10.7 template for report on Ohm's law

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University of Texas *

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1611-004

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Electrical Engineering

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Apr 3, 2024

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Ohm’s law Paul Mac Alevey © Fall 2021 1 REPORT NAME:___________________________ Course & Section: __________________ Questions 1), 2) and 3) ask you to fill this table; [3] 4) Are the values you measured for each resistor within the indicated tolerance according to the color codes on the resistor? Show your calculations. ________________________________________________________________________________ ________________________________________________________________________________ ______________________________________________________________________________ [2] Nominal resistance k Ω Tolerance of the resistor Measured resistance k Ω R 1 R 2 R 3 Shashwat Adhikari 2126-115 8.2 27.0 100.0 5% 5% 5% 8.1 26 101 8.2 +- 5%= 7.79-8.61 kohms. 27+-5%=25.65-28.35 kohms. 100+-5%= 95-105 kohms. As we can see all three values we measured are clearly in range, therefore within the indicated tolerance according to the color codes on the resistor.

Ohm’s law Paul Mac Alevey © Fall 2021 2 5), 6) &7) Complete table 2 below. [1+1+10] Table two Voltage of the Power Supply S V (volts) nominal measured Current out of the Supply (milli amps) Voltage Across Resistor (volts) 12 10 8 6 4 2 8) In circuit 1, why is the voltage measured across the power supply nearly the same as the voltage measured across 𝑅𝑅 1 ? _______________________________________________________________________________ ____________________________________________________________________________ [1] 9) For circuit one, use Excel to make a graph of current through the resistor versus the voltage across the resistor. [4] 12.5 10.0 8.5 6.0 3.5 2.0 1.54 1.23 1.05 0.74 0.43 0.25 12.5 10.0 8.5 6.0 3.5 2.0 R1 is the only resistor in the circiuit, therefore all the voltage drops across the same resistance R1 because its resistence is equal to the total resistence in this circuit.

Ohm’s law Paul Mac Alevey © Fall 2021 3 10) Why don’t we use the nominal voltage for the graph in the previous question? ______________________________________________________________________________ [1] 11) Does your resistor satisfy Ohm’s law? Explain how your graph tells you this. If this ‘law’ is satisfied then find R 1 . ______________________________________________________________________________ [3] 12) Record the currents and voltages from circuit two in the next table. [6] I 1 I 2 V AB V BC V AC S V 13) What is the relation between I 1 and I 2 ? ______________________________________________________________________________ [1] 14) The voltage (or potential difference) between points A and B is 𝑉𝑉 𝐴𝐴𝐴𝐴 ≡ 𝑉𝑉 𝐴𝐴 − 𝑉𝑉 𝐴𝐴 . Use this expression to simplify the expression; AB BC V V  . ______________________________________________________________________________ [1] 15) Do your measurements give the same answer as your simplified version of AB BC V V  ? ______________________________________________________________________________ [1] Nominal voltage is the voltage we are feeding in and is not the actual voltage of the circuit. Yes, since the R^2 value is 1, we can conclude a perfectly linear relationship of I and V in ohms V=IR. If this "law" is satisfied then the slope of the graph is R1 because our y is V and x is I. Y/x is equivalent to V/I which equals to R. The slope is 8.13, therefore R1 equals 8.13 kOhms. 0.34 mA 0.34 mA 2.7 V 8.8 V 11.5 11.5 V They are equal because the current going around a circuit, especially in series is always constant. VAB+VBC= 2.7+8.8= 11.5 which is equal to VAC according to the practive above which then= VS Yes, the measurements agree with the answer above which gives the simplified version of VAB+VBC.

