1 / 76

Basic Electric Circuits: Series and Parallel Connections

Learn about series and parallel connections in electric circuits, including total resistance, current distribution, and voltage drops. Discover how to calculate equivalent resistance, current flowing through each component, voltage drop across each component, and power dissipated by each component.

jobrien
Download Presentation

Basic Electric Circuits: Series and Parallel Connections

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. Chapter 27 Circuits

  2. (Basic) Electric Circuits • Series Connections: • Total Resistance = Sum of individual Resistances. Current the same through every load.

  3. a I R1 The Voltage “drops”: b R2 a c Reffective Whenever devices are in SERIES, the current is the same through both ! This reduces the circuit to: c Hence: Resistorsin Series In general:

  4. Parallel Connection • Different parts of an electric circuit are on separate branches. Voltage: the same Total current: sum of currents in each branch

  5. Current I a I1 I2 • Devices in parallel have the same voltage drop V. V R1 R2 I d I • How is I related to I1 & I2 ? Current (Charge) is conserved! a V R I d Þ Þ Resistors in Parallel In general:

  6. Example Series & Parallel Given: R1=4Ω, R2=6 Ω, R3=12Ω, find: a) Rs b) Rp

  7. Example 2 Series Given: Three bulb s with the resistances R1=4Ω, R2=6 Ω, R3=12Ω, are connected in series with a 6 V battery. Required to find: a) The equivalent resistance of the bulbs Rs b) The current flowing through each bulb c) The voltage drop across each bulb: d) The power dissipated by each bulb:

  8. Example 3 Parallel Given: Three bulb s with the resistances R1=4Ω, R2=6 Ω, R3=12Ω, are connected in parallel with a 6 V battery. Required to find: a) The equivalent resistance of the bulbs Rp b) The current flowing through each bulb and the total current: c) The voltage drop across each bulb: d) The power dissipated by each bulb:

  9. 9 V ConcepTest 18.1a Series Resistors I 1)12 V 2)zero 3)3 V 4) 4 V 5) you need to know the actual value of R Assume that the voltage of the battery is 9 V and that the three resistors are identical. What is the potential difference across each resistor?

  10. 9 V ConcepTest 18.1a Series Resistors I 1)12 V 2)zero 3)3 V 4) 4 V 5) you need to know the actual value of R Assume that the voltage of the battery is 9 V and that the three resistors are identical. What is the potential difference across each resistor? Since the resistors are all equal, the voltage will drop evenly across the 3 resistors, with 1/3 of 9 V across each one. So we get a 3 V drop across each. Follow-up: What would be the potential difference if R= 1 W, 2 W, 3 W ?

  11. R1= 4 W R2= 2 W 12 V ConcepTest 18.1b Series Resistors II 1)12 V 2)zero 3)6 V 4) 8 V 5) 4 V In the circuit below, what is the voltage across R1?

  12. R1= 4 W R2= 2 W 12 V ConcepTest 18.1b Series Resistors II 1)12 V 2)zero 3)6 V 4) 8 V 5) 4 V In the circuit below, what is the voltage across R1? The voltage drop across R1 has to be twice as big as the drop across R2. This means that V1 = 8 V and V2 = 4 V. Or else you could find the current I = V/R = (12 V)/(6 W) = 2 A, then use Ohm’s Law to get voltages. Follow-up: What happens if the voltage is doubled?

  13. R2= 2 W R1= 5 W 10 V ConcepTest 18.2a Parallel Resistors I 1)10 A 2)zero 3)5 A 4) 2 A 5) 7 A In the circuit below, what is the current through R1?

  14. R2= 2 W R1= 5 W 10 V ConcepTest 18.2a Parallel Resistors I 1)10 A 2)zero 3)5 A 4) 2 A 5) 7 A In the circuit below, what is the current through R1? The voltage is the same (10 V) across each resistor because they are in parallel. Thus, we can use Ohm’s Law, V1 = I1 R1 to find the current I1 = 2 A. Follow-up: What is the total current through the battery?

