Download presentation

Presentation is loading. Please wait.

Published byEgbert Tyler Modified over 5 years ago

2
Circuit Symbols: Battery Resistor Light-bulb Switch Wire

3
**Three general types of circuits:**

Closed Circuit - There is a complete loop with wires going from one side of the battery through a resistor(s) to the other side of the battery. Open Circuit - There is not a complete loop. Short Circuit - There is a complete loop, but it does not contain any resistors. Only Working Circuit

4
**There are two ways to put resistors into a circuit.**

1. Resistors can be in series OR 2. Resistors can be in parallel

5
**Resistors in Series is like a trip to Costco**

6
**Resistors in Series (like a trip to Costco)**

Resistors are considered to be in series if the current must go through all of the resistors in order. The current (amps) through all resistors in series is the same. The voltage across resistors in series may be different The rate of electron flow (or current) is determined by which resistor? The resistor with the largest amount of ohms.

7
**Combining (adding) Resistors**

Series Resistors Itotal = I1 = I2 = I3 Req = Rtotal = R1 + R2 + R3 Voltage is calculated with Ohm’s Law Q Amps

8
**Resistors in Parallel is like a trip to Vons**

9
**Resistors in Parallel (like a trip to Vons)**

Resistors are considered to be in parallel if the current is shared between multiple resistors. The current (amps) through all resistors in parallel may be different. The voltage across all parallel resistors is the same. Will a resistor with a large resistance have more or less current through it then a resistor with a small resistance? The resistor with a large resistance will have a smaller current then the resistor with the smaller resistance.

10
**Combining (adding) Resistors**

Parallel Resistors Current is calculated with Ohm’s Law Vtotal = V1 = V2 = V3

11
**Example 1: A circuit has three 8. 0 W, 5**

Example 1: A circuit has three 8.0 W, 5.0 W and a 12 W resistors in series along with a 24 V battery. Draw the circuit. Calculate the total resistance of the circuit. Calculate the total current through the circuit. What is the current through each resistor? Calculate the voltage across each resistor.

12
**Example 2: A circuit has three resistors: 6. 0 W, 4**

Example 2: A circuit has three resistors: 6.0 W, 4.0 W and a 12 W resistors in parallel along with a 24 V battery. Draw the circuit. Calculate the total resistance of the circuit. Calculate the total current through the circuit. What is the voltage across each resistor? Calculate the current across each resistor.

13
Electrical Outlets Electrical outlets provide electric potential (or the voltage) for any appliance plugged in to it. In the United States ALL outlets provide 120 V (in Europe it is 240 V)

14
**Light bulbs are made to be the only appliance plugged into a socket.**

The power rating of a light bulb (25 W or 100 W…) is as if that bulb was the only bulb plugged in to a 120 V power source. The resistance of a light bulb is calculated by knowing the power rating and the voltage (120 V) Current and actual voltage used by a light bulb depends on the circuit.

15
Example 3: What will the power output be if an American-made 45 W light bulb is plugged in to a 310 V power source? Using 120 V, calculate the resistance of the light bulb. Using the resistance and the voltage of the new source, calculate the new power

16
As more identical resistors R are added to the parallel circuit shown, the total resistance between points P and Q … Increases Remains the same Decreases R P Q …

17
As more identical resistors R are added to the parallel circuit shown, the total resistance between points P and Q … 1. Increases Remains the same decreases R P Q … Q

18
**When one bulb is unscrewed, the other bulb will remain lit in which circuit…**

II Both Neither Circuit II Circuit I

19
**When one bulb is unscrewed, the other bulb will remain lit in which circuit…**

1. I II both neither Circuit I Circuit II

20
**A 25W bulb and a 100W bulb are connected in series**

A 25W bulb and a 100W bulb are connected in series. Which bulb will glow brighter?

