Series Circuits and Kirchoff s Voltage Law

Transcription

1 ELEN 236 Series and Parallel Circuits Series Circuits and Kirchoff s Voltage Law Reference All About Circuits->DC->Chapter 5 and Chapter 6 Questions: CurrentVoltageResistance: Qs: (70-85, 86-99, ) Series Circuit Definition Circuits (or portion of a circuit) where components are connected back to back o Share one point Question to get you to start identifying distinguishing characteristics of series circuits In a series circuit, the voltage supplied by the source(s) is all used up by the load(s) (Voltage gains = Voltage drops) In a series circuit, the current is equal through all of the components In a series circuit, the individual resistances add up to the total resistance In a series circuit, the power individual power dissipations equal the total power supplied 1

2 e.g.: Build a circuit using the PHeT CCK. How Would I prove each of the above points? o measure current at all points. Should all be the same o measure voltage across each component. Should sum to zero o Calculate powers for each component. Should sum to zero o Calculate total resistance. Calculate total current. Calculate current for each individual component. Another way to state each of the above rules: 1) Sum of Vrises = Sum of Vdrops 2) Itotal = I1 = I2 = I3 3) Rtotal = R1 + R2 + R3 These statements are true no matter what is in the circuit. e.g.: Compare the direction of current through all components in this circuit with the polarities of their respective voltage drops. What do you notice about the relationship between current direction and voltage polarity for the battery, versus for all the resistors? How does this relate to the identification of these components as either sources or loads? 2

3 Calculate Voltage Across each of the resistors in this circuit (Magnitude and polarity). Also show the direction of current: o Calculate power generated by source o Calculate power dissipated by each component Steps: o Find Total R of circuit o Calculate direction and magnitude of current (I = V/RT) o Calculate polarity and each VR (VR = I*R) o Calculate power delivered by source (Vsrc*I) o Calculate power dissipated by each component (VR*I) o Vsrc*I should equal sum of individual VR*I Example 14 V 780 Ω R 1 R 2 R kω 1.5 kω V I R P R 1 R 2 R 3 Total 780 Ω 1.5 kω 3.3 kω (01957) 3

4 R1 R2 R3 Total V V 8.280V 14V I 2.509mA 2.509mA 2.509mA 2.509mA R 780Ω 1.5kΩ 3.3kΩ 5.58kΩ P 4.910mW 9.442mW 20.77mW 35.13mW What are some ways that you could check your work? Without performing any mathematical calculations, determine the effects on all the component voltage drops and currents if resistor R1 were to fail open. Without performing any mathematical calculations, determine the effects on all the component voltage drops and currents if resistor R1 were to fail shorted. Example showing the effect of changing resistance in a series circuit: SW1 SW2 SW3 (00295) What will happen in this circuit as the switches are sequentially closed starting with #1 and working to #3? Describe how the successive closures will affect o Total Circuit Resistance seen by the battery o Total amount of current drawn from the battery o The current through each resistor o The voltage drop across each resistor Example: Designing a series circuit using LED. LED description: rugged, efficient sources of light Not resistive when turned on the drop a (nearly) constant voltage Naïve LED circuit: What can be put in to the circuit to limit the current. o A resistor A typical LED current is 20mA. Design LED circuit to give you 20mA through LED which has a forward voltage drop of 2Volts. 4

5 Kirchoff s Voltage Law The concept that around any electrical loop the sum of the voltage rises equals the sum of the voltage drops is formalized in Kirchoff s Voltage Law: Another way to put this is that if you start at one point on the circuit and go around the circuit, everytime you go up in voltage, add the voltage, and everytime you go down in voltage you subtract it. o When you return back to where you started, your total should be at zero o Analogous to going for a hike in the mountains and returning to the place you started. During your hike, you will sometimes be climbing upwards, sometimes be climbing downwards, and sometimes walking on the flat. When you return to the start your net elevation gain is 0. o It is basically a conservation of energy equation. It takes energy to boost the voltage and things that dissipate energy drop voltage. Single Loop Examples: Determine the voltage polarity and magnitude of each component in these loops and show that the sum of the voltages in the loop is equal to zero. 5

