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Brooklyn College 1
RC Circuits
Purpose
1. To study transient response of an RC series circuit.
2. To measure and calculate the time constant, .
Introduction
When a battery is applied to a capacitor and a resistor in series (at the instant that we close the switch, S) as shown in
fig. 1, the capacitor takes a small time to charge, after that it becomes
fully charged. The state during the time in which the capacitor is
charging is called the ‘transient state’. When the capacitor reaches its
full charge, the state is then called the ‘steady state’. We are going to
study the transient state.
Initially the capacitor is uncharged (has no net charge). When the
switch is closed, charges flow fast to start charging the capacitor
plates. The initial fast motion of charges causes a large initial current
of the capacitor, . As the charges accumulate on the capacitor, the
capacitor acts as a battery that opposes the flow of charges to it,
therefore the charge flow slows down, and the capacitor current, decreases. The increase of charge of the capacitor
follows an exponential growth curve. The voltage across a capacitor, is
always directly proportional to the charge on the plates of the capacitor, ,
where is called the capacitance of the capacitor. The units of is the Farad,
.
. Therefore, the voltage across a capacitor also
follows a similar exponential grow curve. Since the capacitor current is
decreasing it comes out (from a mathematical derivation) that it follows an
exponential decay curve. If we apply Kirchhoff’s voltage rule, KVL to the loop
of the circuit of figure 1, with the switch closed we get:
where is the source voltage. Using eqns. 1, 2 & 3, it can be shown (see the text book) that the equations, as a function
of time, , for the capacitor charge, and the capacitor voltage, during the
transient charging are inverse exponential:
where, is called the ‘time constant’ and , for the circuit of fig. 1,
And the equation describing the current of the capacitor, during the transient charging is exponential decay:
0
Figure 2: The curve for the capacitor
charge, during transient charging
Figure 1: Series RC circuit
for capacitor charging
S
Figure 3: The curve for capacitor
current decay during transient charging
0
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Where, is the initial current of the capacitor. Notice from eqn. 2 that for fig. 1, during transient charging (and also in
the steady state) is always a constant equal to .
When the capacitor is fully charged, then no further charges flow to the capacitor plates, the charge on the plates of the
capacitor becomes constant and the capacitor current, becomes zero. Then the capacitor is said to have reached the
steady state.
When the time during charging from zero charge, reaches , then from eqn. 4 we get is equal to . The
time constant, is a measure of the charging and discharging time of the capacitor. A large time constant, means
slow charging. Also we can find, using eqn. 7, that for the capacitor current, becomes equal to .
If the battery, is replaced by a wire, then the capacitor discharges through the resistor and the charge and voltage of
the capacitor follow an exponential decay curve. This is the transient
discharge state. The eqn. for the capacitor charge and voltage during
discharging are:
as shown in fig. 4. Since the discharge starts fast then slows, the current
during discharge also starts large then decreases. It follows an
exponential decay curve like the charging case but it has an opposite
direction in the circuit, as shown in fig. 4. The eqn. is:
We can find , & at using equations 8, 9 & 10, and we find
the result as shown in figures 4 & 5. The curve for the capacitor voltage,
is similar to the curve for the capacitor charge except for a
multiplication factor (see eqn. 1 above).
We can cause the charging and the discharging transients by applying a
square wave in place of the battery . We can adjust the square wave
so that the peak is equal to the value of (charging case) and the
lower value of the square wave is equal to zero volts (discharging
case) , as shown in fig. 6.
Figure 4: The curve for capacitor charge
decay during transient discharging
0
Figure 5: The curve for capacitor current
decay during transient discharging
0
0
Figure 6: A square wave voltage source to
cause the charging and discharging transients.
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Running the experiment (The data sheet is on page 6)
Finding the measured time constant for an RC circuit charging and discharging transients
1) Open the simulator http://falstad.com/circuit/ click the Run/Stop to stop the default simulation. Clock circuits from
the top menu and select black circuit. This is to clear the default circuit.
