Morrow Dc Circuit Lab

Physics 210L Fall 2021 Sunderland L Morrow 11-05-21 1 DC Circuit Analysis (40pts) Introduction The principal goal in this experiment is to reinforce the fundamentals of dc circuit analysis and basic wiring skills. In this lab, you will apply your knowledge of series and parallel circuits and Kirchhoff’s laws to a simple bridge circuit. The experimental objectives of this three-part activity are: Part 1,2: To learn how to wire series and parallel circuits. Part 1,2: To learn about Kirchhoff’s current and voltage laws. Set R1=100 W , R2=240 W , R3=470 W , and R4=200 W Part 1 Procedure 1. Measure the value of the resistors with the multimeter. Use multimeter manual to determine the uncertainty in the measurement (see Appendix). Record the measured values in Table 1. 2. Using the “PROTO-BOARD,” wire the circuit shown below. Set the DC power supply voltage to approximately 6.0V. Take a picture of your circuit and paste it before Table 1. Measure and record its value in Table 1. 3. Measure and record, I S , the current through the power supply. Use multimeter manual to determine the uncertainty in the measurement (see Appendix). 4. Measure and record the potential difference across R1, R2, R3, and R4. Use multimeter manual to determine the uncertainty in the measurement (see Appendix). Analysis 1. Calculate, I R1 , the current through R1 and I R2 , the current through R2 along with their uncertainties and record their value in Table 1. Apply Kirchhoff’s current law to calculate I S along with its uncertainty. Finally calculate the percent difference between latter value and measured value I S . Record your results below Table 1. Show all your work below. Experiment 4 A B C D V

2 2. Calculate, I R3 , the current through R3 and I R4 , the current through R4 along with their uncertainties and record their value in Table 1. Apply Kirchhoff’s current law to calculate I S along with its uncertainty. Finally calculate the percent difference between latter value and measured value I S . Record your results below Table 1. Show all your work below. Part 2 Procedure 1. Place the multimeter between points A and C (refer to the figure below). Set the multimeter to the 10A scale, and measure and record the current I AC (along with its direction) through the meter. If this scale setting is too large to give a non-zero value, try a lower scale setting. Take a picture of your circuit and paste it before Table 2. 2. Leave the multimeter between points A and C and measure and record, I S , the current through the power supply. Record your results in Table 2. Do you expect the same current as in Part 1? Why? 3. Measure and record the potential difference across R1, R2, R3, and R4. Use multimeter manual to determine the uncertainty in the measurement (see Appendix). 4. Repeat steps 1 through 3 but this time set the multimeter that is between points A and C to the lowest possible current setting. Do you expect any values to change? Why? Record your results in Table 3. Analysis 1. Calculate, I R1 , the current through R1 and I R2 , the current through R2 along with their uncertainties and record their values in Table 2. Apply Kirchhoff’s current law to calculate I S along with its uncertainty. Finally calculate the percent difference between latter value and measured value I S . Record your results below Table 2. Show all your work below. 2. Calculate, I R3 , the current through R3 along with its uncertainty and record its value in Table 2. Use your calculated value of I R3 and an appropriate current from Step 1 along with Kirchhoff’s current law to calculate I AC and its uncertainty. Finally calculate the percent B A C D V

4 δ I R1 = : ;− & ! # ࠵?࠵?> % + ; " ! ࠵?࠵?> % δ I R1 = :;− ".()) ((.()+,) # 0.0007> % + ; " (.()+, 0.006> % δ I R1 = : ;− ".()) (.((). . % 0.0007> % + (0.0 6 1) % δ I R1 = : ;− (.((( / . (.((). . % > % + (0.0 6 1) % δ I R1 = A(−0.0 7 8) % + (0.0 6 1) % δ I R1 = √0.00 6 0 + 0.00 3 7 δ I R1 = √0.00 9 7 δ I R1 = 0.0 9 8 δ I R1 = 0 .1࠵?࠵? Kirchhoff’s current law for ࠵? 0 : ࠵? !" + ࠵? !% = ࠵? !, + ࠵? !1 ࠵? 0 = ࠵? !, + ࠵? !1 ࠵? 0 = 11.14࠵?࠵? + 14.34࠵?࠵? = 25 .48࠵?࠵? Example calculation for propagation: δI 0 = F G ࠵?࠵? 0 ࠵?࠵? !, ࠵?࠵? !, I % + G ࠵?࠵? 0 ࠵?࠵? !1 ࠵?࠵? !1 I % δI 0 = A(࠵?࠵? !, ) % +(࠵?࠵? !1 ) % δI 0 = A(0.03) % +(0.07) % δI 0 = √0.000 9 + 0.00 4 9 δI 0 = √0.00 5 8 δI 0 = 0.0 7 6 δI 0 = 0 .08࠵?࠵? Percent difference sample calculation: [|25.48࠵?࠵? − 25.33࠵?࠵?|] - 25.48࠵?࠵? + 25.33࠵?࠵? 2 / ∗ 100% = 0.1 5 - 50.8 1 2 / ∗ 100% = 0.1 5 25.4 0 2 ∗ 100%