6 Two Loop Introduction In this case, relative to the source, R2 is in parallel to R3 and R4. We will look at how to solve for the voltages in this type of circuit soon (after we look at parallel circuits). Voltage Dividers One of the reasons for having a resistor that was mentioned in the section on resistors is to split a voltage into proportions. In other words, if you have a voltage from a source, you may want to obtain some fraction of this voltage for a particular purpose (e.g. a Voltage Reference). Another example is in a sensor system as shown here: o this example is a light sensor system. o In this system, if the light level falls below a certain reference an LED is turned on o The circuit looks like this: 6

7 o LDR: Light dependent resistor. The resistance of an LDR varies depending on how much light falls on it. More light means lower resistance o Comparator: an electronic device that outputs a high voltage if the + input is at a higher voltage than the - input. Other wise it outputs a low voltage. In this case assume high voltage means 5 volts and low voltage means 0 volts. o In this circuit are two examples of series circuits where resistors are used to provide proportion of the total (source) voltage to some other part of the overall circuit. o In this case the two Voltage Divider circuits are o The circuit R1 and the LDR in this case the Voltage division is variable because the resistance of the LDR is variable o R2 and R3 in this case the Voltage Division is fixed since R2 and R3 are fixed o Whenever a voltage source is applied across a number of resistors in series, we know that the sum of the voltage drops will be equal to the voltage source o If R2 is 2 kω and R3 is 1 kω. Calculate the voltage that will be applied to the - terminal of the comparator Method 1 Method 2 7

8 o If the light falling on the LDR gave it a resistance of 11kΩ and R1 is 10kΩ. Would the LED be on or off? (Remember if the voltage to the + terminal is higher than the voltage to the - terminal then the comparator output would be 5 volts) Voltage Divider circuits (or voltage dividers ) can consist of more than one resistor: 8

9 A Potentiometer Voltage Divider Referring back to the light sensor circuit, how could you use a potentiometer to make the reference voltage (trigger voltage) that turns the LED on variable? Some More Example Problems 9

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11 Parallel Circuits and Kirchoff s Current Law (KCL) Reference All About Circuits->DC->Chapter 5 and Chapter 6 Questions: CurrentVoltageResistance: Qs: (11-122, , ) Parallel Circuit Definition Circuits (or portion of a circuit) where components share two points- Question to get you to start identifying distinguishing characteristics of parallel circuits A B C D F E 11

12 The main things to keep in mind related to parallel circuits 1) In a parallel circuit, the voltage drop across each of the parallel paths is equal 2) The amount of current flowing in to a node is equal to the amount of current flowing out of the node o A node is simply a junction in a circuit where two or more paths connect o e.g 3) In a parallel circuit, the inverse of the total resistance across the connecting nodes in the circuit is equal to the sum of the inverses of the resistances of each of the individual paths e.g.: Build a circuit using the PHeT CCK. How Would I prove each of the above points? o measure voltage across each component. Should all be the same o measure current in out of each node and current in to each node. Currents in should be equal to currents out o Calculate powers for each component. Should sum to zero (assuming sources are positive and loads are negative) o Calculate total resistance. Calculate total current. Calculate current for each individual component. Proof for point 3 above: 12

13 Some Examples: Qualitatively compare the currents through and voltages across each of these lamps: 6 volt bulb 6 volt bulb 6 volt bulb volt battery According to Ohm s Law, how much current goes through each of the resistors in this circuit: Trace the path of current in the circuit. 13

14 Calculate the total amount of current supplied by the voltage source: Then calculate the total resistance that the voltage source must see in order to create that much current. Explain why the total resistance is less than either of the individual resistances. In this circuit, calculate the current through each resistor. Then use P=VI to calculate the power dissipated by each resistor and the power generated by the voltage source. Do the same thing for this circuit: 14

