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 Introductory Physics 120Hunter CollegeComplex DC Circuits ObjectivesStudy complex circuits using Kirchoff's laws .PreparationPlease review your textbook topics related to series/parallel circuits as well as Kirchoff’s laws. Expect 45 minutes-1 hour of prep time before coming to the lab. To help clarify the math involved in complex circuits please watch and take notes on the following suggested youtube web sites. Introduction to series/parallel circuits (7 minutes). a more complex circuit, see this (15 minutes). a summary of Kirchhoff's Rules see this ( 11 Minutes). one circuit application of Kirchhoff's Rules see this ( 8 minutes). free to skip portions you understand and/or explore other web sites for viewing.Series CircuitFor a series circuit such as this 2961005-33654296227519050Figure 1. Resistors in series296227519050V1 = IR1, V2 = IR2,V3 = IR3V = V1 + V2 + V3 = IR1 + IR2 + IR3 and V = IReq Req = R1 + R2 + R3Thus when we put several resistances in series, the total resistance is the sum of the separate n resistances.( n resistances in series ) Parallel Circuits476250238125Figure 2 Resistors in ParallelThe properties of the parallel circuit are as follows:I = I1 + I2 + I3 When the resistors are connected in parallel, each resistor experiences the same voltage drop. Hence the full voltage of the voltage source V is applied to each resistor, soV = V1 = V2 = V3 Thus IR1 = IR2 = IR3 and 1Req =1R1 +1R2+1R3 Extending this result to the case of several resistances of some number n, the equivalent resistance is ( n resistances in parallel ) Note that when two or more resistances are connected in parallel, the equivalent resistance Req is smaller than any of the combining resistances.Kirchhoff's RulesMore complex circuits are created when two or more voltage sources are included in a circuit with multiple resistors. Finding the potential drops and currents across each component requires the use of Kirchhoff's Rules.Loop rule: Kirchhoff’s Voltage LawThe algebraic sum of the changes in potential encountered in a complete traversal of any loop of a circuit must be zero.Suppose we start at any point in an ideal circuit, and mentally proceed around the circuit in either direction, adding algebraically the potential differences that we encounter. Before actually doing so, we shall formalize this idea in a statement that holds not only for single-loop circuits, but also for any complete loop in a multiloop circuit:This statement is often referred to as Kirchhoff’s loop rule (or Kirchhoff’s voltage law), after German physicist Gustav Robert Kirchhoff. That is, when we return to our starting point, we must also have returned to our starting potential. This then is the electrical reformulation of the Principle of Conservation of Energy. When finding potential differences as we move through a loop, we set down the following ideal rules:Resistance rule: For a move through a resistance R in the same direction of the conventional current, the change in potential is equivalent to -iR; in the opposite direction, is +iR.Voltage device rule: For a move through a voltage device in the direction of its potential (low to high), the change in potential is +V; in the opposite direction (high to low) it is -V 265874554610 When describing a current circuit by convention, a current arrow is drawn in the direction in which positive charge carriers would move, even if the actual charge carriers (electrons) are negative and move in the opposite direction. This convention will be followed in all figures unless otherwise described. Figure 3. A Multiloop circuitJunction rule: Kirchhoff’s Current LawSuppose we start at any point of one of the loops in an ideal multiloop circuit such as shown in Figure 3, and mentally proceed that loop until the current flows to a junction, where it can be split into branches that are connected at these junctions. We arbitrarily label the currents for each branch, and decide the directions of the currents are also assumed arbitrarily – they do not reflect the true direction of the current in that branch yet. Current I1 has the same value everywhere in the branch bad that contains the resistor R1. In a similar fashion, current I2 has the same value everywhere in the branch bcd that contains the resistor R3, andI3 is the current through the central branch bd that contains resistor R2. Consider junction d for a moment: Charge comes into that junction via incoming currents I1 and I3, and it leaves via outgoing current I2. Because there is no variation in the charge at the junction, the total incoming current must equal the total outgoing current. I1 + I3 = I2Applying this condition to junction b leads to exactly the same equation. Thus it suggests the following principle:The sum of the currents entering any junction must be equal to the sum of the currents