Lab 7- Capacitors in Circuits - using PhET simulation
pdf
keyboard_arrow_up
School
Lone Star College System, Woodlands *
*We aren’t endorsed by this school
Course
2426
Subject
Electrical Engineering
Date
Dec 6, 2023
Type
Pages
8
Uploaded by ColonelOwl882
OL-39
CAPACITORS IN CIRCUITS ONLINE LAB
7/1/2020
OBJECTIVES
The purpose of this lab will be to determine how capacitors behave in R-C circuits by measuring
the time for charging and discharging. The manner in which capacitors combine will also be
studied.
EQUIPMENT
PhET interactive simulation tool [Circuit Construction Kit: (AC+DC) - Virtual Lab]
https://phet.colorado.edu/en/simulation/legacy/circuit-construction-kit-ac-virtual-lab
See: 00-PhET Simulation Tool instructions for Electric Circuits Labs, on how to set up the circuit in
PhET.
THEORY
An RC circuit is one in which we have a resistor in series with a capacitor (Figure 1). In this figure,
the battery is not connected to the circuit, and there is no charging of the capacitor. Assume that
the capacitor is initially completely un-charged (which can be done by connecting both ends of
the capacitor with a piece of wire). The swit
ch is then thrown to position ‘a’ at time
t
= 0 s (figure
2). This starts charging the capacitor, and the rate of charging is given by:
𝑞(𝑡) = ? (1 − 𝑒
−
𝑡
𝑅𝐶
)
(1)
Where
q(t)
is the charge at time
t
after charging starts,
R
and
C
are the values of the resistance
and capacitance.
Q
is the maximum charge that can be stored on the capacitor, and is given by
? = 𝐶 ∆𝑉
𝐶
(2)
𝜏 = ? 𝐶
(3)
Where
C
is the Capacitance of the capacitor, and
ΔV
is the applied potential across the
capacitor (=
ε
volts). The term
RC
is called the time constant (
τ
) of the
RC
circuit, and is the time
taken for the capacitor to charge to 63.2% of its maximum possible value.
The capacitor charges with time as shown in Figure 3. Even though equation (1) and this curve
indicate that the capacitor will never be fully charged, for most practical purposes we consider
that the capacitor is fully charges after a time that is more than 5 time-constants have passed.
To discharge the capacitor through the resistor, we move the switch to position ‘b’, as in Figure 4.
The capacitor discharges according to equation (3). The charge decreases exponentially as
shown in Figure 5.
𝑞(𝑡) = ? 𝑒
−
𝑡
𝑅𝐶
(4)
In this experiment we will determine the time that it takes for the capacitor to charge and
discharge to some value.
From Eqn (2) we see that since the capacitance ‘C’ is constant for a
given capacitor, the charge ’Q’ is proportional to the voltage ‘V’ across the capacitor. So,
instead of measuring the charge, we will measure the voltage across the capacitor, which will be
indicative of the charge. Then the charge will be
𝑞(𝑡) = 𝐶∆𝑉
𝐶
(𝑡)
(5)
Where
∆𝑉
𝐶
(𝑡)
is the voltage drop across the capacitor at time
t
. While charging, when the
potential difference across the capacitor reaches the battery potential, the capacitor is fully
charged. At that point, the potential drop across the resistor is zero and there is no current in the
circuit.
PROCEDURE
1.
Open the simulation by ctrl+click the link, or copy paste the link to the browser. The simulation
should look like that shown in Fig.6
2.
Since this simulation is in java (and not web based as some of the others), you may have to
download the simulation. If you cannot run the simulation, you may need to follow the
following PhET help guidelines: https://phet.colorado.edu/en/help-center/running-sims
Then click
“Why can I
run some of the simulations but not all?”
3.
Run the simulation, and you will see a page like that shown in Fig.7.
4.
You would now set up the circuit. For assistance in setting up the circuit, see the manual on
Ohm’s Law or Kirchhoff’s Rules
.
5.