2) Click ‘Draw’ in the top menu and move the mouse to Inputs and Sources and select ‘Add A/C Voltage Source (2-
terminal). Click somewhere at the left of the black space and drag up to place your AC voltage source (make sure the +
sign is on the top of the circle of the voltage source. Move the mouse over the source till its color turns light blue and
right click and select edit. Set the Max voltage to 0.5 (the simulator already knows all the units so here do not write v, it
already knows that the unit is volts). Click the drop down menu of the waveform and select square wave. Set the DC
offset to 0.5 (this will cause the upper peak to be at 1v and the lower at 0v). The DC offset is a constant value of voltage
that is added to all point of the voltage of the wave. Set the frequency to 12 Hz (do not write Hz, it already knows the
unit). Keep the rest of the values at their default: phase offset = 0 (meaning the wave will start at 0 time) and the Duty
cycle = 50 (meaning the square wave is symmetrical. Meaning it has equal time duration for the max voltage as for the
time duration for its minimum voltage, as in fig 6 above). Click ‘Apply’ then click ‘OK’.
3) Click Draw from the top menu and select ‘Add Resistor’. Click and drag to place it as shown in fig. 7 below. Edit it to
change its value in a similar way as you did for the voltage source (moving the mouse over it till its color is light blue
then right click and select edit). Set it to 180 (it already knows the unit is ohms). Click Draw in the top menu, move the
mouse to ‘Passive Components’ and select ‘Add Capacitor’. Click and drag to place it as shown in fig. 7 below. Change its
value to 33u. We will call this capacitor C1.The simulator accepts ‘u’ for μ. 33μF means 33 micro and the simulator
already knows that the unit is Farad so do not write the unit Farad, F. The milli, m is , the micro, μ is and the
nano, n is .
4) To complete the circuit connections, click Draw and select Add wire. And click and drag to connect the components
with wires as shown in fig. 8 above. You need to click the square terminal of the wire (that shows when you place the
mouse over the round terminal of the circuit element (say the voltage source)) and drag a segment of wire (in the case
of the top terminal of the voltage source, two segments, one small vertically up from the top terminal of the voltage
source(as shown in fig. 8) and the other small segment towards the right to the resistor’s left terminal). Make sure not to
run a wire through a circuit element, otherwise you short circuit that element. After completing the circuit, click ‘Draw’
in the top menu and select ‘Select/Drag Sel’. This is to stop giving you more wires.
5) Move the mouse to the square wave voltage source and right click to select ‘View in Scope’. This will give a graph
below to display the square wave voltage as a function of time. Click the gear icon on the graph and choose’ Properties’.
Adjust the ‘Scroll Speed’ to 5ms/div, then click OK (do not click set as default). Again, repeat this step for the capacitor
Figure 7 (Left): Building the circuit. Figure 8 (Right): Completing wire connections.
Brooklyn College 4
first, and then for the resistor so we can display the voltage across the capacitor and the voltage across the resistor. This
way the middle graph is for the capacitor.
6) Now you are ready to simulate. Click Run/ Stop. Keep it running for some time and watch the scopes graphs and
notice in the middle graph (for the capacitor) how the voltage (green curve) is increasing in charging and decreasing in
discharging. Then Click ‘Run/Stop’ again when you have half a cycle in each graph displayed with the charging part and
with the zero voltage (green) near the left corner of the graph. The green curves are voltages and the yellow curves are
currents. Notice the middle graph is for the capacitor. What do you notice concerning the capacitor current (yellow), the
Resistor current (yellow) and the source current (yellow) are they the same? Why?
7) Now focus on the capacitor voltage. Note that if the voltage comes out negative (below the time axis) it means that
the capacitor is upside down (this is specific to the simulator we are using), if this is the case, then right click the
capacitor and select ‘Swap terminals’. Move the mouse on the green graph of the capacitor (middle graph) and on the
charging (increasing part) of it, till the voltage displays a value of 0.63v (630 mv) or as close as possible. Record the time
value. Call it t1. If you do not have the charging part of the voltage displayed, then click ‘Run/Stop’ and stop when you
get the charging part displayed with the zero at the left corner of the graph. You will need also to have the staring zero
voltage within the displayed part. Move the mouse to the previous zero value of voltage on the same green voltage
graph and for same cycle and record the time call it t0. Calculate t1-t0. This is your measured time constant, .
Compute RC, which is your calculated value for the time constant, . Calculate the percentage error. Take a screen shot
and include it in your report. Record all time values: t1 and t0 in the data sheet, as well as measured & calculated time
constant, . Find the percentage error.