  15. ConcepTest 18.2b Parallel Resistors II 1)increases 2)remains the same 3)decreases 4) drops to zero Points P and Q are connected to a battery of fixed voltage. As more resistors R are added to the parallel circuit, what happens to the total current in the circuit?

  16. ConcepTest 18.2b Parallel Resistors II 1)increases 2)remains the same 3)decreases 4) drops to zero Points P and Q are connected to a battery of fixed voltage. As more resistors R are added to the parallel circuit, what happens to the total current in the circuit? As we add parallel resistors, the overall resistance of the circuit drops. Since V = IR, and V is held constant by the battery, when resistance decreases, the current must increase. Follow-up: What happens to the current through each resistor?

  17. ConcepTest 18.3a Short Circuit I 1) all the current continues to flow through the bulb 2) half the current flows through the wire, the other half continues through the bulb 3)all the current flows through the wire 4) none of the above Current flows through a lightbulb. If a wire is now connected across the bulb, what happens?

  18. ConcepTest 18.3a Short Circuit I 1) all the current continues to flow through the bulb 2) half the current flows through the wire, the other half continues through the bulb 3)all the current flows through the wire 4) none of the above Current flows through a lightbulb. If a wire is now connected across the bulb, what happens? The current divides based on the ratio of the resistances. If one of the resistances is zero, then ALL of the current will flow through that path. Follow-up: Doesn’t the wire have SOME resistance?

  19. ConcepTest 18.3b Short Circuit II 1) glow brighter than before 2) glow just the same as before 3) glow dimmer than before 4) go out completely 5) explode Two lightbulbs A and B are connected in series to a constant voltage source. When a wire is connected across B, bulb A will:

  20. ConcepTest 18.3b Short Circuit II 1) glow brighter than before 2) glow just the same as before 3) glow dimmer than before 4) go out completely 5) explode Two lightbulbs A and B are connected in series to a constant voltage source. When a wire is connected across B, bulb A will: Since bulb B is bypassed by the wire, the total resistance of the circuit decreases. This means that the current through bulb A increases. Follow-up: What happens to bulb B?

  21. ConcepTest 18.4aCircuits I 1) circuit 1 2) circuit 2 3) both the same 4) it depends on R The lightbulbs in the circuit below are identical with the same resistance R. Which circuit produces more light? (brightness  power)

  22. ConcepTest 18.4aCircuits I 1) circuit 1 2) circuit 2 3) both the same 4) it depends on R The lightbulbs in the circuit below are identical with the same resistance R. Which circuit produces more light? (brightness  power) In #1, the bulbs are in parallel, lowering the total resistance of the circuit. Thus, circuit #1 willdraw a higher current, which leads to more light, because P = I V.

  23. A C B 10 V ConcepTest 18.4b Circuits II 1) twice as much 2) the same 3) 1/2 as much 4) 1/4 as much 5) 4 times as much The three lightbulbs in the circuit all have the same resistance of1 W . By how much is the brightness of bulb B greater or smaller than the brightness of bulb A? (brightness  power)

  24. A C B 10 V ConcepTest 18.4b Circuits II 1) twice as much 2) the same 3) 1/2 as much 4) 1/4 as much 5) 4 times as much The three lightbulbs in the circuit all have the same resistance of1 W . By how much is the brightness of bulb B greater or smaller than the brightness of bulb A? (brightness  power) We can use P = V2/R to compare the power: PA = (VA)2/RA = (10 V) 2/1 W= 100 W PB = (VB)2/RB = (5 V) 2/1 W= 25 W Follow-up: What is the total current in the circuit?

  25. R1 S R3 V R2 ConcepTest 18.5aMore Circuits I 1) increase 2) decrease 3) stay the same What happens to the voltage across the resistor R1 when the switch is closed? The voltage will:

  26. R1 S R3 V R2 ConcepTest 18.5aMore Circuits I 1) increase 2) decrease 3) stay the same What happens to the voltage across the resistor R1 when the switch is closed? The voltage will: With the switch closed, the addition of R2 to R3decreases the equivalent resistance, so the current from the battery increases. This will cause an increase in the voltage across R1 . Follow-up:What happens to the current through R3?