This experiment requires you to measure the voltage as a function of time. The timer can be
easily controlled by using the
Pause/Play button (►) and/or the step button (|►)
(these are at
the bottom of the page).
Figure 6.
Figure 7.
Case-A: charging the capacitor.
1.
Set up the circuit as shown in figure 8. Once set up, it should look something like that shown in
figure 9.
2.
Set the resistance to 100
Ω, capacitance to
0.05 F, and Battery to 10.0 V.
3.
Before charging the capacitor, make sure that it has no charge (the voltmeter reads zero).
Otherwise you need to discharge the capacitor first until the voltage across the capacitor
becomes zero.
4.
Put switch S
1
in the ON state and switch S
2
to the OFF state.
5.
Set the Pause/Play button (►) to pause and the timer to zero. Before 5 seconds, use the step
button (|►) to increase time by 0.5 second intervals and record the voltage values in Table I.
After 5 seconds, use the Pause/Play button (►/||) to record the volt
age at around 7.00, 10.0,
15.0, 20.0, and 25.0 seconds.
6.
Using equation (5), obtain the charge at each time, and enter in Table 1.
7.
Draw a graph between charge on y-axis and time on x-axis. It should look like Fig. 3.
8.
Use the known values of resistance and capacitance to calculate the time constant and the
maximum charge by using eqn. (2) and eqn. (3), and enter in Table 2.
9.
Calculate the charges equal to one time constant, two time constants, and five time
constants and enter in Table 2. Compare these with the experimental values using % error. Put
your calculation in the table II.
Figure 9.
Case-B: Discharging capacitor
1.
Set up the circuit as shown in figure 8.
2.
Set the resistance to 100 Ω, capacitance to 0.05 F, and Battery to 10.0 V.
3.
Before discharging the capacitor, make sure the capacitor has been fully charged (the
voltmeter reading is very close to 10.0 V).
4.
Set switch
?
1
to off and switch
?
2
to on.
5.
Set the Pause/Play button (►) to pause, and the stopwatch to zero.
For time less than 5
seconds, use the step button (|►) to increase time by 0.5 second intervals. Record the
voltage values in Table 3
. After 5 seconds, use the Pause/Play button (►/
||) to record the
voltage at about 10.0, 15.0, 20.0, and 25.0 seconds.
6.
Using equation (5), obtain the charge at each time, and enter in Table 3.
C
?
∆V
Figure 8
?
1
?
2
Volt-
meter
Your preview ends here
Eager to read complete document? Join bartleby learn and gain access to the full version
- Access to all documents
- Unlimited textbook solutions
- 24/7 expert homework help
7.
Draw a graph between charge on y-axis and time on x-axis. It should look like Fig. 5.
8.
Use the known values of resistance and capacitance to calculate the time constant and the
maximum charge by using eqn. (2) and eqn. (3), and enter in Table 4.
9.
Calculate the charges equal to one time constant, two time constants, and five time
constants and enter in Table 4.
Compare these with the experimental values using % error. Put
your calculation in the table II.
Case-C: Capacitors in Series.
1.
Set up the circuit as shown in figure 10.
2.
Set the resistance to 100
Ω,
each capacitance to 0.05 F, and Battery to 10.0 V.
3.
Before charging the capacitor, make sure that it has no charge (the voltmeter reads zero).
Otherwise you need to discharge the capacitor first until the voltage across the capacitor
becomes zero.
4.
Put switch S
1
in the ON state and switch S
2
to the OFF state.
5.
Now calculate the value of the time constant by using the equation for sum of capacitors in
series.
6.
Start charging the capacitors and note the voltage difference across both capacitors. Note
the time it takes for the voltage to reach 63.2 % of V
max
. This is the measured value of time
constant. Note this in Table 5.
7.
Now charge the capacitors to full charge, and by using proper switching, measure the time
for the voltage across them to fall BY 63.2% of V
max
. This is the measured time constant for
discharging the capacitors.
8.
Compare the measured and calculated values of the time constant for capacitors in series.