Physics 210L 5 = 0.005 9 0 ∗ 100% = 0 .59% Calculated I S from I R1 and I R2 along with its uncertainty = ( 25 .54 ± 0 .1)࠵?࠵? Percent Difference: 0 .59% Calculated I S from I R3 and I R4 along with its uncertainty = ( 25 .48 ± 0 .08)࠵?࠵? Percent Difference: 0 .59% Questions 1. Do your results verify Kirchhoff’s current law? In other words, is the calculated value of I S obtained from I R1 and I R2 in agreement with the measured value of I S ? Discuss and explain. Don’t forget to discuss sources of random and systematic errors. ࠵?࠵?࠵?࠵?࠵?࠵?࠵?࠵?࠵?࠵? ࠵?࠵?࠵?࠵?࠵?: ࠵? ࠵? = ࠵? !" + ࠵? !% = 11.2࠵?࠵? + 14.34࠵?࠵? = 25 .5࠵?࠵? Measured value: I S ±δ I S = ( 25 .3 ± 0 .2 )࠵?࠵? The calculated value (from ࠵? !" + ࠵? !% ) agrees with the measured value of I S . These results verify Kirchhoff’s current law. The law states that the sum of all currents at a junction will equal zero. Another way of looking at this law is that if a given current splits up into multiple other currents, the original current is equal to the numerical sum of the those new currents. Hence, ࠵? ࠵? = ࠵? ࠵?࠵? + ࠵? ࠵?࠵? is true for the above calculations. The error in this calculation stems from systematic error. No matter how many times the experiment is repeated and measured, similar uncertainty in the measured values will always be present due to the multimeter itself. 2. Is the calculated value of I S obtained from I R3 and I R4 in agreement with the measured value of I S ? Discuss and explain. Don’t forget to discuss sources of random and systematic errors. ࠵?࠵?࠵?࠵?࠵?࠵?࠵?࠵?࠵?࠵? ࠵?࠵?࠵?࠵?࠵?: ࠵? ࠵? = ࠵? !, + ࠵? !1 = 11.14࠵?࠵? + 14.34࠵?࠵? = 25 .48࠵?࠵? Measured value: I S ±δ I S = ( 25 .3 ± 0 .2 )࠵?࠵? The calculated value (from ࠵? !, + ࠵? !1 ) agrees with the measured value of I S since the measured value is within the bounds of the calculated value. Once again, the source of error is due to systematic error. The multimeter is an instrument with documented uncertainty written in its manual. No matter how many times this experiment is measured, there will consistently be similar uncertainty exhibited.

Physics 210L 7 Part 2 Table 2 R±δR (kΩ) V±δV (Volts) I ±δ I (mA) R1 0.0983 ± 0.0007 2.084 ± 0.006 21.2 ± 0.2 R2 0.239 ± 0.001 2.083 ± 0.006 8.72 ± 0.04 R3 0.463 ± 0.001 4.143 ± 0.007 9.00 ± 0.02 R4 0.1980 ± 0.0009 4.151 ± 0.007 21.0 ± 0.1 Power Supply Voltage= ( 6 .291 ± 0 .008)࠵? Scale Setting of Ammeter for I AC : 10A I S ±δ I S = ( 29 .7 ± 0 .3)࠵?࠵? I AC ±δ I AC = ( 0 .01 ± 0 .01)࠵? to the right or to the left , circle one Calculated I S from I R1 and I R2 along with its uncertainty = ( 30 .0 ± 0 .2 )࠵?࠵? Percent Difference: 1% Calculated I AC from Step 2 along with its uncertainty: → → ࠵? !" − ࠵? !, = ࠵? 78 I 78 = 21.2࠵?࠵? − 9.00࠵?࠵? = 12 .2࠵?࠵?

