Lab 4

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Apr 3, 2024

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Physics Lab (Online Simulation) 1 OHM’S LAW - KIRCHOFF’S LAW Electricity and Light TA name: Madhab Pokhrel Due Date: 02/27/24 Student Name: Uyen Do Student ID: 1002080329 Theory: A D.C. circuit consists of sources of direct current (EMFs), connected to a network of elements. If the elements are ohmic (obey Ohm’s Law) then the currents through the elements are directly proportional to the voltages applied across the elements. For ohmic elements, the relation between the current through an element (I), in amps, and the voltage across the element (V), in volts, is V = IR [1] where R is the resistance of the element in ohms. Resistors, which are commonly used in electronic devices, are ohmic. If two or more resistors are connected in series with a voltage source, as illustrated in Figure 18-1, they collectively behave as an equivalent single resistor with resistance, R ser , where R ser = R 1 + R 2 + R 3 +  + R n [2] Figure 18-1 the resistors are connected in parallel to the voltage source, as illustrated in Figure 18-2, then they collectively behave as an equivalent single resistor with a resistance, R par , where R 1 R 2 R 3 R n V

Physics Lab (Online Simulation) 2 Figure 18-2 n 3 2 1 par R 1 R 1 R 1 R 1 R 1 + + + + =  [3] The circuits illustrated in Figures 18-1 and 18-2 can be reduced to a single loop containing a single voltage source and a single resistor. And Ohms Law can be used to determine the currents in the circuit. If however the circuit contains more than one voltage source in a network of resistors, the circuit usually cannot be reduced to a single loop. Kirchhoff’s rules are useful in analyzing a multi -loop circuit. This Prelab is worth 15 points Type all your answers in Blue (1) Write down an equation and solve for the total resistance of three resistors R 1 =100Ω, R 2 =150 Ω, R 3 =350Ω in the resistors were arranged in the following combination: a) All three in series (1 point) R= R1 + R2 + R3 = 100 + 150 + 350 = 60 0 Ω b) All three in parallel (1 point) 1/R = (1/R1) + (1/R2) + (1/R3) = (1/100) + (1/150) + (1/350) = 0.0195 • R = 5 1.28 Ω c) How would you arrange these three resistors to get a net resistance of 410 Ω. (1 point) To get a net resistance o f 410 Ω , let R1 and R2 connected in Parallel. So 1/ R’ = 1/150 + 1/100 R’ = 60 Ω Now this connected with R3 in series. Rnet = R’ + R3 = 60 + 350 = 410 Ω V R n R 3 R 2 R 1

Physics Lab (Online Simulation) 3 (2) If you have an experimental setup which has three unknowns, how many linearly independent equations do you need to determine the unknowns to find one unique solution? (1 point) - To determine the unknowns to find one unique solution, we need 3 linearly independent equations: (1) a1x+ b1y + c1z = d1 (2) a2x+ b2y + c2z = d2 (3) a3x+ b3y + c3z = d3 (x,y,z are the unknowns) (3) What is Kirchoff’s law of current? What does it conserve? (1 point) • The Kirchhoff's current law (1st Law) states that the current flowing into a node (or a junction) must be equal to the current flowing out of it. This is a consequence of charge conservation. (4) What is Kirchoff’s law of voltage? What does it conserve? (1 point) • The Kirchhoff's voltage law (2nd Law) states that in any complete loop within a circuit, the sum of all voltages across components which supply electrical energy (such as cells or generators) must equal the sum of all voltages across the other components in the same loop. This law is a consequence of both charge conservation and the conservation of energy. (5) Electrical circuits have two main problems: “Short” and “Open”. Define these two conditions with diagram and an example showing the consequence of each of these faults. (Use back of the page if necessary). (2+2 = 4 points) - Short circuit is the processes of bypassing the current through another path, it means no current flows in particular branch of the circuit. This is the diagram for open circuit. In this circuit the current cannot flow from one end of the power source to the other end of power source.