Case-D: Capacitors in parallel.
1.
Set up the circuit as shown in figure 11.
2.
Set the resistance to 100
Ω, each capacitance to 0.05 F, and Battery to 10.0 V.
3.
Repeat the steps needed to measure the time constant while charging and while discharging,
and compare with the calculated value for capacitors in parallel.
4.
Enter the results in Table 5.
?
∆V
Figure 11
?
1
?
2
?
∆V
?
1
?
2
C
1
C
2
Figure 10
C
1
C
2
Volt-
meter
Volt-
meter
DATA
Case-A: Data for charging a single capacitor
Table-1
Resistance R = ___100 ohm______
Capacitance C = __0.05F______
Time
(s)
Measured
Voltage
(V
C
)
Charge
on
Capacitor
q(t)
(eqn. (5)
Time
(s)
Measured
Voltage
(V
C
)
Charge
on
Capacitor
q(t)
(eqn. (5)
Time
(s)
Measured
Voltage
(V
C
)
Charge
on
Capacitor
q(t)
(eqn. (5)
0.50
0.968
0.0484
3.00
4.511
0.22555
7.00
7.529
0.37645
1.00
1.795
0.08975
3.50
5.043
0.25215
10.0
8.644
0.4322
1.50
2.59
0.1295
4.00
5.497
0.27485
15.0
9.802
0.4751
2.00
3.309
0.16545
4.50
5.933
0.29665
20.0
9.817
0.49085
2.50
3.921
0.19605
5.00
6.328
0.3164
25.0
9.932
0.4966
Make a graph between q(t) and time.
1s =0.05F* 1.795=0.08975 C
1.5s =0.05F* 2.59=0.1295 C
2s =0.05F* 3.309=0.16545 C
2.5s =0.05F* 3.921=0.19605 C
3s =0.05F* 4.511=0.22555 C
3.50s =0.05F* 5.043=0.25215 C
4s =0.05F* 5.497=0.27485 C
4.5s =0.05F* 5.933=0.29665 C
5s =0.05F* 6.328=0.3164 C
7s =0.05F* 7.529=0.37645 C
10s =0.05F* 8.644=0.4322 C
15s =0.05F* 9.802=0.4751 C
20s =0.05F* 9.817=0.49085 C
25s =0.05F*9.932=0.4966 C
Table 2
Maximum Charge from eqn (2) =
Q
= _____0.05C______
RC time constant from eqn (3) =
τ
= ______5.00_____
Calculated value
eqn (1)
Experimental value
eqn (5)
% error
Charge at
t = 1 τ
0.0316 C
0.316 C
0.095%
Charge at
t = 2 τ
0.432 C
0.432 C
0.007%
Charge at
t = 3 τ
0.4966 C
0.496 C
0.006%
Your preview ends here
Eager to read complete document? Join bartleby learn and gain access to the full version
- Access to all documents
- Unlimited textbook solutions
- 24/7 expert homework help
Case-B: Data for Discharging a single capacitor
Table-3
Resistance R = ___100 Ohm______
Capacitance C = ____0.05____
Time
(s)
Measured
Voltage
(V
C
)
Charge
on
Capacitor
q(t)
(eqn. (5)
Time
(s)
Measured
Voltage
(V
C
)
Charge
on
Capacitor
q(t)
(eqn. (5)
Time
(s)
Measured
Voltage
(V
C
)
Charge
on
Capacitor
q(t)
(eqn. (5)
0.50
9.031
0.45155
3.00
5.489
0.27445
7.00
2.471
0.12355
1.00
8.205
0.41025
3.50
4.957
0.24785
10.0
1.356
0.0678
1.50
7.409
0.37045
4.00
4.503
0.22515
15.0
0.498
0.0249
2.00
6.691
0.33455
4.50
4.066
0.2033
20.0
0.183
0.00915
2.50
6.078
0.3039
5.00
3.672
0.1836
25.0
0.068
0.0034
Make a graph between
q(t)
and time.
