This week Electrical Circuits Series or parallel that s the question. Current, Power and Energy Why does my laptop battery die? Transmission of power to your home Why do we have big transmission towers? Household Appliances Why do fuses blow? 6/2/2015 Physics 214 Summer 2015 1
Electrical power and energy In order to separate charge we have to do work and energy is stored. The simplest example is a battery. The quantity we call voltage is related to the energy stored V = PE/q (joules/coulomb = volt) When positive charge moves to the negatively charged plate or vice versa then the stored energy is released. In order for this to happen the two sides of the battery must be joined by a conductor so that the charge can move easily. This is a simple circuit. 6/2/2015 Physics 214 Summer 2015 2
Electrical circuits All circuits are basically the same. There is an external source of energy which produces a voltage. In a charged battery there is separation of charge. When the circuit is made positive charge flows from high to low voltage or negative charge flows from low to high voltage releasing the stored energy. Normally it is electrons which flow. As they move they collide with the atoms of the wire and lose some of their energy in the form of heat. There is resistance to the flow. 6/2/2015 Physics 214 Summer 2015 3
Current and resistance The rate of flow of charge determines how much energy is released per unit time I = q/t amperes (coulombs/sec) The direction of I is the flow of positive charge or opposite to the flow of negative charge. OHM s Law R = V/I ohms Every part of a circuit has resistance including an internal resistance in the battery. The higher the resistance the lower is the current for a given voltage difference 6/2/2015 Physics 214 Summer 2015 4
Electromotive force The electromotive force of a battery, ε, is the voltage difference between the two terminals when no current is being drawn. When it is connected to a simple circuit I = ε/(r circuit + R battery ) A voltage difference is the energy stored or the energy released per unit charge. So if charge +q goes from high to low voltage then the energy released is qε or q V 6/2/2015 Physics 214 Summer 2015 5
Series circuit If we add more light bulbs in the circuit in series the total resistance increases and the current will be reduced. The current is the same in all parts of the circuit and the voltage difference across each light is the same ε = I(R battery + R + R + R) The voltage difference across any light = IR 6/2/2015 Physics 214 Summer 2015 6
Parallel circuit The current q at A divides into three and then recombines at B In the circuit shown the voltage difference across each light is the same and the total current is the sum of the three currents I = I 1 + I 2 + I 3 and since a current = V/R V AB /R circuit = V AB /R 1 + V AB /R 2 + V AB /R 3 and 1/R circuit = 1/R 1 + 1/R 2 + 1/R 3 That is the total resistance of the circuit is smaller than any of the single resistances. It is also true that if one bulb fails the other two will stay lit at the same brightness. 6/2/2015 Physics 214 Summer 2015 7
What is voltage? We have seen that the definition of voltage is V = PE/q when a charge q is moved in an electrical force field. So energy is stored as potential energy as a positive charge is moved in the opposite direction to E or a negative charge is moved in the same direction as E. If we move a positive charge toward a positive charge potential energy and V increase or if we move a negative charge away from a positive charge. Just as in the gravitational field there are only differences in PE. So normally we use the term V. But very often for circuits we choose one point, usually the negative terminal, to be zero and then instead of V we just use V. When charge is free to move, that is positive charge moving to a lower voltage or negative charge moving to a higher voltage the PE will transform into KE just like dropping something. In a simple circuit with resistance this KE is turned into heat and light so there is a voltage drop across every element in the circuit. 6/2/2015 Physics 214 Summer 2015 8
