PHYS 122 Lab 7 Draft Report

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Name(s): Kirsten North Date: 10/18/2023 RC Circuits Lab Purpose The lab's purpose involves constructing an RC circuit and investigating the time-dependent voltage dynamics across the capacitor in both charging and discharging RC circuits. Furthermore, the objective includes creating graphical representations of the V(t) function for charging and discharging RC circuits, which will be compared to experimental data, enabling a comprehensive analysis of their behavior. Apparatus (include photo of built circuit) ● EM 8675 PASCO kit ● 1F Capacitor, 47 ࠵? resistor, wires with alligator clips ● 1 x 1.5 V batteries ● Multimeter ● Capstone software for plotting data Procedure Charging Cycle: 1. Build a series RC circuit by connecting the 1F Capacitor in series with the 47 Ω resistor and a single 1.5 V battery. 2. Using a multimeter and a timer, measure the voltage across the capacitor at approximately 10-second intervals until the capacitor is fully charged. 3. Include a data table in your report (data section) to record these measurements. Plot the measured voltages vs. time (data evaluation section). On the same plot, use the 'user- defined' function in Capstone to plot the expected charging function: ࠵?(࠵?) = ϵ(1 − e ! ! " ) where τ is the time constant. 4. Calculate the time constant (τ) for your RC circuit and include the formula for τ along with your calculation in your report. 5. Using the equation for the voltage across the charging capacitor, calculate the expected voltage at times t = τ, 2τ, 3τ, 4τ, and 5τ, and include these values in your report as a table in the data evaluation section. 6. Calculate the ratio of the expected voltage drop across the capacitor to the battery voltage at t = τ, 2τ, 3τ, 4τ, and 5τ, and add this column to the table in your data evaluation section. 7. Qualitatively describe what happens to the voltage across the capacitor during the charging portion of the experiment.

Name(s): Kirsten North Date: 10/18/2023 Discharging Cycle: 1. Using a multimeter and a timer, measure the voltage across the capacitor at approximately 10-second intervals until the capacitor is fully discharged. 2. Include a data table in your report to record these measurements in the data section. Plot the measured voltages vs. time (data evaluation section). On the same plot, use the 'user- defined' function in Capstone to plot the expected discharging function: ࠵?(࠵?) = V " ∗ e ! ! " where V₀ is the initial voltage. 3. Using the equation for the voltage across the discharging capacitor, calculate the expected voltage at times t = τ, 2τ, 3τ, 4τ, and 5τ, and include these values in your report as a table in the data evaluation section. 4. Calculate the ratio of the expected voltage drop across the capacitor to the initial voltage at t = τ, 2τ, 3τ, 4τ, and 5τ, and add this column to the table in your data evaluation section. 5. Qualitatively describe what happens to the voltage across the capacitor during the discharging portion of the experiment. Data Charging graph: Discharging graph:

Name(s): Kirsten North Date: 10/18/2023 Evaluation of Data Given values from the lab: R= 47 Ω V " = 2V C= 1.0F Charging Cycle: The equation for the voltage across the charging capacitor is given below: ࠵?(࠵?) = ϵ(1 − e ! # $ ) It can also be written differently to substitute and solve easier with the given values as seen below: ࠵?(࠵?) = V " (1 − e ! # %& ) To calculate expected voltage across the capacitor at specific times, substitute values of t in the equation and solve for V(t). Time Substitution in Equation Expected Voltage (V)

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Name(s): Kirsten North Date: 10/18/2023 ࠵? ࠵?(࠵?) = 2(1 − e ! ’( ’( ) 1.26V 2 ࠵? ࠵?(࠵?) = 2(1 − e ! )’ ’( ) 1.73V 3 ࠵? ࠵?(࠵?) = 2(1 − e ! *’* ’( ) 1.90V 4 ࠵? ࠵?(࠵?) = 2(1 − e ! *++ ’( ) 1.96V 5 ࠵? ࠵?(࠵?) = 2(1 − e ! ,-. ’( ) 1.99V To calculate the ratio of the expected voltage drop, divide the voltage across the capacitor by the initial voltage across the capacitor. V & V " Time Substitution in Equation Expected Voltage Drop (V) ࠵? V & V " = 1.26 2 0.63V 2 ࠵? V & V " = 1.73 2 0.87V 3 ࠵? V & V " = 1.90 2 0.95V 4 ࠵? V & V " = 1.96 2 0.98V 5 ࠵? V & V " = 1.99 2 0.99V During the initial phase of the experiment focused on charging, the voltage across the capacitor initiates from a baseline of 0 volts and progressively goes up towards the battery voltage. How fast this voltage goes up depends on something called the "time constant." The time constant is figured out by looking at the resistance and capacitance in the circuit, and it tells us how quickly the voltage goes up. At first, when charging, the voltage on the capacitor goes up really quickly. This is because there is a big difference in voltage between the battery and the capacitor. But as the capacitor gets closer to the battery's voltage, it goes up more slowly. Eventually, the voltage on the capacitor stops changing and stays the same as the battery's voltage, which means the charging is finished. Discharging Cycle: The equation for the voltage across the discharging capacitor is given below: ࠵?(࠵?) = V " ∗ e ! # $ To calculate expected voltage across the capacitor at specific times, substitute values of t in the equation and solve for V(t).

