LAB253 Troy University Resistance and Capacitance in A Circuit Analysis Please answer all questions in the RC circuit document. include the question in Ana

LAB253 Troy University Resistance and Capacitance in A Circuit Analysis Please answer all questions in the RC circuit document. include the question in Analysis part.

Resistance and capacitance in a circuit

When resistance and capacitor are connected in series, capacitor acts as a device for storing charge. The amount of charge stored is directly proportional to voltage applied, Q = CV. When battery in turned on, capacitor starts charging until it reaches to maximum voltage supplied by battery. Since capacitor is storing the charge so amount of current in the circuit keeps decreasing across the resistance. Once the capacitor is fully charged and disconnected from battery, it will discharge through the resistance. When the amount of charge stored in capacitor decreases, the amount of current through the resistance is also decreased. During charging and discharging of the capacitor, the current will decrease exponentially as the function of time. The time in which the current reduced to 1⁄2 of its maximum value is defined as half life. The half life can be calculated using the formula

τ1/2 =ln2RC

You will measure be two RC decay half-lives. The first will be long enough to measure using a multimeter and a stopwatch.Step 1

Set up the circuit below with the power supply off.

Step 2

Turn on the power supply, adjust it to 2-3 volts and start the stopwatch. Record the time and 5000Ω and 20000μF voltage every 15 seconds for 5 minutes.

Step 3

Disconnect the wires from the power supply and short them together. Again, record the time and voltage every 15 seconds for 5 minutes.

Step 4

Plot the data from step 1 and 2 on separate plots. Find an experimental half-life from both plots.

Step 5

Using that resistance and your measured half-life (from step #4) get an experimental value for capacitance of C which should match with theoretical value given on the capacitor( 0.2 F). 

Setup

1. Construct the circuit shown in Figure 2. The voltage source is Signal Generator #1 on the 850 Universal Interface. C = 3900 pF and R = 47 kΩ.

Figure 2. RC Circuit Diagram

2. Click on Signal Generator #1 to connect the internal Output Voltage-Current Sensor. Set the signal generator to a 350 Hz square wave with 2 V amplitude and 2 V offset. This will make the square wave all positive with an amplitude of 4 V. Set the signal generator on Auto.

3. Plug the Voltage Sensor into Channel A. Connect the Voltage Sensor across the capacitor.

Procedure

1. Set up an oscilloscope display with the Voltage Ch.A and the Output Voltage on the same axis. Click Monitor and adjust the scale on the oscilloscope so there is a complete cycle, so the capacitor fully charges and discharges.

2. Increase the number of points (using the tool on the scope toolbar) to the maximum allowed. Then take a snapshot of both voltages shown. Rename the snapshots “3900pF”.

Analysis

1. Create a graph with Voltage Ch.A and the Output Voltage vs. time. Select the voltages for the 3900 pF run on the graph.

2. Using the Coordinates Tool, measure the time it takes for the voltage to decay to half of its maximum. This time is the half-life. It may be necessary to reduce the snap-to-pixel distance to 1 in the properties of the Coordinates Tool (right click on the tool to access the properties).

3. Measure the time it takes for the voltage to decay to one-quarter of its maximum. This is two half-lives. Then divide this time by two to find the half-life.

4. Measure the time it takes for the voltage to decay to one-eighth of its maximum. This is three half-lives. Then divide this time by three to find the half-life. Take the average of the three measured values of the half-life. Estimate the precision of the measurement and state it as {half-life ± precision}.

5. Calculate the theoretical half-life given by Equation (12) and compare it to the measured value using a percent difference.

Conclusions

1. Summarize how changing the voltage and capacitance changes the half-life.

2. Include the values found for the half-lives and the % differences. Does the theoretical value lie within the range of precision of your measurements? Explain what causes the differences.

