LAB 2 - Systems (Circuits)

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LAB # 2 SYSTEMS OF EQUATIONS (Circuits) ENGR 190 1 Terry W. Armstrong, PhD (2021) Safety Requirements Instructor/TA must confirm power sufficiency of resistors Power (i 2 R) must be less than 0.25 for 1/4W resistors Equipment Requirements Electrical circuit set-up with the following: - Breadboard - Three jumper wires - 9v battery - Two resistors in parallel - One resistor in series Multimeter Extra multimeter fuses Calculator Alligator clips Procedural Requirements Work in teams of 3 together to - Complete the lab activities - Complete Appendix A - Complete required calculations and hand-written conclusions Each student individually submits a copy of this report with all sections complete Clean-up Requirements Return equipment to proper location Emergency Actions Hot or smoking resistors – disconnect circuit; do not touch hot resistors Mild burns – run under cool water; call 911 if necessary

LAB # 2 SYSTEMS OF EQUATIONS (Circuits) ENGR 190 2 Terry W. Armstrong, PhD (2021) Introduction A mathematical "system of equations" establishes a dependent relationship between elements within a physical system. These equations “work together” to fully describe the defined system. Complex systems may require more than one equation to fully resolve the unknowns within the system. These additional equations are often developed by applying different known laws to the same problem or applying the same law to different parts of the problem. A system of equations can have any number of unknown variables and any number of equations. If the number of equations equal the number of unknowns, the system is generally solvable, if a solution actually exists. A variety of techniques can be used to solve a system of equations. In the simplest case of two equations and two unknowns, the method of substitution is usually the easiest solution approach. This lab considers the electric current traveling via two paths within a circuit of parallel and series resistors. The resistances are known. However, with two possible current paths, two equations are necessary to resolve the current along each path. Objective Use method of substitution to solve a system of equations for unknown amperages. Background Consider the electrical circuit in Figure 1. The total available current driven by the battery will split and travel through both parallel resistors, R 1 & R 2 and also through the series resistor R 3 . Since R 1 and R 2 have different values, the current values, I 1 & I 2 , within each path will also be different.

LAB # 2 SYSTEMS OF EQUATIONS (Circuits) ENGR 190 3 Terry W. Armstrong, PhD (2021) Figure 1. Parallel and series circuit. One approach to determining the unknown currents, I 1 & I 2 , along each path is to begin with an easy visualization of Kirchhoff’s Voltage Law in terms of “voltage drop” around the circuit. It’s easy to imagine a voltmeter measuring a 9v potential directly across a 9v battery as shown in Figure 2. Measuring the voltage drop across all the resistors of interest is more helpful. This voltage drop across all resistors is also the exact same voltage drop across the battery since the measurement locations are actually the same, just with some extra wire in place. Figure 2. Setting up Kirchhoff’s Voltage Law . Dividing the circuit into sections A & B, the voltage drop across the two parallel resistors in A and voltage drop across the series resistors in B must still add up to 9v: V A +V B =9v. Since we’re interested in current rather than voltage, use Ohm’s law (V=IR) to replace the voltage values to arrive at Equation 1.

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LAB # 2 SYSTEMS OF EQUATIONS (Circuits) ENGR 190 4 Terry W. Armstrong, PhD (2021) (IR) ? + (IR) ? = 9𝑣 (1) Referring back to Figure 1, consider only the current flow along the "upper" path through R 1 and R 3 . Since we've already identified 9v must be measured across this distance, the voltage across the upper path and the lower path must therefore separately measure 9v. Using Equation 1 for only the upper path provides the first equation with two unknowns (I 1 and I 2 ) in the form of Equation 2. (R 1 I 1 ) ? + R 3 (I 1 + I 2 ) ? = 9𝑣 (2) Even with R 1 and R 3 known, both I 1 and I 2 are unresolved with only one equation, so an additional equation is necessary. The same logic is used along the "lower" path through R 2 to develop Equation 3. (R 2 I 2 ) ? + R 3 (I 1 + I 2 ) ? = 9𝑣 (3) Taken together, Equations 2 and 3 provides a system of two equations and two unknowns as given in Equation 4a & 4b. With known resistances, the two unknown currents can be resolved. The easiest approach is to solve for either I 1 or I 2 and then substitute back into one of the two equations to solve for the other unknown. A single brace is often used to identify equations working together to form a system. R 1 I 1 + R 3 (I 1 + I 2 ) = 9 v (4a) R 2 I 2 + R 3 (I 1 + I 2 ) = 9𝑣 (4b) Experimental Method Do NOT connect the circuit yet; measure each component value independently before placing in the circuit shown in Figure 3.