Name(s): Kirsten North Date: 10/18/2023 Time Substitution in Equation Expected Voltage (V) ࠵? ࠵?(࠵?) = 2 ∗ e ! ’( ’( 0.74V 2 ࠵? ࠵?(࠵?) = 2 ∗ e ! )’ ’( 0.27V 3 ࠵? ࠵?(࠵?) = 2 ∗ e ! *’* ’( 0.09V 4 ࠵? ࠵?(࠵?) = 2 ∗ e ! *++ ’( 0.04V 5 ࠵? ࠵?(࠵?) = 2 ∗ e ! ,-. ’( 0.01V To calculate the ratio of the expected voltage drop, divide the voltage across the capacitor by the initial voltage across the capacitor. V & V " Time Substitution in Equation Expected Voltage Drop (V) ࠵? V & V " = 0.74 2 0.37V 2 ࠵? V & V " = 0.27 2 0.14V 3 ࠵? V & V " = 0.09 2 0.05V 4 ࠵? V & V " = 0.04 2 0.02V 5 ࠵? V & V " = 0.01 2 0.01V In the beginning of the discharging cycle within the experiment, the voltage across the capacitor starts from the same level it reached at the end of the charging phase. This voltage is equal to the battery voltage which is 2V. The rate at which this voltage decreases is also influenced by the time constant. Just like in the charging cycle, the time constant helps us understand how quickly the voltage changes. At the beginning of the discharging process, the voltage on the capacitor drops rapidly. This fast decrease is because there is a considerable voltage difference between the charged capacitor and the battery. However, as the capacitor's voltage approaches zero, it decreases more slowly. Ultimately, the voltage on the capacitor stabilizes at zero, signifying the completion of the discharging cycle. Conclusion Charging cycle:

Name(s): Kirsten North Date: 10/18/2023 - Using an equation editor, write the formula for the voltage across the capacitor, current through the resistor, and charge on the capacitor as a function of time for the charging cycle. Describe each formula using a complete sentence(s). The voltage (V) across the charging capacitor as a function of time (t) is given by the formula: ࠵?(࠵?) = ϵ(1 − e ! ! " ) , where V(t) is the voltage at time t, ϵ is the battery voltage, and τ is the time constant of the RC circuit. The formula for voltage (V(t)) across the charging capacitor is an exponential growth function. It represents the buildup of voltage across the capacitor as it charges. The current (I) through the resistor during the charging cycle can be expressed as: ࠵?(࠵?) = / % ∗ e ! ! " , where I(t) represents the current at time t, ϵ is the battery voltage, R is the resistance, and τ is the time constant. The equation for current (I(t)) during the charging cycle is also exponential, and it describes how the current rapidly decreases as the capacitor charges. The formula indicates that the current starts high and gradually decreases as the capacitor charges. The charge (Q) on the capacitor as a function of time is represented by: ࠵?(࠵?) = ϵ ∗ C ∗ (1 − e ! ! " ) , where Q(t) denotes the charge at time t, ϵ is the battery voltage, C is the capacitance, and τ is the time constant. It represents how the charge on the capacitor increases over time, showing that the capacitor accumulates charge as it charges. - Why does the current eventually stop flowing in a charging RC circuit? In a charging RC circuit, the current eventually stops flowing because, as time progresses, the voltage across the capacitor (V(t)) gradually approaches the battery voltage (ϵ). As V(t) gets closer to ϵ, the potential difference across the resistor decreases, leading to a decrease in the rate of change of current (dI/dt) through the resistor. Discharging cycle: - Using an equation editor, write the formula for the voltage across the capacitor, current through the resistor, and charge on the capacitor as a function of time for the discharging cycle. Describe each formula using a complete sentence(s). The voltage (V) across the discharging capacitor as a function of time (t) is given by the formula: ࠵?(࠵?) = V " ∗ e ! ! " , where V(t) is the voltage at time t, V₀ is the initial voltage across the capacitor, and τ is the time constant. As time (t) progresses, V(t) diminishes exponentially, approaching zero. This formula demonstrates the decrease in voltage on the capacitor as it discharges.

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Name(s): Kirsten North Date: 10/18/2023 The current (I) through the resistor during the discharging cycle can be expressed as: ࠵?(࠵?) = 1 # % ∗ e ! ! " , where I(t) represents the current at time t, V₀ is the initial voltage, R is the resistance, and τ is the time constant. It represents how the current decreases rapidly, starting with a high value and gradually approaching zero as the capacitor discharges. The charge (Q) on the capacitor as a function of time is represented by: ࠵?(࠵?) = Q " ∗ e ! ! " , where Q(t) denotes the charge at time t, Q₀ is the initial charge on the capacitor, and τ is the time constant. The formula for the charge on the capacitor during the discharging cycle follows an exponential decrease. It shows how the charge on the capacitor decreases over time, reaching zero as the discharging process completes.