Voltage drop If we have a circuit with many different resistors then there is a voltage drop across each resistor and there is also a voltage drop for the whole circuit. Current only flows if there is a voltage difference and in a time t charge q passes through the resistor. I = q/t and V = IR Case1 R = 15+20+25 = 60 I = 6/60 = 1/10 V 15 = 15/10 V 20 = 20/10 V 25 = 25/10 Case 2 1/R = 1/24 +1/24 +1/24 R = 8 I = 12/8 I 24 = 12/24 Case 3 R = 1.5 + 3 + 1 = 5.5 I = V AB /5.5 V 31 = 3I/2 V 3 = 3I V 33 = 3I/3 6/2/2015 Physics 214 Summer 2015 9
Series plus parallel A V I R 1 R 2 R 3 A B V B R parallel I R 4 B R 4 V AB = I 1 R 1 V AB = I 2 R 2 V AB = I 3 R 3 I = I 1 + I 2 + I 3 V 4 = IR 4 V= V AB + V 4 V = I(R 4 + R parallel ) 6/2/2015 Physics 214 Summer 2015 10
Measuring current and Voltage It is often very important to know the current in a circuit or the voltage difference between two points. A hand held meter is very useful to test batteries or a circuit. An ammeter is a device inserted into a circuit. The resistance of an ammeter is very small so as to minimize the effect on the circuit. A voltmeter is attached in parallel and V is found by measuring the current and V = I meter R meter. The resistance has to be much larger than the circuit resistance so that the current is very small and does not disturb the main circuit. Both meters actually measure current 6/2/2015 Physics 214 Summer 2015 11
Power Electromotive force or voltage difference between two points is the difference in potential energy/unit charge. So the energy delivered if charge q is transferred is energy = Vq power = Vq/t = VI watts For any voltage difference V and current I. power = VI For circuits that obey OHM s law V = I R P = VI = I 2 R = V 2 /R watts The power used appears as heat or light Practical Unit 1 kilowatt Energy Unit 1 kilowatt hour = 1000 x 3600 joules 6/2/2015 Physics 214 Summer 2015 12
Summary Chapter 13 I = q/t amperes (coulombs/sec) OHM s Law R = V/I ohms I = ε/(r circuit + R battery ) P = VI = I 2 R = ( V) 2 /R watts 6/2/2015 Physics 214 Summer 2015 13
Transmission In the distribution of electric power the goal is to deliver to the user as large a fraction as possible of the generated power. Practical cables have a specific resistance so the power losses will be I 2 R cable and we need I to be as small as possible. But we also need the delivered power P = V source I source to be as high as possible, therefore, the electrical power is distributed at very high voltage and low current. The voltage is reduced from 250,000volts to 220volts for households by using a transformer. The current increases by the same factor since for an ideal transformer no power is lost. Transformers are the dominant reason electrical transmission is alternating current i V source R cable i V user R user V user = V source IR cable P user = iv user = iv source i 2 R cable 6/2/2015 Physics 214 Summer 2015 14
Household appliances Household circuits are wired in parallel so that when more than one appliance is plugged in each sees the same voltage and can get the required current. As we plug in more and more appliances the current in the circuit increases and the I 2 R losses could cause a fire. This is why we have fuses and why major appliances use 220 volts and many parts of the world use 220 volts for all household use. As many people turn on appliances (air conditioners) the grid has to supply more power by increasing the current. P user = iv user = iv source i 2 R cable This results in a higher fraction of the power being lost in the cable In cases of very heavy load the power station reduces the transmission voltage resulting in a brown out and in extreme cases there are rolling blackouts. 6/2/2015 Physics 214 Summer 2015 15
Mini Review In order for a current to flow there must be a voltage difference. The voltage difference is the amount of potential energy/unit charge. As the charge, usually electrons move along a conductor they lose energy continuously so the voltage along the conductor decreases continuously. For a simple circuit we ignore losses in the conductor but there is a voltage drop across every resistor. V V I = q/t Energy liberated is qv. So the power is qv/t = IV For a resistor V = IR and power = I 2 R = VI = V 2 /R 6/2/2015 Physics 214 Summer 2015 16
