LAB-7 physics

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103

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Electrical Engineering

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May 3, 2024

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Lab-7 Combined Series and Parallel Lab Name__________________________________ Instructions : Construct each of the circuits below using the PhET Circuit Simulation. Each light bulb/resistor is 10 Ω by default. The battery has a potential difference of 9 V by default. Complete the RVIP charts mathematically and check your answers with the “Non-Contact Ammeter” and “Voltmeter.” Then answer the questions following each diagram. A. Series Circuit in a Parallel Circuit R V I P 1 10 9 9 8.1 2 10 4.5 0.45 2.025 3 10 4.5 0.45 2.025 12 20 9 0.45 battery 6.67 9 1.35 12.15 1. On a separate sheet of paper, draw a schematic version of this circuit ( and ) and then draw simplified versions to solve. 2. Explain which part of the circuit is in series. Explain which part of the circuit is in parallel. In the circuit, Resistors 2 and 3 are placed one after another on the same branch, which means they are connected in series. This arrangement causes the same current to flow through both resistors. On the other hand, Resistor 1 is located on a separate branch that is parallel to the branch containing Resistors 2 and 3. In this configuration, Resistor 1 shares the same two endpoints as Resistors 2 and 3, which means that the voltage across Resistor 1 is the same as the voltage across Resistors 2 and 3. However, the current through Resistor 1 is different because it is not on the same branch as Resistors 2 and 3. 3. Compare the current in the top branch to the current in the middle branch. Explain why. The current flowing through the top branch of a circuit is twice as much as the current flowing through the middle branch. This is because the top branch has half the resistance of the middle branch. As per Ohm's law, when the R 2 S 3 S 2 S 1 R 3 R 1

resistance of a circuit decreases, the current through it increases proportionally. Thus, the current loop for the top branch will be twice as strong as the middle branch due to its lower resistance. 4. Rank the light bulbs in order of brightness. In terms of current flow and resistance, explain why. Less resistance means a higher current going through it. The greater current will increase the brightness of the bulb. If we consider R1=R2=R3 then the large current will pass through R3 as compared with R1 and R2. So R3 will have the greatest brightness and R1&R2 will have the same brightness as each other, both less than R3. 5. If bulb R 2 were removed (right-click to remove), explain what happens to the other two bulbs and why. R 1 would stay on and would be the exact same brightness because its current loop was completely unaffected by R 2 . However, R 3 would turn off because the current flow within the middle branch would come to a complete stop. 6. Determine which bulbs are affected by each of the switches (S 1 , S 2 , S 3 ). Explain why. B. Parallel Circuit in a Series Circuit R (Ω) V (V) I (A) P (W) 1 10 3 0.3 0.9 2 10 3 0.3 0.9 3 10 6 0.6 3.6 12 5 3 0.6 battery 15 9 0.6 5.4 7. Explain which part of the circuit is in series. Explain which part of the circuit is in parallel. 8. Rank the light bulbs in order of brightness. In terms of current flow and resistance, explain why. R1 and R2 are said to be in parallel because they are connected to separate branches of the circuit. This means that the current flowing through R1 is separate from the current flowing through R2. On the other hand, R3 is connected in series with both R1 and R2. This means that if R3 were to be removed from the circuit, the current flow in both R1 and R2 would stop, which would, in turn, turn off the entire circuit. Therefore, R3 plays a crucial role in the functioning of the circuit. 9. Compare the potential difference across R 3 to the potential difference across the other two bulbs. Explain why. R 3 > R 1 = R 2 R 3 receives all of the current coming out of the battery, while R 1 and R 2 have the current split between them. 10. If R 3 were removed (right-click to remove), explain what happens to the other two bulbs and why. This circuit has three resistors: R1, R2, and R3. R3 has twice the resistance of R1 and R2 combined. In a series circuit, the potential difference is divided among the resistors. However, a resistor with a higher resistance will draw more potential difference. R3, having double the resistance, will draw a greater percentage of the potential R 3 R 2 R 1

difference. This helps predict the distribution of potential differences and how it affects the circuit's overall operation. 11. If R 2 were removed, what kind of circuit does this become? The other bulbs would both turn off because R 3 controls the entire flow of electrons through the rest of the circuit. 12. After R 2 is removed, determine what happens to the brightness of each bulb and explain why. (Hint: Complete the chart if you get stuck. Due to the changes in the circuitry, Resistor 1 now appears brighter as it no longer has to share any current with Resistor 2. This is because the current flow through it has increased. On the other hand, Resistor 3 appears dimmer because the overall resistance of the circuit has increased. As a result, the current's value has gone down, making it appear as if Resistor 3 has decreased in brightness. This is because the brightness of a bulb is directly proportional to the current flow through it, and in this case, the current has decreased due to the increase in resistance. R (Ω) V (V) I (A) P (W) 1 10 4.5 0.45 2.025 3 10 4.5 0.45 2.025 battery 20 9 0.45 4.05

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