  27. R1 S R3 R4 V R2 ConcepTest 18.5b More Circuits II 1) increases 2) decreases 3) stays the same What happens to the voltage across the resistor R4 when the switch is closed?

  28. ConcepTest 18.5b More Circuits II R1 B A S R3 R4 V R2 C 1) increases 2) decreases 3) stays the same What happens to the voltage across the resistor R4 when the switch is closed? We just saw that closing the switch causes an increase in the voltage across R1 (which is VAB). The voltage of the battery is constant, so if VAB increases, then VBC must decrease! Follow-up:What happens to the current through R4?

  29. R5 R2 R3 R1 R4 V 1) R1 2) both R1and R2 equally 3) R3and R4 4) R5 5) all the same ConcepTest 18.6Even More Circuits Which resistor has the greatest current going through it? Assume that all the resistors are equal.

  30. R5 R2 R3 R1 R4 V 1) R1 2) both R1and R2 equally 3) R3and R4 4) R5 5) all the same ConcepTest 18.6Even More Circuits Which resistor has the greatest current going through it? Assume that all the resistors are equal. The same current must flow through left and right combinations of resistors. On the LEFT, the current splits equally, so I1 = I2. On the RIGHT, more current will go through R5 than R3+ R4 since the branch containing R5 has less resistance. Follow-up:Which one has the smallest voltage drop?

  31. Kirchhoff’s First Rule“Junction Rule” or“Kirchhoff’s Current Law (KCL)” • In deriving the formula for the equivalent resistance of 2 resistors in parallel, we applied Kirchhoff's First Rule (the junction rule). "At any junction point in a circuit where the current can divide (also called a node), the sum of the currents into the node must equal the sum of the currents out of the node." • This is just a statement of the conservation of charge at any given node. • The currents entering and leaving circuit nodes are known as “branch currents”. • Each distinct branch must have a current, Iiassigned to it

  32. KVL: e1 R1 e2 R2 I =0 Kirchhoff’s Second Rule“Loop Rule” or “Kirchhoff’s Voltage Law (KVL)” "When any closed circuit loop is traversed, the algebraic sum of the changes in potential must equal zero." • This is just a restatement of what you already know: that the potential difference is independent of path! -e2 +e1 -IR1 -IR2

  33. e1 R1 e2 R2 I =0 Rules of the Road Our convention: • Voltage gains enter with a + sign, and voltage drops enter with a - sign. • We choose a direction for the current and move around the circuit in that direction (cw or ccw). • When a battery is traversed from the negative terminal to the positive terminal, the voltage increases, and hence the battery voltage enters KVL with a + sign. • When moving across a resistor, the voltage drops, and hence enters KVL with a - sign. -e2 +e1 -IR1 -IR2

  34. Þ Loop Demo R1 R4 e1 b f a I I d c e R2 e2 R3 Þ KVL:

  35. Example, Single loop circuit with two real batteries:

  36. 12V R I a 12V 12V R b Example Switch • Consider the circuit shown. • The switch is initially open and the current flowing through the bottom resistor is I0. • Just after the switch is closed, the current flowing through the bottom resistor isI1. • What is the relation betweenI0 and I1? (c) I1 > I0 (b) I1 = I0 (a) I1 < I0 • The key here is to determine the potential (Va-Vb) before the switch is closed. • From symmetry, Va-Vb = +12V. • Therefore, when the switch is closed, NO additional current will flow! • Therefore, the current before the switch is closed is equal to the current after the switch is closed.

