallel • The capacitors can be replaced with one capacitor with a capacitance of Ceq • The equivalent capacitor must have exactly the same external effect on the circuit as the original capacitors • Capacitors in parallel all have the same voltage differences as does the equivalent capacitance

  36. Capacitors in Parallel • The equivalent capacitance of several capacitors in parallel is the sum of all the individual capacitors. C = C1 + C2 + … • The equivalent capacitance of a parallel combination of capacitors is greater than any of the individual capacitors

  37. + + + - - - 20 mF 30 mF 12 V Example 5 • The figure below shows two capacitors attached to a 12-V battery. Determine the equivalent capacitance and the charge it would carry. What is the charge on each of the capacitors in the figure? • Given: C1 = 20 mF, C2 = 30 mF, and V = 12 V • Find: C, Q, Q1, and Q2

  38. Example 5 • Solution: Capacitors are in parallel and the potential across each capacitor is 12 V Q = CV = (50 x 10-6 F)(12 V) = 6.0 x 10-4 C

  39. Capacitors in Series • When a battery is connected to the circuit, electrons are transferred from the left plate of C1 to the right plate of C2 through the battery • As this negative charge accumulates on the right plate of C2, an equivalent amount of negative charge is removed from the left plate of C2, leaving it with an excess positive charge • All of the right plates gain charges of –Q and all the left plates have charges of +Q

  40. More About Capacitors in Series • An equivalent capacitor can be found that performs the same function as the series combination • The potential differences add up to the battery voltage • Capacitors in series all have the same charge, Q, as does their equivalent capacitance

  41. Capacitors in Series • The equivalent capacitance of several capacitors in series • The equivalent capacitance of a series combination is always less than any individual capacitor in the combination

  42. C1 C2 + 2.0 mF 2.0 mF C3 12 V 5.0 mF – Example 6 • The circuit shown in the figure consists of a 12-V battery and three capacitors. It is redrawn from Fig. 12.27a in the book. Determine both the voltage across and charge on each capacitor after the switch S is closed and electrostatic equilibrium is established. Find the equivalent capacitance of the network. • Given: C1 = 2.0 mF, C2 = 2.0 mF, C3 = 5.0 mF,and V = 12 V • Find: C, V1, V2, V3, Q1, Q2, and Q3

  43. + C3 C1 + C2 12 V 5.0 mF 1.0 mF – Example 6 • Combining C1 and C2 which are in series

  44. + C 12 V 6.0 mF – Example 6 • Combining C3 and (C1 + C2) which are in parallel • The equivalent capacitance of the network

  45. C1 C2 + 2.0 mF 2.0 mF C3 12 V 5.0 mF – Example 6 • There is 12 V across C3 so Q3= C3V3 = (5.0 mF)(12 V) = 60 mC • There is 12 V acrossthe combination of the two 2.0 mF capacitors so there is a potential difference of 6.0 V across each Q1= Q2 = (2.0 mF)(6.0 V) = 12 mC

  46. Problem-Solving Strategy • Be careful with the choice of units • Combine capacitors following the formulas • When two or more unequal capacitors are connected in series, they carry the same charge, but the potential differences across them are not the same • The capacitances add as reciprocals and the equivalent capacitance is always less than the smallest individual capacitor

  47. Problem-Solving Strategy, cont • Combining capacitors • When two or more capacitors are connected in parallel, the potential differences across them are the same • The charge on each capacitor is proportional to its capacitance • The capacitors add directly to give the equivalent capacitance

  48. Problem-Solving Strategy, final • Repeat the process until there is only one single equivalent capacitor • A complicated circuit can often be reduced to one equivalent capacitor • Replace capacitors in series or parallel with their equivalent • Redraw the circuit and continue • To find the charge on, or the potential difference across, one of the capacitors, start with your final equivalent capacitor and work back through the circuit reductions

  49. Electric Power, AC and DC Electricity 20.3

  50. Household Circuits • The utility company distributes electric power to individual houses with a pair of wires • Electrical devices in the house are connected in parallel with those wires • The potential difference between the wires is about 120V

More Related