21
25W 100W 120V

22
**The Light Bulbs are really Resistors**

A) Calculate the resistance for each resistor shown. B) Calculate the total resistance of the circuit. C) Calculate the current through each resistor. D) Calculate the power used by each resistor. E) Calculate the voltage across each resistor. 25W 100W 120V

23
Part A. 25W Bulb 100W Bulb

24
**144 576 120V B) Calculate the total circuit resistance Rtotal**

Rtotal = R1 + R Series Resistors = = 720

25
720 120V C) Calculate the total circuit current (I)

26
144 576 120V D) Calculate the Power used by each resistor. 25 W Bulb P = I2R = x 576 = 16 watts 100 W Bulb P = I2R = x 144 = 4 watts

27
**120V 25W Bulb 100W Bulb E) Calculate the Voltage across each resistor.**

96 volts 24 volts 144 576 120V E) Calculate the Voltage across each resistor. 25W Bulb V = IR = .167 x 576 = 96 volts 100W Bulb V = IR = .167 x 144 = 24 volts 120 volts

28
**E) Consider the Percent Power Needed to Light Each Bulb**

100 W Bulb 25 W Bulb Q

29
The circuit below consists of two identical light bulbs burning with equal brightness and a single 12V battery. When the switch is closed, the brightness of bulb A… A Increases Decreases Remains unchanged

30
The circuit below consists of two identical light bulbs burning with equal brightness and a single 12V battery. When the switch is closed, the brightness of bulb A… Increases decreases 3. remains unchanged When the switch is closed, bulb B goes out because all of the current goes through the wire parallel to the bulb. Thus, the total resistance of the circuit decreases, the current through bulb increases, and it burns brighter. A Q

31
**Bird 1 2) Bird 2 3) Neither 4) Both 1 2**

Which bird is in trouble when the switch is closed? Bird 1 2) Bird 2 3) Neither 4) Both 1 2

32
**1) Bird 1 2) bird 2 3) neither 4) both**

Which bird is in trouble when the switch is closed? 1) Bird 1 2) bird 2 3) neither 4) both 1 2

33
**Charge flows through a light bulb**

Charge flows through a light bulb. Suppose a wire is connected across the bulb as shown. When the wire is connected… All the charge continues to flow through the bulb, and the bulb stays lit. Half the charge flows through the wire, the other half continues through the bulb. Essentially all the charge flows through the wire and the bulb goes out. None of these. Q

34
**Analyze the circuit: Parallel:**

A) Calculate Rtotal B) Calculate the current through each resistor. C) Calculate the voltage through each resistor. 16 Parallel: 120V 32 32 16

35
Series: R123-4=8+16 R1234=24 16 120V 8

36
**All these numbers will be the same.**

These are in parallel so their voltage is the same along with the total voltage Make chart: R I V 1 16 2 3 32 4 234 8 1234 24 120 16 120V 32 32 16 All these numbers will be the same.

37
**All these numbers will be the same.**

These are in series so their current is the same along with the total current Make chart: R I V 1 16 2 3 32 4 234 8 1234 24 120 16 120V 8 All these numbers will be the same.

38
**Fill out the chart with V=IR**

I = 5 A V = IR V = (5) (16) V = 80 V R I V 1 16 5 80 2 2.5 40 3 32 1.25 4 234 8 1234 24 120 V = IR V = (5) (8) V = 40 V V = IR 40 = I (16) I = 2.5 A V = IR 40 = I (32) I = 1.25 A

39
**Another way to do the problem (without the chart)**

I=V/R I=120v/24 I=5 amps 120V 5amps 24

40
**V=IR V=(5)(16) V=80volts V=IR V=(5)(8) V=40volts 16 80volts 5amps**

8 40volts 120volts

41
I=V/R =40volts/16 =2.5 amps 5 amps 16 80volts 120V I=V/R =40volts/32 =1.25 amps 32 40 volts 32 16 5 amps

42
HIDDEN 16 80 ft 8 each 16 40 ft Q

43
When the series circuit shown is connected, Bulb A is brighter than Bulb B. If the positions of the bulbs were reversed… Bulb A would again be brighter Bulb B would be brighter They would be equal brightness

44
When the series circuit shown is connected, Bulb A is brighter than Bulb B. If the positions of the bulbs were reversed… Bulb A would again be brighter Bulb B would be brighter They would be the same The bulbs are connected in series, so the same current passes through both of them. Different brightnesses indicate different filament resistances. Bulb A is NOT brighter because it is “first in line” for the current of the battery! After all, electrons deliver the energy, and they flow from negative to positive --- in the opposite direction!