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Physics Lab (Online Simulation) 4 - Short circuit: The current flow to the least resistance path, it destroyes the powet supply - This is the diagram for short circuit, in this case the light bulb will not turn on because all the current will flow throught the path 1. (6) Apply Kirchoff’s law of current and voltage in loop 1 and loop 2 of the above circuit and write down the corresponding equation. (2.5 + 2.5 = 5 points) R 1 R 2 R 3 I 1 I 2 I 3 V 1 V 2 I 2 I 1 Loop 1 Loop 2

Physics Lab (Online Simulation) 5 Loop I: Current equation: I1 + I2 = I3 Voltage equation: V1 = (R1+R3)*I1 + I2R3 Loop 2: Current equation: I1 + I2 = I3 Voltage equation: V2 =(R1 + R3)*I2 + I1R3 Part A: Ohm’s Law This lab uses the Ohm’s Law and Circuit Construction Kit DC simulation from PhET Interactive Simulations at University of Colorado Boulder, under the CC-BY 4.0 license. https://phet.colorado.edu/sims/html/ohms-law/latest/ohms-law_en.html https://phet.colorado.edu/sims/html/circuit-construction-kit-dc/latest/circuit-construction-kit-dc_en.html Develop your understanding: Open Ohm’s Law , then explore to develop your own ideas about how resistance, current, and battery voltage are related. 1. As you change the value of the battery voltage, how does this change the current through the circuit and the resistance of the resistor? If the current or resistance remains constant, why do you think it is? → The relation for the voltage and current: V=IR If the value of the battery voltage is changed, the current will change in direct proportion to the voltage.

Physics Lab (Online Simulation) 6 Thus, if the voltage is increased, the current will also increase and if the voltage is decreased, the current will also decrease. Now, since, the resistance is a property of the resistor, the resistance will not change with a change in the battery voltage. 2. As you change the value of the resistance of the resistor, how does this change the current through the circuit and the battery voltage? If the current or voltage remains constant, why do you think? → If the resistance is changed, the current and voltage will change. The current will increase upon decreasing the resistance and it will decrease upon increasing the resistance. The voltage is a property of the battery, so, the voltage will remain constant. 3. Use understanding to make predictions about a circuit with lights and batteries. → If another battery is connected in series with the circuit, the brightness will increase. On adding more batteries, the brightness will increase even more. In the case of a circuit with one bulb and a battery and if more and more bulbs are being added, the brightness of each bulb will go on decreasing Demonstrate your understanding: Directions: As you answer the questions, explain in your own words why your answer makes sense and provide evidence from your #1 experiments. Add more experiments to #1 if you need to get better evidence. 2. If you change the value of the battery voltage: a. How does the current through the circuit change? (answer, explain) - When the value of the battery voltage is increased, the current through the circuit also increases, assuming the resistance remains constant. According to Ohm's Law, if the voltage (V) increases while the resistance (R) remains constant, the current (I) flowing through the circuit will also increase.

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Physics Lab (Online Simulation) 7 Explain : When set up a simple circuit with a fixed resistor and a variable voltage source. Gradually increase the voltage and measure the current flowing through the circuit using an ammeter. I observe that as the voltage increases, the current also increases. b. How does the resistance of the resistor change? (answer, explain) - Changing the value of the battery voltage does not directly affect the resistance of the resistor. The resistance remains constant unless explicitly modified. 3. If you change the resistance of the resistor: a. How does the current through the circuit change? (answer, explain) - When the resistance of the resistor is increased, the current through the circuit decreases, assuming the voltage remains constant. Explain: Using Ohm’s law, If the resistance (R) increases while the voltage (V) remains constant, the current (I) flowing through the circuit will decrease b. How does the voltage of the battery change? (answer, explain) - Changing the resistance of the resistor does not directly affect the voltage of the battery. The battery voltage remains constant unless explicitly modified . 4. Consider the two circuits below. Use your understanding of voltage, resistance, and current to answer these questions: a. What do you think will happen when the switches are closed? (answer, explain) . As the circuit is closed when switches are turned closed, the current will flow in the circuit and therefore bulbs in both circuits will glow. -