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Ohm’s law Paul Mac Alevey © Fall 2021 4 16) Use your measured data to show that V AB = I 1 R 1 and V BC = I 2 R 2 . Show your calculations. (These relations will only have the correct sign if you got the ammeter the right way around!) _______________________________________________________________________________ ______________________________________________________________________________ [2] 17) …draw a circuit that is equivalent to circuit two but with the two resistors replaced by one resistor with their equivalent resistance. What is the value of this equivalent resistance? Show the necessary calculations. [3] 18) Record the eight currents and voltages: I 1 , I 2 , I 3 , I 4 , V s , V AB , V CD and V BE (from circuit three in the second column of the following table . ) [6] Table four Circuit 3 Measured Values Circuit 3 (With Ground connection) Measured Values I 1 I 1 I 2 I 2 I 3 I 3 I 4 I 4 V BE V BE V CD V CD V AB V AB S V S V VAB=I1R1=(0.34)(8.1)=2.75 V which agrees with the measured value in table 3. VBC=(0.34)(26)=8.84 V which again agrees with the value measured above. If we are to only have one resistor instead of two resistors such as in circuit two, but we want the equivalent resistence to be equal between AC. The two resistors in series add up to give us an equivalent resistance of R1+R2=8.1+26=34.1 kOhms. Therefore, the circuit needs one resistor with 34.1 kOhms resistence. 0.40 mA 0.32 mA 0.08 mA 0.40 mA 8.3 volts 8.3 volts 3.2 volts 11.5 volts 0.40 mA 0.32 mA 0.07 mA 0.41 mA 8.3 volts 8.3 volts 3.2 volts 11.5 volts 34.1 kOhms

Ohm’s law Paul Mac Alevey © Fall 2021 5 19) (Use the preceding table to answer the next question.) How is V BE related to V CD ? Explain. ________________________________________________________________________________ ______________________________________________________________________________ [2] 20) Is the current flowing into the negative terminal of the power supply equal to the current flowing out of the positive terminal? ______________________________________________________________________________ [1] 21) How is V AB + V BE related to any other voltage that you have measured? ________________________________________________________________________________ ______________________________________________________________________________ [2] 22) Are the following true (up to sign)? V AB equals I 1 R 1 ; V BE equals I 2 R 2 ; and V CD equals I 3 R 3 . Show the necessary calculations. ________________________________________________________________________________ ________________________________________________________________________________ ________________________________________________________________________________ ______________________________________________________________________________ [3] VBE and VCD are connected in parallel and according to our established model of electricity the potential difference/ Voltage across them should be equal which holds true in our observations. Yes, the current might split up in the middle when it reaches the parallel branch but it will come back together at the end and be equal to the current flowing out of the positive terminal. This is conservation of charge coming into play. VAB+VBE equals the total voltage supplied by the battery VS. We know there is voltage drop on R1 and equal voltage drop on VBE and VCD. Therefore, the total voltage =VBE/VCD+VAB=Vs. This holds true once again in our measurements as 8.3 volts + 3.2 volts= Vs= 11.5 volts. VAB= I1R1 which should be 3.2 V according to measurements. VAB=(0.40)(8.1)=3.24 V. They hold true. VBE=I2R2 which should be 8.3 V according to measurements. VBE=(0.32)(26)=8.32 V. They hold true. VCD=I3R3 which should be 8.3 V according to measurements. VCD=(0.08)(101)=8.08. They hold true. Since the measured and calculated values of all three voltages match up, we can conclude that they are all true.

Ohm’s law Paul Mac Alevey © Fall 2021 6 23) …draw a circuit that is equivalent to circuit three and contains only one resistor. What is the resistance of the only resistor? Show your calculations. [3] 24) Record the eight currents and voltages: I 1 , I 2 , I 3 , I 4 , V s , V AB , V CD and V BE in the third column of table 4 (with question 12). [6] 25) Are the currents the same as before you included the connection to ground? Explain why or why not. ________________________________________________________________________________ ______________________________________________________________________________ __ ______________________________________________________________________________ [2] Yes, the currents are the same as before including the connection to ground. This is because the current is still continuing to flow from high potential to low potential and since the ground is zero, the current is not deflected to the ground. If only one resistor with the same equivalent resistence is what is desired we would get a circuit with one resistor with resistence that is equal to equiv. resistence of previous circuit. That would be equal to Req= 8.1 + (1/26 + 1/101)^-1 = 28.78 kOhms. 28.78 kOhms

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