LAB # 2 SYSTEMS OF EQUATIONS (Circuits) ENGR 190 5 Terry W. Armstrong, PhD (2021) Turn on the multimeter and set it to measure DC volts ( V̅ ). Verify the actual voltage of the battery. Record the battery value in both Table 1 (in the results section) and also Appendix A. Switch the multimeter to measure resistance in Ohms (  ). Verify each of the resistor values BEFORE placing into the breadboard circuit shown in Figure 3. Record the resistance values in Figure 3, Table 1 (in the results section) and Appendix A. Insert the resistors in the breadboard as shown, but do not connect the battery in the circuit yet. NOTE: Take care to note the leftmost placement of both R1 and R2 are NOT aligned in the same column of connections (note the internal breadboard connections). The jumpers are used to complete the desired circuit. Figure 3. Circuit set-up. Solve the system of equations. With both the battery and resistance values verified, use Appendix A to solve the system of equations given in Equation 4a & b. Substitute the measured resistor values into the equations and solve for either I 1 or I 2 . Then, substitute in the solved value (I 1 or I 2 ) and solve for the other amperage value. Record these values in Table 1 in the results and discussion section. R3 R1 R2 9V BATT (ACTUAL: _____V) J1 J2 Internal connections R 1 = ______  R 3 = ______  R 2 = ______ 

LAB # 2 SYSTEMS OF EQUATIONS (Circuits) ENGR 190 6 Terry W. Armstrong, PhD (2021) CAUTION: Have the TA assist with this check. Determine the total amperage: I 1 +I 2 =I TOTAL . Verify ITOTAL is LESS than the amperage of the multimeter setting to avoid blowing out the internal fuse. Also, verify total power, I TOTAL 2 R 3 , is less than 1.0W (resistor power rating). Next, switch the multimeter to measure milli-amperage (mA). To measure amperage, it is necessary to "break into" the circuit and place the meter in series. The jumpers will be used for this purpose. Complete the circuit diagram as shown in Figure 3 with the battery and jumper wires. The polarity of the battery wire does not matter. Measure the current associated with I 1 as shown in Figure 4a by removing the RIGHT end of Jumper #1 from the breadboard and inserting another jumper wire #3 as shown to connect the multimeter probes. Record the amperage, I 1 , in Table 1. Replace Jumper #1 back into the circuit as originally shown in Figure 3. Next, measure I 2 by removing ONLY the right end of Jumper #2 and moving Jumper #3 to R2 as shown in Figure 4b. Measure amperage with the multimeter between Jumper #2 and Jumper #3, similar for measuring I 1 in Figure 4b. Record the value of I 2 in Table 1. Figure 4a & 4b. Measuring I 1 and I 2 . R3 R1 R2 9v J1 J2 m A J3 R3 R1 R2 9v J1 J2 mA J3

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LAB # 2 SYSTEMS OF EQUATIONS (Circuits) ENGR 190 7 Terry W. Armstrong, PhD (2021) Remove the battery from the circuit and disassemble the circuit. Turn off the multimeter. Return all items to their proper locations. Results and Discussion Record all calculated and measured values in Table 2. Compute the error percentage between calculated amperage and measured amperage. Table 1. Circuit set-up. MEASURED VALUES: CALCULATED AMPERAGE VALUES FROM APPENDIX A (Amps): MEASURED AMPERAGE VALUES (Amps): % ERROR IN AMPERAGE: R 1 (  )= I 1 (A)= I 1 (A)= R 2 (  )= I 2 (A)= I 2 (A)= R 3 (  )= I 1 + I 2 (A)= BATT (V)= Conclusion Write a paragraph here to discuss the following: - A system of linear equations is "inconsistent" if there is no solution, and it is "consistent" if there is a solution. How do you classify the system in this lab? - Can substitution always be used to solve a system of two unknowns and two equations? - Do the jumpers, probes and connections contribute to the overall error?

Appendix A ENGR 190 8 Terry W. Armstrong, PhD (2021) Solving the system of equations R 1 I 1 + R 3 (I 1 + I 2 ) = 𝑉 (4a) R 2 I 2 + R 3 (I 1 + I 2 ) = 𝑉 (4b) Beginning with the above equations, use actual MEASURED values for R 1 , R 2 , R 3 and V. Rewrite the above equations with the actual values and solve for both I 1 and I 2 using substitution. Record I 1 and I 2 in Table 1 in the results and discussion section. Show all work here. ( ______ ) I 1 ( ______ ) ( I 1 + I 2 ) = ( ______ ) R 1 R 3 V ( ______ ) I 2 ( ______ ) ( I 1 + I 2 ) = ( ______ ) R 2 R 3 V