8 δI 78 = F G ࠵?I 78 ࠵?࠵? !" ࠵?࠵? !" I % + G ࠵?I 78 ࠵?࠵? !, ࠵?࠵? !, I % δI 78 = A(࠵?࠵? !" ) % +(࠵?࠵? !, ) % δI 78 = A(0.2) % +(0.02) % δI 78 = √0.04 + 0.0004 δI 78 = √0. 4 0 δI 78 = 0 .2࠵?࠵? → ( 12 .2 ± 0 .2)࠵?࠵? Percent Difference: 4% Calculated I AC from Step 3 along with its uncertainty: δI 78 = F G ࠵?I 78 ࠵?࠵? !% ࠵?࠵? !% I % + G ࠵?I 78 ࠵?࠵? !1 ࠵?࠵? !1 I % δI 78 = A(࠵?࠵? !% ) % +(࠵?࠵? !1 ) % δI 78 = A(0.04) % +(0.1) % δI 78 = 0 .1࠵?࠵? → ( 12 .3 ± 0 .1)࠵?࠵? Percent Difference: 3% Table 3 R±δR (kΩ) V±δV (Volts) I ±δ I (mA) R1 0.0983 ± 0.0007 2.035 ± 0.006 20.7 ± 0.2 R2 0.239 ± 0.001 2.154 ± 0.006 9.02 ± 0.04 R3 0.464 ± 0.001 4.202 ± 0.007 9.06 ± 0.02 R4 0.1978 ± 0.0009 4.085 ± 0.007 20.6 ± 0.1 Power Supply Voltage= ( 6 .294 ± 0 .008)࠵? Scale Setting of Ammeter I AC : 20࠵?࠵? I S ±δ I S = ( 29 .53 ± 0 .3 )࠵?࠵? I AC ±δ I AC = ( 11 .66 ± 0 .07 )࠵?࠵? to the right or to the left , circle one

10 With the addition of the ammeter, the circuit’s overall current changed slightly. With a decrease in resistance (from point A to C) there should exhibit a decrease in overall current. This is notable when comparing the two I S values. They however still agree with each other. The two I AC values disagree with each other and that is to be expected. The scale of the ammeter was changed from 10A to 20mA, regarding the Table 2 and 3 respectively. Because of this adjustment, it is expected that the measurement will now change as the internal resistance of the ammeter has changed. 5. Do your results in Table 3 verify Kirchhoff’s voltage law? Choose a loop in the circuit that contains the ammeter between points A and C to verify your claim. Show all work and don’t forget to quote uncertainties. Discuss and explain. Don’t forget to discuss sources random and systematic errors. Yes, the results in Table 3 verify Kirchhoff’s voltage law. Loop example (counter-clockwise): ࠵? !" + ࠵? !# + ࠵? !$% ࠵? !& = 0 −2.035࠵? − 4.202࠵? + 2.154 + 4.085 = 0 .002࠵? Uncertainty: F G ࠵?࠵? ࠵?࠵? !" ࠵?࠵? !" I % + G ࠵?࠵? ࠵?࠵? !, ࠵?࠵? !, I % + G ࠵?࠵? ࠵?࠵? !% ࠵?࠵? !% I % + G ࠵?࠵? ࠵?࠵? !1 ࠵?࠵? !1 I % = A(࠵?࠵? !" ) % + (࠵?࠵? !, ) % + (࠵?࠵? !% ) % + (࠵?࠵? !1 ) % = A(0.006) % + (0.006) % + (0.007) % + (0.007) % = 0 .01࠵? Therefore, the numerical sum of the potential differences within the specified loop is: (0 .002 ± 0 .01 )࠵? of which zero falls within the bounds of. The above calculations clearly verify that Kirchhoff’s voltage law holds true with Table 3’s data when the ammeter is present in the circuit.

Physics 210L 11 6. Derive an expression for the current through R1 in terms of R1 and the other resistor values and the power supply voltage, V, when an ideal ammeter is connected across points A and C. ࠵? 9:" = G 1 ࠵? " + 1 ࠵? % I ࠵? 9:% = G 1 ࠵? , + 1 ࠵? 1 I ࠵? 9:%; ࠵? 9:" = ࠵? 9:, ࠵? < = ࠵? ࠵? 9: ࠵? ";% = ࠵? 0 ∗ ࠵? 9:" = ࠵? 0 ࠵? 9:, ∗ ࠵? 9:" ࠵? !9:" = ࠵? ";% ࠵? " With an idea ammeter, the internal resistance is zero. The above calculations take into account for that. Appendix-Understanding Quoted Uncertainties from Multimeter Manual Some manuals quote uncertainties in the following manner: ± (1%+1). The 1 digit means that the least significant digit can be off by ± 1. If you are on a scale that reads to 0.001 units, then 1 digit would mean ± 0.001V . If the manual specifies ± (0.5%+4) and you are on a scale that reads to 0.01 units and you measured a value of 10.23 units, 4 digits means that the displayed 10.23 units could be 10.27 units or 10.19 units (not including the 0.5%). If you include the 0.5%, the total uncertainty in the measured value of 10.23 units is 0.05units+0.04units =0.09units, and you would report the measurement as (10.23 ± 0.09) units.