8) Repeat step 7 but this time find the measured time constant from the discharging part (decreasing voltage part) of
the graph. Click ‘Run/Stop’ and stop when the discharging part is displayed. Which values of discharging voltage should
you use? (Hint: see eqn. 9). Finally, find the measured time constant from the current graph (yellow curve). Which values
of the current should you use? (Hint: see eqns. 7 & 10). Notice that
. Find the percentage error for
.
9) Change the value of the capacitor to 30μF. We will call this capacitor C2. Click ‘Reset’ and then repeat steps 6 and 7.
Record the values of t1 and t0 and find the measured . Compute RC. This your calculated value for the time constant,
. Find the percentage error.
10) Click ‘Reset’. Now we want to see the effect of connecting both capacitors in parallel. We know that the equivalent
capacitance of capacitors connected in parallel is the sum of the individual capacitances. So the parallel equivalent of
C1= 33μF and C2 = 30μF is 63μF. Do you expect a smaller or larger time constant in this case? Why? Compute the
calculated time constant in this case and record in the data sheet. Now let’s try it. Set the value of the capacitor to 63μF
(this represents the parallel equivalent of both capacitors). Repeat steps 6 and 7 to find the measured & calculated time
constant, .
Figure 9: Circuit with C equivalent representing the parallel equivalent of C1= 33μF and C2 = 30μF
Brooklyn College 5
11) Now we want to connect both capacitors in series. So to do that, first change the value of the capacitor to 33μF then
you can delete the wire at the bottom shown in light blue in fig. 9, and replace it with the capacitor C= 30μF. To achieve
that, click ‘Draw’ and select ‘Passive Components’ and choose ‘Add Capacitor’. Click and drag to place it in place of the
wire you deleted. For the series capacitors do you expect a larger or smaller time constant compared to the parallel
case? Why?
Also we want to read the voltage across the series combination of the two capacitors. To achieve that, click ‘Draw’,
select ‘Outputs and Labels’ and choose ‘Add voltmeter/ scope probe’. Place it as shown in figure 10 and connect it to the
terminals of the capacitors using wires as shown in fig. 11. From ‘Draw’ choose ‘Select/Drag Sel’. Move the mouse to
the voltmeter and right click it and select ‘View in Scope’. This adds a 4th graph. You can delete the second graph, from
the left (right click it and select the 1st option at the top: ‘remove scope’) since it was for only one of the two capacitors.
12) Now with the capacitors in series, repeat steps 6 and 7 to find the measured time constant. Compute the calculated
time constant and find the percentage error.
13) If you change the value of the resistor to 200 ohms, do you expect the time constant to increase or decrease? Why?
Try it using the simulator only for your last circuit of step 12. Find the calculated time constant, in this case & % error.
Figure 10 (Left): Adding a scope probe. Figure 11 (Right): Connecting the terminals of the scope probe.
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Data Sheet
Name: Group: Date experiment performed:
Step 6) What do you notice concerning the capacitor current (yellow), the Resistor current (yellow) and the source
current (yellow) are they the same?
Why?
Step 7) C1= 33 μF
t1 (ms) t0 (ms) t1-t0= measured (ms) RxC= calculated (ms) % error
Step 8) Which values of discharging voltage should you use?
t1 (ms) t0 (ms) t1-t0= measured (ms) RxC= calculated (ms) % error
Which values of current should you use?
t1 (ms) t0 (ms) t1-t0= measured (ms) RxC= calculated (ms) % error
Step 9) C2= 30 μF
t1 (ms) t0 (ms) t1-t0= measured (ms) RxC= calculated (ms) % error
Step 10) C parallel equivalent = 63μF. Do you expect a smaller or larger time constant in this case?
Why?
t1 (ms) t0 (ms) t1-t0= measured (ms) RxC= calculated (ms) % error
Step 11) C1 & C2 in series. For the series capacitors do you expect a larger or smaller time constant compared to the
parallel case?
Why?
Step 12) C1 & C2 in series
t1 (ms) t0 (ms) t1-t0= measured (ms) RxC= calculated (ms) % error
Step 13) R= 200 ohm., with series capacitors C1 & C2. Do you expect the time constant to increase or decrease?
Why?
t1 (ms) t0 (ms) t1-t0= measured (ms) RxC= calculated (ms) % error