Physics Lab (Online Simulation) 8 b. How do you think the lights’ brightness will compare? - The single bulb in the second circuit will glow brighter compared to bulbs in the first circuit. c. Open the Intro screen of Circuit Construction Kit DC. Build 2 circuits. Turn on “values”. An ammeter is used to measure current in a circuit. Use the ammeter to compare the current in the two circuits. Compare and contrast the two circuits. Explain the difference in brightness from the two circuits by relating it to Ohm’s Law. Insert a capture of the circuits with the switch closed for supporting evidence.

Physics Lab (Online Simulation) 9 5. Consider the two circuits below.

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Physics Lab (Online Simulation) 10 Use your understanding of voltage, resistance, and current to answer these questions: • What do you think will happen when the switches are turned closed? (answer, explain, evidence) - When the switches closed, the circuit will get connected because of current will start flowing from battery to the bulb • How do you think the lights’ brightness will compare? - The bulb in first circuit will be brighter than second circuit • Open the Intro screen of Circuit Construction Kit DC. Build the 2 circuits and check your answers. Use the values obtatined from the ammeter and Ohm’s Law to explain the difference in brightness from the two circuits. Insert a capture of the circuits with the switch closed for supporting evidence.

Physics Lab (Online Simulation) 11 Part B: Kirchoff’s Law Objectives: To investigate Kirchhoff’s Laws: Kirchhoff’s Current Law (KCL) and Kirchhoff’s Voltage Law (KVL). Simulation Tools: DC ‐ Power supply, voltmeter, ammeter, resistors, and connecting wires. Theory and Background:

Physics Lab (Online Simulation) 12 Kirchhoff’s laws follow from the laws of conservation of energy and conservation of charge. These laws are used to analyze electrical circuits, which contain combinations of batteries, resistors and capacitors. In this experiment, we are interested in inve stigating Kirchhoff’s laws for a direct current (DC) circuit, for which the electrical currents are constants in magnitude and direction. The two Kirchhoff’s laws are referred to as Kirchhoff’s Current Law (KCL) , also called Junction Rule and Kirchhoff’s Voltage Law (KVL) , also called Loop Rule. Kirchhoff’s Current Law (KCL): The sum of the currents entering any junction, in a closed circuit, must equal the sum of the currents leaving it; or the algebraic sum of all currents at that point is zero ,  I = 0 ……..………….. (1) This law is a restatement of charge conservation. Kirchhoff’s Voltage Law (KVL): The algebraic sum of the changes in potential around any closed path of a closed circuit is equal to zero . In mathematical terms, this statement can be expressed as:  +  IR = 0 ……………… (2) Procedure Circuit 1: 1) Click on the following link from PHET Colorado Simulation to open the lab https://phet.colorado.edu/sims/html/circuit-construction-kit-dc/latest/circuit-construction-kit- dc_en.html 2) Choose Conventional Current 3) Use the components in the left side to build the circuit shown below: Note* there is a larger resistor in the components menu. R 1 R 2 R 3  1  2

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Physics Lab (Online Simulation) 13 4) Click on the resistor (R 1 ) and fix it at 1000Ω. That is R 1 = 1000Ω. 5) R 2 = 1000Ω and R 3 = 100Ω 6) Click on the Battery to the left (  1 ) and fix it at 9V,  1 = 9V. 7) Click on the Battery to the right (  2 ) and fix it at 6V,  2 = 6V 8) Click on the Voltmeter from the right side and drag it to measure (V 1 ) the voltage across R 1 , (V 2 ) the voltage across R 2 and (V 3 ) the voltage across R 3 . 9) Click on the Ammeter from the right side and drag it and put it in series with R 1 to measure (I 1 ), with R 2 to measure (I 2 ) and with R 3 to measure (I 3 ) 10) Record the values (I 1 , I 2 , I 3 , V 1 , V 2 and V 3 ) into table 1. Table 1 Show your mathematical calculations here: - Va -9/1000 + Va/100 + Va-6/1000 = 0 (Va-9) + 10Va + (Va-6) = 0 Va = 1.25V  1 = 9V ……………………………  2 = 6V …… ......................... Experimental results Calculated results (use the loops) R (  ) I (A) V (volt) R (  ) I (A) V (volt) 1000 0.008 7.75 1000 0.00775 7.75 1000 0.005 4.75 1000 0.00475 4.75 100 0.013 1.25 100 0.0125 1.25

