Lab04SeriesandParallelResistorscon

Korsen Miller and David Morgan Physics 212.19 February 10, 2022 Lab 04 - Series and Parallel Resistors Analysis of Circuits 1 and 2: 1. Looking at the two circuits and knowing the definition of parallel and series circuits, it seems that circuit one will be the series circuit and circuit two will be the parallel circuit. Circuit 1 - 2. The Value of the current through each resistor can be calculated using the following equations: V = I ( R 1 + R 2 + R 3 )−→ 3 V = I ( 10 + 3 + 15 )−→ I = 0.11 A For series circuit to calculate the equivalent resistor is the sum of the individual resistors. For a series connection, the total current flowing through the power supply into the entire circuit is constant throughout. 3. To find the voltage across each resistor, we would multiply the current by the resistance of each resistor. So for the 15 ohm resistor, we would have a voltage of 1.61 V, for the 3 ohm the voltage would be 0.32 V, and for the 10 ohm resistor, the voltage would be 1.07 V. 4. Since this is a series circuit and the amps remain constant, the total current flowing shall remain 0.11 A throughout the entire circuit. 5. To find the power dissipated in each resistor, we can use the following equation: ΔV ¿ 2 / R = IΔV P = I 2 R = ¿ in these equations we see that the power in watts is equal to the current (amps) squared times resistance (ohms) which is the same as voltage (V) squared divided by resistance (ohms) which is the same as current (amps) times voltage(V). The power dissipated through the 15 ohm resistor would be 0.1815 W, the 3 ohm resistor would have 0.0363 W dissipated, and the 10 ohm resistor would have 0.121 W dissipated. Circuit 2 - 1. The value of current through each resistor on a parallel series can be calculated by multiplying the current by the resistance for that resistor. V = IR

The value of the current for the 12 ohm resistor would be 3V/12 ohms= 1/4 A, the value of the current for the 30 ohm resistor would be 1/10 A, and the value of the current for the 20 ohm resistor 3/20 A. 2. The voltage across each resistor, as it being a parallel circuit, is constant throughout the circuit, meaning constant as the value from the power supply, so 3 volts. 3. To find the total current flowing through the power supply into the entirety of a parallel circuit, we would ass the current from each resistor and that total is the total current through the circuit, which would be 0.5 A. 4. To find the power dissipated at each resistor, we would square the current and multiply it by the resistance, similar to that of the series connection. The 12 ohm resistor would dissipate 0.75 W, the 20 ohm resistor would dissipate 0.45 W, and the 30 ohm resistor would dissipate 0.3 W. Construction of Circuits 1 and 2: We first constructed circuit 1 by connecting our power supply to the resistor, without any meters present. Once we confirmed our wiring and configuration, we connected our meters to the circuit. We used the capstone software to set a power supply voltage of 3V for the whole experiment. The ammeter and the voltmeter were both connected to the 10 ohm resistor. We then measured the current through the resistor, the current through the circuit, the voltage through the resistor and the total voltage through the circuit. This process was repeated and these values were recorded for each resistor in both a series setup and a parallel setup. For the parallel setup, the voltmeter was set up in the parallel style and the ammeter was set up in a series style. The circuit setups for each resistor in each configuration are as follows: Overall series circuit: