# Physics 214 Spring

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1 Lecture 23 March The elation between Voltage Differences V and Voltages V? Current Flow, Voltage Drop on esistors and Equivalent esistance Case 1: Series esistor Combination and esulting Currents Case 2: Parallel esistor Combination and esulting Currents Case 3: Mixed Series and Parallel esistor Combination, Corresponding Equivalent esistance and esulting Currents Current, Power and Energy Transmission of Power to Your Home Household Appliances Questions 3/3/2016 Physics 214 Spring

2 3/3/2016 Physics 214 Spring What is the elation between Voltage V and Voltage Difference V? We have seen that the definition of voltage is ΔV = Δpotential energy/q when a charge q is moved in an electrical force field. Δis the notation for difference of some quantity. 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 ΔV 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 ΔV. But very often for circuits we choose one point, usually the negative terminal, to be zero and then instead of ΔV 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.

3 Voltage drop, Current and Equivalent esitance 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. Case1 I = q/t and ΔV = I 6V 1 V I 0.1A V A, V A, V A Case 2 12V I24, Itotal 3 I A 1.5A 24 Case 3 3/3/2016 total , total VAB , I 5.5 Physics 214 Spring 2016 V V V VAB 3 3V I VAB 3 VAB I VAB 3V I AB AB 3

4 Series plus parallel A I 1 V I I 4 I 1 I 2 I 3 A B I 2 I 3 B V B I 4 I parallel V AB = I 1 1 V AB = I 2 2 V AB = I 3 3 I = I 1 + I 2 + I 3 V 4 = I 4 V = V AB + V 4 V = I( 4 + parallel ) 3/3/2016 Physics 214 Spring

5 Simple circuits The current is the same everywhere = V = V 1 + V 2 +V 3 V V is the same across all resistors V = I circuit 1/ circuit = 1/ 1 + 1/ 2 + 1/ 3 I 1 = V/ 1 I 2 = V/ 2 I 3 = V/ 3 = V ab = I /3/2016 Physics 214 Spring

6 Power and Energy 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 = ΔVq power = ΔVq/t = ΔVI watts Eq.1 Eq.2 For any voltage difference ΔV and current I, the delivered power is power = ΔVI Eq.3 For circuits that obey OHM s law ΔV = I we substitute for ΔV in Eq.3, we have Power = ΔVI = I 2 = ΔV 2 / watts The power used appears as heat or light Practical Unit for power 1 kilowatt Energy Unit 1 kilowatt hour = 1000 x 3600 joules 3/3/2016 Physics 214 Spring

7 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 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. 3/3/2016 Physics 214 Spring

8 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 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 cable i V user user V user = V source I cable P user = iv user = iv source i 2 cable 3/3/2016 Physics 214 Spring

9 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 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 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. 3/3/2016 Physics 214 Spring

10 Summary Chapter 13 I = q/t amperes (coulombs/sec) OHM s Law = ΔV/I ohms I = ε/( circuit + battery ) P = ΔVI = I 2 = (ΔV) 2 / watts 3/3/2016 Physics 214 Spring

11 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 3/3/2016 Physics 214 Spring

12 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. 3/3/2016 Physics 214 Spring

13 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) In a) the battery is not connected. In b) the bulb is lit (b) 3/3/2016 Physics 214 Spring

14 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 3/3/2016 Physics 214 Spring

15 Q13 Two resistors are connected in a series with a battery as shown in the diagram. 1 is less than 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. 1 Є 2 The current is the same in both. Since V = I the greatest voltage drop will be across 2 3/3/2016 Physics 214 Spring

16 Q14 In the circuit shown below, 1, 2, and 3 are three resistors of different values. 3 is greater than 2, and 2 is greater than 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 I 3 = I 1 + I 2 so I 3 is the largest 3/3/2016 Physics 214 Spring

17 Q15 In the circuit shown in question 14, which of the three resistors, if any, has the largest voltage difference across it? V = I and both I and are the largest for 3 Q16 If we disconnect 2 from the rest of the circuit shown in the diagram for question 14, will the current through 3 increase, decrease, or remain the same? The resistance of the circuit will increase so the current through 3 will decrease. Є 1 2 3/3/2016 Physics 214 Spring

18 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. B. No current will flow through the meter, so it will have no effect. C. The meter will draw a large current. Є + + 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. 3/3/2016 Physics 214 Spring

19 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. 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 + + A The meter is in the wrong position. A large current will flow because the meter resistance is very low. 3/3/2016 Physics 214 Spring

20 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 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 3/3/2016 Physics 214 Spring

21 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. 3/3/2016 Physics 214 Spring

22 Ch 13 E 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 = I I = V/ = 3/24 = 0.125A b) P = IV = V 2 / = (3) 2 /24 = 0.375W 3/3/2016 Physics 214 Spring

23 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 = = 770 W b) V = I, = V/I = 110/7 = V I = 7A 3/3/2016 Physics 214 Spring

24 Ch 13 CP 2 Three 30- light bulbs connected in PAALLEL 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 = 30 3/3/2016 Physics 214 Spring

25 Ch 13 CP2 cont. a) 1/ p = 1/ 1 + 1/ 2 + 1/ 3 = 1/30 + 1/30 + 1/30 = 1/10 p = 10 V = I t p, I t = V/ p = 1.5/10 = 0.15 A b) V = I, I = V/ = 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. 3/3/2016 Physics 214 Spring

26 CH 13 CP 4 A B = 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/ P2 = 1/ + 1/ = 1/3 + 1/3 = 2/3, P2 = 3/2 b) Three resistor parallel combination I/ P3 = 1/ + 1/ + 1/ = 1/3 + 1/3 + 1/3 = 1, P3 = 1 3/3/2016 Physics 214 Spring

27 CH 13 CP 4 cont A C B = 3 A 3/2 3 C 1 B A 9/2 C 1 B c) V = I T, I = V/ T = 6/(11/2) = 12/11 = 1.09 A 3/3/2016 Physics 214 Spring

28 CH 13 CP 4 After accounting for AC part of circuit we have left C d) V AC = I AC = (1.09) (9/2) = 4.91 V V CB = V V AC = 1.09 V I = V/ = 1.09/3 = 0.37 A V=1.09 V 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 3/3/2016 Physics 214 Spring

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