Table 4
Maximum Charge from eqn (2) =
Q
= ___0.50C________
RC time constant from eqn (3) =
τ
= ______5.00s_____
Calculated value
eqn (4)
Experimental value
eqn (5)
% error
Charge at
t = 1 τ
0.1839
0.1836
0.163%
Charge at
t = 2 τ
0.067668
0.0678
0.195%
Charge at
t = 3 τ
0.003369
0.0034
0.92%
Case C and D: Data for Two Capacitors in Series and Parallel:
Table 5:
Resistance: ____100 ohm________
Capacitance 1: ____0.05C_________ Capacitance 2:
____0.05_________
Type of Circuit
Capacitors in:
Calculated
values of
τ
C
and
τ
D
Measured
Charging
time
τ
C
Measured
Discharging
time
τ
D
Percent
error in time
of charging
Percent
error in time
of
discharging
Series
0.03161
5.00
10.00
0.92%
0.095%
Parallel
0.4323
5.00
10.00
0.0062%
0.0062%
τ
C
: Time constant for charging
τ
D
:
Time constant for discharging
Related Documents
Related Questions
Pls. write solution neatly
arrow_forward
Describe the behavior of the voltage across the resistor with time as thecapacitor charges. Explain the reason for this behavior.
arrow_forward
1.
Which of the following answers best describes discontinuous conduction mode?
A. The inductor voltage ripple goes negative
B. The inductor current ripple stays positive
C. The inductor current ripple falls to zero
D. The output current ripple falls to zero
arrow_forward
For direct current (dc), a capacitor acts like
(a) a closed switch.
(b) an open.
(c) a short.
(d) a small resistance.
arrow_forward
v. What is the purpose of a capacitor in electronic circuits?
arrow_forward
How does the internal resistance of the multimeter (in voltage setting) affect the charging/discharging rate of the capacitor. Explain.
arrow_forward
Don't use ai to answer I will report you answer
arrow_forward
1.
arrow_forward
Describe the electrochemical double layer capacitor (EDLC) using carbon electrode.
arrow_forward
Question 1:
In the circuit shown below, the output (Vo = 10V Max.) Unipolar. The frequency of Primary is 60 Hz. The
diodes are Silicon with VD = 0.7V.
a. Sketch the output without a Capacitor.
b. Determine Voc without a Capacitor.
c. Sketch Vs (at the Secondary).
d. Determine Voc with a Capacitor of 10 uF across RL.
e. Determine the RMS Value of Vp (at the Primary).
f. PIV (Peak Inverse Voltage).
10:1
Output
C.
22 k1
All diodes are IN4001.
|
00000
arrow_forward
SEE MORE QUESTIONS
Recommended textbooks for you

Delmar's Standard Textbook Of Electricity
Electrical Engineering
ISBN:9781337900348
Author:Stephen L. Herman
Publisher:Cengage Learning
Related Questions
- Pls. write solution neatlyarrow_forwardDescribe the behavior of the voltage across the resistor with time as thecapacitor charges. Explain the reason for this behavior.arrow_forward1. Which of the following answers best describes discontinuous conduction mode? A. The inductor voltage ripple goes negative B. The inductor current ripple stays positive C. The inductor current ripple falls to zero D. The output current ripple falls to zeroarrow_forward
- For direct current (dc), a capacitor acts like (a) a closed switch. (b) an open. (c) a short. (d) a small resistance.arrow_forwardv. What is the purpose of a capacitor in electronic circuits?arrow_forwardHow does the internal resistance of the multimeter (in voltage setting) affect the charging/discharging rate of the capacitor. Explain.arrow_forward
arrow_back_ios
arrow_forward_ios
Recommended textbooks for you
- Delmar's Standard Textbook Of ElectricityElectrical EngineeringISBN:9781337900348Author:Stephen L. HermanPublisher:Cengage Learning

Delmar's Standard Textbook Of Electricity
Electrical Engineering
ISBN:9781337900348
Author:Stephen L. Herman
Publisher:Cengage Learning