Simple circuits The current is the same everywhere R = R 1 +R 2 + R 3 V = V 1 + V 2 +V 3 V V is the same across all resistors V = IR circuit 1/R circuit = 1/R 1 + 1/R 2 + 1/R 3 I 1 = V/R 1 I 2 = V/R 2 I 3 = V/R 3 R = R 1 + R 2 + R 3 V ab = IR R 1 R 2 R 3 6/2/2015 Physics 214 Summer 2015 17
Questions Chapter 13 Q3 In a simple battery-and-bulb circuit, is the electric current that enters the bulb on the side nearer to the positive terminal of the battery larger than the current that leaves the bulb on the opposite side? The current is the same Q4 Are electric current and electric charge the same thing? No. electric current is the flow of charge I = q/t 6/2/2015 Physics 214 Summer 2015 18
Q6 Consider the circuit shown. Could we increase the brightness of the bulb by connecting a wire between points A and B? Explain. A Є B No. The voltage drop from A to B is stiil the same so the current through the bulb does not change. 6/2/2015 Physics 214 Summer 2015 19
Q7 Two circuit diagrams are shown. Which one, if either, will cause the light bulb to light? Explain you analysis of each case. Open Switch 1.5 V 1.5 V (a) (b) In a) the battery is not connected. In b) the bulb is lit 6/2/2015 Physics 214 Summer 2015 20
Q11 A dead battery will still indicate a voltage when a good voltmeter is connected across the terminals. Can the battery still be used to light a bulb? A battery has internal resistance so although one can measure an open circuit voltage when connected to a circuit the voltage will drop and the current flow will be very low. Q12 When a battery is being used in a circuit, will the voltage across its terminals be less than that measured when there is no current being drawn from the battery? Explain. The voltage will be less because of the voltage drop due to the internal resistance 6/2/2015 Physics 214 Summer 2015 21
Q13 Two resistors are connected in a series with a battery as shown in the diagram. R 1 is less than R 2. A. Which of the two resistors, if either, has the greater current flowing through it? Explain. B. Which of the two resistors, if either, has the greatest voltage difference across it? Explain. R 1 Є R 2 The current is the same in both. Since V = IR the greatest voltage drop will be across R 2 6/2/2015 Physics 214 Summer 2015 22
Q14 In the circuit shown below, R 1, R 2, and R 3 are three resistors of different values. R 3 is greater than R 2, and R 2 is greater than R 1. Є is the electromotive force of the battery whose internal resistance is negligible. Which of the three resistors has the greatest current flowing through it? Є R 1 R 2 R 3 I 3 = I 1 + I 2 so I 3 is the largest 6/2/2015 Physics 214 Summer 2015 23
Q15 In the circuit shown in question 14, which of the three resistors, if any, has the largest voltage difference across it? V = IR and both I and R are the largest for R 3 Q16 If we disconnect R 2 from the rest of the circuit shown in the diagram for question 14, will the current through R 3 increase, decrease, or remain the same? The resistance of the circuit will increase so the current through R 3 will decrease. Є R 1 R 2 6/2/2015 Physics 214 Summer 2015 24 R 3
Q18 In the circuit shown, the circle with a V in it represents a voltmeter. Which of the following statements is correct? Explain. A. The voltmeter is in the correct position for measuring the voltage difference across R. B. No current will flow through the meter, so it will have no effect. C. The meter will draw a large current. Є + + R V The voltmeter is in the correct position. Current will flow through the meter but the current will be very small because the resistance is very high. 6/2/2015 Physics 214 Summer 2015 25
