Lab Report 1_ DC Circuits (2)

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Apr 3, 2024

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Lab 1 Report Date: January 31 st , 2022 Name: Faraz Nazemi Group members’ names: 1. 2. 3. 4. Lab Activity 1 – Power Supply/Ohm’s Law. a. Results for a commercial resistor ΔV (V) I (A) Resistance, based on Ohm’s Law R = ΔV/I (Ω) R (Ω) Measured with DMM as an ohmmeter R (Ω) Based on colour code Tolerance based on colour code % and in (Ω) 1.0 0.02 50 65.56 No colour as it starts a 1 V and becomes very light colour as it reaches 4 V 5% 2.0 0.03 66.67 4.0 0.05 80 Is this resistor linear or non-linear element? This resistor is a linear element because as voltage is added to the circuit, the resistance I constant rather than staying the same. Does the resistance obtained using Ohm’s law, R = ΔV/I, agree with the value specified by the colour code (within the precision limit)? The resistance obtained using Ohm’s law does agree with the value specified by the colour code within the precision limit. The resistance obtained using Ohm’s law is 65.56 and the colour code is light yellow which gives us a 5% code tolerance. What is the resistance measured with ohmmeter? The meter which measures the resistance and the continuity of the electrical circuit and their component. This is known as the ohmmeter. It shows the flow of resistance, throughout the whole circuit and the different parts of it. Which calculates how much resistance it maintains. b. Summary of results for the electric bulb

ΔV (V) I (A) R = ΔV/I (Ω) Linear or non-linear 1.0 0.07 14.29 This circuit is nonlinear because the resistant is not constant 2.0 0.1 20 4.0 0.3 13.33 Lab Activity 2 – Load Resistors in Series and Parallel Summary of results for the circuit containing a resistor and four bulbs. I (A) ΔV (V) Power P = (I)(ΔV) (W) Order bulbs by light intensity Bulb 1 5 20 100 Highest, 1 Bulb 2 4 20 80 High, 2 Bulb 3 5 20 100 Lowest, 4 Bulb 4 4 20 80 High, 3 Which bulb shines brightest? Was your prediction correct? I personal thought that bulb number 1 would be shine the brightest because it is the first bulb close to the source, therefore I thought bulb 1 would shine the brightest. Hence, my prediction were correct for this circuit. What is the relationship between currents in light bulbs 1, 2, 3 and 4? Hint: Think about the Kirchhoff’s junction rule. Kirchhoff’s junction law stated that the sum of all currents entering a junction must equal the sum of all currents leaving the junction. Kirchhoff’s second rule stated that the loop rule is the algebraic sum of changes in potential around any closed circuit path (loop) must be zero. I believe the relationship between all 4 bulbs is that for a parallel path the total current entering a circuit’s junction is exactly equal to the total current leaving the same junction. This is because it has no other place to go as no charge is lost. This means that all 4 bulbs have the same amount of current. What is the relationship between the potential difference across bulbs 2, 3 and 4? The electric potential difference across each resistor (ΔV) is the same. The current in a resistor follows Ohm's law. Since the ΔV is the same for each resistor, the current will be smallest where the resistance is greatest. Therefore, all 3 bulbs gain the same amount of voltage in order to process and turn on. Is the sum of potential differences across the resistor, bulb 1 and bulb 4 equal to the potential difference at the output of the power supply? The sum of potential differences across each resistors equals to the total potential differences across the whole chain. Therefore they are both equal. Hence, the potential difference at the output of the power supple is maintained equally throughout the whole circuit.

For bulb 1, determine absolute errors for current I and voltage ΔV. Voltage errors; 20V(100%) =0.3+- therefore this is the absolute error for Voltage Current error: 5(100%)= 0.5+- therefore this is the absolute error for current

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