Physics Lab (Online Simulation) 14 For current: I1 = 9 – 1.25/1000 = 0.00775A I2 = 6 – 1.25/1000 = 0.00475A I3 = 1.25/100 = 0.0125A Voltage: V1 = 0.00775*1000 = 7.75V V2 = 0.00475*1000 = 4.75V V3= 0.0125*100 = 1.25V Questions: 1) Using your experimental results of I 1 , I 2 and I 3 from Table 1 and Circuit 1, verify the Kirchhoff’s current law (KCL) . - Verify the Kirchhoff’s current law: - At node A, current I3= I1 + I2 0.013A = 0.008A+ 0.005A 0.013A = 0.013A KCL is satisfied 2) Using your experimental results of V 1 , V 2 and V 3 from Table 1 and Circuit 1, verify Kirchhoff’s voltage law (KVL) . - I1R1 – I2R2 = 3 0.008*1000-0.005*1000= 3 8-5=3 KVL is satisfied Conclusions:

Physics Lab (Online Simulation) 15 • (for Ohm’s Law) • (for Kirchoff’s laws) - From observing above results ,Ohm's law is verified and KVL is verified.(ohm's law V=IR is constant). Ohm’s law draws a relationship between voltage, current, and resistance. The ohm’s law formulas can be used to deduce the current flowing in a conductor or the resistance or the voltage if any of the two values is known. -Each circuit setup in the experiment clearly showed Kirchhoff's Laws in a physical example. By taking measurements of values and computing the same values, it was easy to see how Kirchhoff's Laws relate to the circuits. Circuit 2: (Extra credit: 15 points) Build the follwing cicuit. Note* you can reset the screen if it’s easeier by clicking the reset button on the bottom right corner. R 1 R 3 R 2  1  2

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Physics Lab (Online Simulation) 16 1) Click on the resistor (R 1 ) and fix it at 5600Ω. That is R 1 = 5600Ω. 2) R 2 = 3300Ω and R 3 = 1800Ω 3) Click on the Battery to the left (  1 ) and fix it at 120V,  1 = 120V. 4) Click on the Battery to the right (  2 ) and fix it at 100V,  2 = 100V 5) Click on the Voltmeter from the right side and drag it to measure (V 1 ) the voltage across R 1 , (V 2 ) the voltage across R 2 and (V 3 ) the voltage across R 3 . 6) Click on the Ammeter from the right side and drag it and put it in series with R 1 to measure (I 1 ), with R 2 to measure (I 2 ) and with R 3 to measure (I 3 ). Record the values (I 1 , I 2 , I 3 , V 1 , V 2 and V 3 ) into table 2. Table 2  1 = 120V  2 = 100V Experimental results Calculated results (use the loops) R (  ) I (A) V (volt) R (  ) I (A) V (volt) 5600 0.016 88.63 5600 0.0158 88.48 3300 0.01 31.37 3300 0.0095 31.35 1800 0.006 11.37 1800 0.0063 11.34

Physics Lab (Online Simulation) 17 Show your calculations here: Mesh analysis in part A 5,600i – 1 – 3300(i1-i2) + 120 = 0. (1) 8900i – 3300i2 = -120 (2) Apply mesh analysis in part B: -120 + 3300(i2-i1) +1800i2 + 100 = 0. (1) -3300i1 + 5100i2 = 20. (2) Solving equation (2) from both part We got: i1 = 0.0158A. And i2 = 0.0063A, i3 = 0.0095A

Lab 4

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