  37. (c) I1 >I0 (b)I1 = I0 (a) I1 < I0 • Write a loop law for original loop: 12V +12V -I0R-I0R = 0 I0 = 12V/R • Write a loop law for the new loop: 12V -I1R = 0 I1 = 12V/R Example Switch • Consider the circuit shown. • The switch is initially open and the current flowing through the bottom resistor is I0. • After the switch is closed, the current flowing through the bottom resistor is I1. • What is the relation betweenI0 andI1? 12V R a I 12V 12V R b

  38. Summary • When you are given a circuit, you must first carefully analyze circuit topology • find the nodes and distinct branches • assign branch currents • Use KVL for all independent loops in circuit • sum of the voltages around these loops is zero!

  39. I I R r I a I V R e b C e R More on Kirchhoff’s Laws

  40. R1 I3 e 2 I2 I1 R2 R3 e 1 How to use Kirchhoff’s Laws A two loop example: • Analyze the circuit and identify all circuit nodes and use KCL. (1)I1 = I2 + I3 • Identify all independent loops and use KVL. (2)e1-I1R1-I2R2 = 0 (3) e1-I1R1-e2-I3R3 = 0 (4) I2R2-e2-I3R3 = 0

  41. R1 I3 I2 I1 e 2 R2 R3 e 1 From eqn. (2) From eqn. (3) Þ How to use Kirchoff’s Laws • Solve the equations for I1, I2, and I3: Firstfind I2 and I3 in terms of I1: Now solve for I1using eqn. (1):

  42. R1 I3 I2 I1 e 2 R2 R3 e 1 Let’s plug in some numbers e1 = 24 V e 2 = 12 V R1= 5W R2=3W R3=4W Then,and I1=2.809 A I2= 3.319 A, I3= -0.511 A

  43. Outside loop: e1 R Top loop: I1 I2 e2 R Junction: I3 e3 R Junction Demo

  44. 50W a b I2 I1 20W 80W 12V c d (a)(Va-Vd) < (Va-Vc) (b)(Va-Vd) = (Va-Vc) (c)(Va-Vd) > (Va-Vc) 1B 1B • What is the relation between I1 and I2? (c)I1 > I2 (b)I1 = I2 (a)I1 < I2 Potential Difference, SET 1 • Consider the circuit shown: • What is the relation between Va-Vd and Va -Vc ? 1A

  45. 50W a b I2 I1 20W 80W 12V c d (a)(Va-Vd) < (Va-Vc) (b)(Va-Vd) = (Va-Vc) (c)(Va-Vd) > (Va-Vc) Example Multiloop • Consider the circuit shown: • What is the relation between Va -Vd and Va -Vc ? 1A • Do you remember that thing about potential being independent of path? Well, that’s what’s going on here !!! (Va-Vd) = (Va-Vc) Point d and c are the same, electrically

  46. 50W a b I2 I1 20W 80W 12V c d (a)(Va-Vd) < (Va-Vc) (b)(Va-Vd) = (Va-Vc) (c)(Va-Vd) > (Va-Vc) 1B • What is the relation between I1 and I2? (c)I1 > I2 (b)I1 = I2 (a)I1 < I2 • Note that: Vb-Vd = Vb-Vc • Therefore, Example Multiloop • Consider the circuit shown: • What is the relation between Va -Vd and Va -Vc ? 1A 1B

  47. Summary of Simple Circuits • Resistors in series: Current thru is same; Voltage drop across is IRi • Resistors in parallel: Voltage drop across is same; Current thru is V/Ri

  48. Household Circuits • Connections are made in parallel. • Ground wire. • Fuses: made of a strip of metal that melts and breaks the circuit if the current becomes to high. • Circuit breakers: a switch that flips open if the current becomes to high.

  49. 5 A P 8 A 2 A ConcepTest 18.7Junction Rule 1) 2 A 2) 3 A 3) 5 A 4) 6 A 5) 10 A What is the current in branch P?

  50. ConcepTest 18.7Junction Rule 5 A P 8 A 6 A junction 2 A 1) 2 A 2) 3 A 3) 5 A 4) 6 A 5) 10 A What is the current in branch P? The current entering the junction in red is 8 A, so the current leaving must also be 8 A. One exiting branch has 2 A, so the other branch (at P) must have 6 A. S

More Related