Q19 In the circuit shown, the circle with an A in it represents an ammeter. Which of these statements is correct? Comment on each. A. The meter is in the correct position for measuring the current through R. B. No current will flow through the meter, so it will have no effect. C. The meter will draw a significant current from the battery. Є + + R A The meter is in the wrong position. A large current will flow because the meter resistance is very low. 6/2/2015 Physics 214 Summer 2015 26
Q21 Is electric energy the same as electric power? Power is the rate at which energy is used. Your electrical bill is for the total energy you use. Q22 If the current through a certain resistance is doubled, does the power dissipated in that resistor also double? P = I 2 R so the power increases by a factor of 4 Q23 Does the power being delivered by a battery depend on the resistance of the circuit connected to the battery? Yes because increasing the resistance lowers the current 6/2/2015 Physics 214 Summer 2015 27
Q29 Suppose that the appliances connected to a household circuit were connected in series rather than in parallel. What disadvantages would there be to this arrangement? First if one failed then all the appliances will not function. Secondly the current will be determined by the whole string of appliances by the voltage drop from end to end and will depend on the number of appliances. Each appliance requires the same voltage drop and this is what happens when they are connected in parallel. 6/2/2015 Physics 214 Summer 2015 28
Ch 13 E 14 24 resistor has voltage difference 3V across leads. a) What is the current through the resistor? b) What is the power dissipated in resistor? 3V 24 a) V = IR I = V/R = 3/24 = 0.125A b) P = IV = V 2 /R = (3) 2 /24 = 0.375W 6/2/2015 Physics 214 Summer 2015 29
Ch 13 E 16 A toaster draws current = 7A on a 110-V AC line a) What is the power consumption of the toaster? b) What is the resistance of the heating element in the toaster? a) P = IV = 7.110 = 770 W b) V = IR, R = V/I = 110/7 = 15.7 110 V I = 7A R 6/2/2015 Physics 214 Summer 2015 30
Ch 13 CP 2 Three 30- light bulbs connected in PARALLEL to 1.5 V battery with negligible internal resistance. a) What is the current through the battery? b) What is the current through each bulb? c) If one bulb burns out, does the brightness of the other bulbs change? 1.5 V R R R R = 30 6/2/2015 Physics 214 Summer 2015 31
Ch 13 CP2 cont. a) 1/R p = 1/R 1 + 1/R 2 + 1/R 3 = 1/30 + 1/30 + 1/30 = 1/10 R p = 10 V = I t R p, I t = V/R p = 1.5/10 = 0.15 A b) V = IR, I = V/R = 1.5/30 = 0.05 A Notice that total current, I t, through the battery is the sum of currents through each bulb. I t = 3(I) = 0.15 A c) Brightness of remaining two bulbs do not change. Instead, load on the battery is reduced. Each remaining bulb still feels 1.5 V. So, each remaining bulb still draws 0.05 A of current. Since, P=IV, each remaining bulb still outputs same power. This is benefit of hooking circuits in parallel. 6/2/2015 Physics 214 Summer 2015 32
CH 13 CP 4 A R R R R R R B R = 3 a) What is the resistance of each parallel combination? b) What is the total resistance between A and B? c) 6V voltage difference b/w A and B. What is the current through the entire circuit? d) What is the current through each resistor in three-resistor parallel combination? a) Two-resistor parallel combination I/R P2 = 1/R + 1/R = 1/3 + 1/3 = 2/3, R P2 = 3/2 b) Three resistor parallel combination I/R P3 = 1/R + 1/R + 1/R = 1/3 + 1/3 + 1/3 = 1, R P3 = 1 6/2/2015 Physics 214 Summer 2015 33
CH 13 CP 4 cont A R R R C R R R B R = 3 A 3/2 3 C 1 B A 9/2 C 1 B c) V = I R T, I = V/R T = 6/(11/2) = 12/11 = 1.09 A 6/2/2015 Physics 214 Summer 2015 34
CH 13 CP 4 After accounting for AC part of circuit we have left C d) V AC = I R AC = (1.09) (9/2) = 4.91 V V CB = V V AC = 1.09 V I = V/R = 1.09/3 = 0.37 A V=1.09 V R R R B This answer could have been easily noted as follows: We calculated 1.09 A flowing from A to B. That means, at point C, there is 1.09 A. Now this current must make it to point B, but there are 3 different paths. Since each path is of equal resistance, the current will equally choose all three paths. thus any one path has 1.09/3 A = 0.37 A of current 6/2/2015 Physics 214 Summer 2015 35