15 Note that the lowest valued resistor dissipates more power. This may seem paradoxical, but the lower the value of the resistor, the larger the load that it will place on the source. Another way to look at this is that a heavy load draws more current, since for the same voltage source, a lower resistance will draw more current, a lower resistance represents a heavier load. What will happen to the brightness of the bulb when the switch is closed: Calculate the total resistance between A and B: Without doing any calculations, which would have more effect on RAB, adding a fourth parallel resistor with R=1Ω or R=1MI Ω? Suppose you needed to create a resistance of precisely 197Ω for a precision measurement device, but all you had were a 1000 Ω resistor and a rheostat 15

16 (variable resistor) with a resistance range of Ω. How would you create your precise resistance? What is wrong with the measurements in this circuit? 4 A + A - 3 A + A - 5 A + A - Kirchoff s Current Law Pretty much already covered: CurrentInt othenode CurrentsOutOfTheNode Another way to look at KCL: What goes in must come out! 16

17 Series and Parallel Circuits o Generally a circuit is not going to be only a series circuit or only a parallel circuit. It will usually be a combination of some parts of the circuit in series and some part of the circuit in parallel. o to analyze such circuits, just keep all of the rules about circuits in mind: 1) Elements in series will have the same current going through them 2) Elements in parallel will have the same voltage drop across them. 3) In a circuit loop, the voltage rises will equal the voltage drops (KVL) 4) At a node, the currents in will equal the currents out (KCL) Identify which of these components are connected directly in series with each other and which are connected directly in parallel with each other: Figure 1 Figure 2 Figure 3 R1 R1 R1 R2 R3 R2 R3 R2 R3 Figure 4 Figure 5 Figure 6 R1 R1 R3 R1 R3 R2 R4 R2 R4 R2 R3 17

18 Figure 1 Figure 2 Figure 3 SW1 R2 R4 C1 C1 R1 R1 R3 R2 L1 R1 Figure 4 Figure 5 Figure 6 C1 X1 L1 R2 R4 R2 L1 L2 C1 R1 R3 R1 Some equivalent resistance questions: For these questions, just go through and combine series R s into one equivalent R and parallel R s in to one equivalent R. Continue doing this until you have one equivalent R left. 18

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20 Rank these 5 light bulb circuits in order from least resistance to greatest resistance: A D B C E (00039) Calculate the resistance between points A and B (RAB) for each of the following resistor networks: Class does these ones: 20

21 Figure 1 Figure 2 Figure 3 All resistors 500 Ω All resistors 1 kω B A A 2 kω 5 kω A B 100 Ω 470 Ω B Figure 4 Figure 5 Figure Ω All resistors 2.2 kω B A A 220 Ω 470 Ω 470 Ω 100 Ω A 940 Ω B 330 Ω B (01757) 1: 500 Ω 2: 750 Ω 3: 1.511k Ω 4: 940 Ω 5: 880 Ω 6: 80.54Ω Think about how you approached each of these problems, and let the entire class participate in the reasoning process. The point of this question, like most of the questions in the Socratic Electronics, is not merely to obtain the correct answers, but to stimulate understanding of how to solve problems such as these. Complete the table of values for this circuit: 2.2 kω 6.8 kω 1 kω R 1 R Ω R 2 R 4 6 volts V I R P R 1 R 2 R 3 R 4 Total 1 kω 2.2 kω 470 Ω 6.8 kω (01756) 21

22 V I R P R 1 R 2 R 3 R 4 Total V V 725 mv 725 mv ma ma ma 107 μa 1 kω 2.2 kω 470 Ω 6.8 kω 6 V ma 3.64 kω mw mw mw μw mw Think about and outline your problem-solving strategies for this kind of problem. I may give you an opportunity to explain your methods and to see other methods of solution. An especially good point to think about how would you go about checking your work to see if any mistakes were made. Practice Problems: 149, 151, 152, 154, 156, 158, 159, 163,