45
**Example: Find the voltage and current for each resistor.**

6 3 6 3 4 2 12 18 volts

46
6 3 6 3 4 2 12 18 volts

47
3 3 3 4 2 12 18 volts

48
3 3 3 4 2 12 18 volts

49
6 3 4 2 12 18 volts

50
6 3 4 2 12 18 volts

51
6 3 3 2 18 volts

52
6 3 3 2 18 volts

53
6 3 5 18 volts

54
6 3 5 18 volts

55
2.73 3 18 volts

56
2.73 3 18 volts

57
5.73 18 volts

58
**Now, find the total current flowing**

5.73 18 volts

59
6 9.42volts 3 V=IR 6 3 V=(3.14)(3) 4 V=9.42 2 3.14 amp 12 18 volts

60
8.57volts 6 9.42volts 3 6 3 4 3.14 amps 2 12 18 volts

61
8.57volts 9.42volts 6 3 4 3.14 amps 2 12 18 volts

62
8.57volts 9.42volts 6 3 4 3.14 amps 2 12 18 volts

63
8.57volts 9.42volts 6 3 3.14 amps 5 18 volts

64
8.57volts 9.42volts 6 3 5 18 volts

65
8.57volts 9.42volts 6 1.43 amps 3 3.14 amps 5 1.71 amps 18 volts

66
**V=IR V=(1.71)(2) V=3.42volts 18-9.42 8.57volts 9.42volts 6 1.43 amps**

3 3.42Volts 4 3.14 amps 2 12 1.71 amps 1.71 amps V=IR V=(1.71)(2) V=3.42volts 18 volts

67
8.57volts 9.42volts 6 1.43 amps 5.15 volts 3 3.42Volts 4 3.14 amps 2 12 1.71 amps 1.71 amps 18 volts

68
**I=V/R I=5.15volts/12 I= 0.43 amps 18-9.42 8.57volts 9.42volts 6**

3 3.42Volts 4 3.14 amps 2 12 1.71 amps 1.71 amps 0.43 amps 18 volts

69
**Or… 1.71 amps – 0.43 = I=V/R 1.28 amps I=5.15volts/4 I= 1.28 amps**

8.57volts Or… 1.71 amps – 0.43 = 1.28 amps I=V/R I=5.15volts/4 I= 1.28 amps 9.42volts 6 5.15 volts 3 3.42Volts 4 3.14 amps 2 12 1.71 amps 1.71 amps 0.43 amps 18 volts

70
6 3 6 3 4 2 12 18 volts

71
6 3 6 3 4 2 12 18 volts

72
6 3 6 3 4 2 12 18 volts Q

73
**Given: R1=1; R2=2 ; R3=3 . Rank the bulbs according to their relative brightness**

74
**Given: R1=1; R2=2 ; R3=3 . Rank the bulbs according to their relative brightness**

15 R1 R2 R3 R1 > R2 > R3 R1 > R2 = R3 R1 = R2 > R3 R1 < R2 < R3 R1 = R2 = R3 Q

75
**If the four light bulbs in the figure below are identical, which circuit puts out more total light?**

Circuit II 1. I II Same

76
**If the four light bulbs in the figure below are identical, which circuit puts out more total light?**

Circuit II 1. I II Same The resistance of two light bulbs in parallel in smaller than that of two bulbs in series. Thus the current through the battery is greater for circuit I than for circuit II. Since the power dissipated is the product of current and voltage, it follows that more is dissipated in circuit I.

Similar presentations

© 2021 SlidePlayer.com Inc.

All rights reserved.

To make this website work, we log user data and share it with processors. To use this website, you must agree to our Privacy Policy, including cookie policy.

Ads by Google