leaving that junction.This statement is often called Kirchhoff’s junction rule (or Kirchhoff’s current law). It is simply a statement of the conservation of charge for a steady flow of charge— there is neither a buildup nor a depletion of charge at a junction.Using Kirchoff's Laws for Complex Circuit AnalysisGiven the resistances and potential differences for a complex circuit, such as the circuit of Figure 3, if the currents are all unknown, we can use Kirchoff's Laws to compute those values. The steps below generally apply to any complex DC circuit: 1. Label the current in each branch, where a branch is a segment between two junctions. Choose the direction for each labeled current by using an arrow. If the current in one of the branches is actually in the opposite direction, it will come out with a negative sign with the solution.2. Identify the unknowns. You will need as many independent equations as there are unknowns. You may write down more equations than this, but you will find that some of the equations are redundant. You may use V = IR for each resistor, which sometimes will reduce the number of unknowns.3. Apply Kirchoff's junction rule at one or more junctions.For the circuit of Figure 3, the equation at junction b or d is the same:I1 + I3 = I24. Apply Kirchoff's loop rule for one or more loops: follow each loop in one direction only. Pay careful attention to the labeled currents, voltage sources, their respective directions, and to signs (–/+). In the circuit of Figure 2, there are three loops to choose from. Which two loops we choose do not matter. the top loop (badb), clockwise from point b, the equation isV1 – I1R1 + I3R3 = 0the bottom loop (bcdb), counterclockwise from point b, the equation is– I2R2 – I3R3 – V2 = 0the big loop (badcb), clockwise from point b, the equation isV1 – I1R1 – I2R2 – V2 = 0(This equation is the sum of the above two loops).5. Solve the equations algebraically for the unknowns. Be careful when manipulating equations not to err with signs. At the end, check your answers by plugging them into the original equations or even by using any additional loop or junction rule equations not used previously.Now go to this Simulation. Navigate to the Circuit Construction Kit: DC Lab available here and launch the simulation. Select the “Lab” experiment to begin. 2. The experiment defaults to electrons for the Current setting. Select the Conventional setting instead. 1394335The checkbox for 'Values' can be check-marked to always display the values of each element of the circuit as they are placed in the blue workspace from the left-window pane. At the bottom-right of the simulation, the display of the circuit can be switched between the literal elements of the circuits (default) to the electronic symbols that correspond to each element based on traditional conventions.After placing a circuit element onto the workspace, two red-dashed circles will appear on each end of it. Dragging one of these red circles will rotate the element. To join circuit elements and create junctions, overlap one of the circles of an element to a circle of the other element(s): the overlap will change to a black-dashed circle with a smaller grey inner circle. 19526253162300Tap a circuit element to edit the value, delete it, or other options as they apply to each unique element, such as changing the voltage and orientation of potential for a battery, the resistance for a resistor or a light bulb, or deleting a wire. When including a switch element, tapping the switch will open or close it as desired. Finally, take note of the three diagnostic instruments available to you below the display settings: you will find a voltmeter and two different kinds of ammeters. The single voltmeter measures voltage by placement of the two leads across any part of a circuit, such an individual circuit element, or a group of circuit elements.Following conventional current, a voltmeter is arranged in parallel around a circuit element.334327538100the red lead is to be placed on a part of the circuit that flows from high potential,the black lead is to be placed on a part of the circuit that flows to low potential.The two kinds of ammeters available vary based on their usage. The single instantaneous ammeter on the left can measure the current at any particular point in a circuit. It does not require inclusion into the circuit.The ammeter on the right can measure current at a particular point or even a particular arrangement of various elements in a circuit, but does require inclusion into the circuit.Although it acts very much like a regular wire, the ammeter must be arranged in series. Unlike the other diagnostic tools, up to six ammeters of this type can be placed into a circuit as desired. Take some time to familiarize yourself with the simulation parameters by shifting the checkboxes of circuit display settings and