3. Did your answers to the Pre-Lab Questions agree with the results? Page #1 Snapshot
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Discharge Recharge
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Discharge plot
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Volts(V)
T(sec)
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Time(s)
Recharge plot
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PHY 253/263
Troy University
Dept. of Chemistry and Physics
RC Circuit
Introduction
The manner by which the voltage on a capacitor decreases is studied. The half-life for the decay
is measured directly and also calculated using the capacitive time constant.
Theory
Capacitors are circuit devices that can store charge. The capacitance (size) of the capacitor is a
measure of how much charge it can hold for a given voltage.
Q = CVC
(1)
where C is the capacitance in Farads, Q is the charge in Coulombs, and VC is the voltage across
the capacitor in Volts.
Figure 1. RC Circuit Voltages
To determine how the charge on a capacitor decays in time, use Kirchhoff’s Loop Rule for Figure
1:
Vo = VC + VR
(2)
Solving Equation (1) for the voltage across the capacitor gives
VC = Q⁄C
(3)
The voltage across the resistor is given by Ohm’s Law:
VR = IR
Therefore,
(4)
PHY 253/263
Troy University
Vo = Q⁄C + IR
Dept. of Chemistry and Physics
(5)
Since the applied voltage, V₀, is zero when the capacitor is discharging, Equation (5) reduces to
Q⁄C + IR = 0
(6)
Since the current is
dQ
I = dt
(7)
Equation (6) becomes the differential equation
dQ
dt
1
+ ( )Q = 0
(8)
RC
Solving Equation (8) for Q gives
t
Q = Qmax e−(RC)
(9)
Plugging Q into Equation (2) gives the voltage across the capacitor as a function of time
t
V(t) = Vo e−(RC)
(10)
where Vo = Qmax/C. The rate that voltage across a capacitor (and the charge stored in the
capacitor) decreases depends on the resistance and capacitance that are in the circuit. If a
capacitor is charged to an initial voltage, Vo, and is allowed to discharge through a resistor, R,
the voltage, V, across the capacitor will decrease exponentially.
The half-life, t1/2 is defined to be the time that it takes for the voltage to decrease by half:
V(t ½ ) = Vo ⁄2 = Vo e−
t½⁄
RC
Solving for the half-life gives t½ = RC ln 2.
(11)
(12)
The product RC is called the capacitive time constant and has the units of seconds.
PHY 253/263
Troy University
Dept. of Chemistry and Physics
Pre-lab Questions
1. Show that the capacitive time constant RC has units of seconds.
2. If the capacitance in the circuit is doubled, how is the half-life affected?
3. If the resistance in the circuit is doubled, how is the half-life affected?
4. If the charging voltage in the circuit is doubled, how is the half-life affected?
5. To plot the equation V(t) = Voe− RC so the graph results in a straight line, what quantity do
you have to plot vs. time? What is the expression for the slope of this straight line?
t
Resistance and capacitance in a circuit
When resistance and capacitor are connected in series, capacitor acts as a device for storing
charge. The amount of charge stored is directly proportional to voltage applied, Q = CV. When
battery in turned on, capacitor starts charging until it reaches to maximum voltage supplied by
battery. Since capacitor is storing the charge so amount of current in the circuit keeps
decreasing across the resistance. Once the capacitor is fully charged and disconnected from
battery, it will discharge through the resistance. When the amount of charge stored in capacitor
decreases, the amount of current through the resistance is also decreased. During charging and
discharging of the capacitor, the current will decrease exponentially as the function of time. The
time in which the current reduced to 1⁄2 of its maximum value is defined as half life. The half life
can be calculated using the formula
τ1/2 =ln2RC
You will measure be two RC decay half-lives. The first will be long enough to measure using a
VR
multimeter
multimeter and a stopwatch.
Step 1
Set up the circuit below with the power supply off.
Step 2
Turn on the power supply, adjust it to 2-3 volts and start the stopwatch. Record the time and
5000Ω and 20000μF voltage every 15 seconds for 5 minutes.
PHY 253/263
Troy University
Dept. of Chemistry and Physics
Step 3
Disconnect the wires from the power supply and short them together. Again, record the time and
voltage every 15 seconds for 5 minutes.
Step 4
Plot the data from step 1 and 2 on separate plots. Find an experimental half-life from both plots.
Step 5
Using that resistance and your measured half-life (from step #4) get an experimental value for
capacitance of C which should match with theoretical value given on the capacitor( 0.2 F).
PHY 253/263
Troy University
Dept. of Chemistry and Physics
Setup
1. Construct the circuit shown in Figure 2. The voltage source is Signal Generator #1 on the 850
Universal Interface. C = 3900 pF and R = 47 kΩ.
Figure 2. RC Circuit Diagram
2. Click on Signal Generator #1 to connect the internal Output Voltage-Current Sensor. Set the
signal generator to a 350 Hz square wave with 2 V amplitude and 2 V offset. This will make the
square wave all positive with an amplitude of 4 V. Set the signal generator on Auto.
3. Plug the Voltage Sensor into Channel A. Connect the Voltage Sensor across the capacitor.
Procedure
1. Set up an oscilloscope display with the Voltage Ch.A and the Output Voltage on the same
axis. Click Monitor and adjust the scale on the oscilloscope so there is a complete cycle, so
the capacitor fully charges and discharges.
2. Increase the number of points (using the tool on the scope toolbar) to the maximum allowed.
Then take a snapshot of both voltages shown. Rename the snapshots “3900pF”.
Analysis
1. Create a graph with Voltage Ch.A and the Output Voltage vs. time. Select the voltages for
the 3900 pF run on the graph.
2. Using the Coordinates Tool, measure the time it takes for the voltage to decay to half of its
maximum. This time is the half-life. It may be necessary to reduce the snap-to-pixel distance
to 1 in the properties of the Coordinates Tool (right click on the tool to access the properties).
3. Measure the time it takes for the voltage to decay to one-quarter of its maximum. This is two
half-lives. Then divide this time by two to find the half-life.
PHY 253/263
Troy University
Dept. of Chemistry and Physics
4. Measure the time it takes for the voltage to decay to one-eighth of its maximum. This is three
half-lives. Then divide this time by three to find the half-life. Take the average of the three
measured values of the half-life. Estimate the precision of the measurement and state it as
{half-life ± precision}.
5. Calculate the theoretical half-life given by Equation (12) and compare it to the measured
value using a percent difference.
Conclusions
1. Summarize how changing the voltage and capacitance changes the half-life.
2. Include the values found for the half-lives and the % differences. Does the theoretical value
lie within the range of precision of your measurements? Explain what causes the differences.
3. Did your answers to the Pre-Lab Questions agree with the results?

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