placement of the circuit elements, as well as using the diagnostic instruments. When you feel that you have a good understanding of how to use the simulation, move on to the next section to begin data collection.3595370102870Procedure A3. Connect three resistances in series as illustrated in circuit of Figure 4. The letter A enclosed in a circle represents placement of an ammeter which should be included in the circuit. The letter V enclosed in a circle represents the arrangement of the voltmeter tool to measure the voltage drop across part of the circuit, such as the voltage drop across resistor R1. Edit the voltage of the battery to 75 V and edit the values of the resistors to the respective values: R1 = 50 Ω, R2 = 25 Ω, and R3 = 10 Ω.3340100165100Figure 4. Series circuit arrangement3340100165100The circuit should look like this4. Using the voltmeter tool, measure the voltage drop across each resistor. Then measure the voltage drop across all three resistors for the total circuit.5. Record the total current of the circuit that is passed into the resistors. Then rearrange the circuit with additional ammeters:one ammeter placed between R1 and R2, one ammeter placed between R2 and R3, andan additional ammeter between R3 and the low potential end of the battery. Measure and record the currents in the circuit as the current flows through each resistor. Calculate and record power across each resistor and for the circuit.SERIES123TotalR ohms502510I ampsV voltsPower wattsProcedure B7. Connect three resistances in parallel as illustrated in the circuit of Figure 5. Edit the voltage of the battery to 75 V and edit the resistors to the following values: R1 = 10 Ω, R2 = 25 Ω, and R3 = 50 Ω. Arrange them as shown below.Note that voltage is read across each resistor and current is recorded when the ammeter is in series with the resistor in each branch as well as to the right of the battery.8. Using the voltmeter tool, measure the voltage drop across all three resistors at the junctions. Then measure the voltage drop across each resistor.9. Record the current of the circuit that is passed into the resistor R1. Then rearrange the circuit with additional ammeters: like the arrangement for resistorR1, place one after resistor R2, as well as one ammeter directly after resistor R3,one ammeter directly before the junction of incoming current, andan additional ammeter directly after the junction of outgoing current.Measure and record the currents in the circuit as the current flows through each part of the circuit. Calculate and record power across each resistor and for the circuit.PARALLEL123TotalR ohms502510I ampsV voltsPower wattsProcedure C11. Connect two resistances in parallel and a third resistance in series as illustrated in the circuit of Figure 6. Edit the voltage of the battery to 75 V. Readjust the resistors to to the following values: R1 = 50 Ω, R2 = 25 Ω, R3 = 10 Ω.297180012382512. Using the voltmeter tool, measure the voltage drop across each resistor. Then measure the voltage drop across all three resistors for the total circuit.2857500876300Figure 6. Series-parallel circuit arrangement2857500876300It should look like this13. Record the total current of the circuit that is passed into the resistors. Then rearrange the circuit with additional ammeters:one ammeter directly after resistor R1,one ammeter directly after resistor R2,one ammeter directly after resistor R3, andan additional ammeter directly after the junction of outgoing current towards resistor R3.Measure and record the currents in the circuit as the current flows through each part of the circuit. Calculate and record power across each resistor and for the circuit.SERIES/PARALLEL123TotalR ohms502510I ampsV voltsPower wattsCalculations and ConclusionsNow for each of the three circuits calculate equivalent values of resistance, current voltage and power. SHOW YOUR WORK! How well do your results match the equivalent values you obtained with the ammeter and voltmeter? Explain any discrepancies. G. Kirchhoff Circuits:For the given complex circuits below, use Kirchoff's Laws to write the corresponding junction equations, and show the sum of all voltages around the circuit is zero for all 3 possible loops.Let R1= 3 ohms, R2= 16 ohms and R3 = 8 ohms, V1 = 24 volts, V2 = 12 V.Calculate the values of I1, I2 and I3 . SHOW ALL YOUR WORK NEATLY!Now using the simulation , construct this circuit using the resistance and values given. Including the ammeters ,in the circuit show that the values measured match the calculated values of current. You must include a screenshot of the current you created. Finally In this last one R1 = 3 ohms, R 2 = 3 ohms and R3 =30 ohms V1=12 Volts and V2 = 6 Volts Find the current in each of the branches. SHOW YOUR WORK!Also provide a screenshot of the circuit. ................
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