ElectronicsLab9.nb
1
9. Capacitor and Resistor
Circuits
Introduction
Thus far we have consider resistors in various combinations with a power supply or battery which
provide a constant voltage source or direct current (voltage) DC. Now we start to consider various
combinations of components and much of the interesting behavior depends upon time so we will also
consider AC or alternating current (voltage) sources which are signal generators. The first combination
we consider is a resistor in series with a capacitor and a battery.
The RC Circuit
Consider the resistor-capacitor circuit indicated below:
When the switch is closed, Kirchoff's loop equation for this circuit is
V=
Q
C
+ iR
(1)
for t>0 where both Q[t] and i[t] are functions of time. There are two unknown quantities Q[t] and i[t] in
equation (1) and we need an additional equation namely
ElectronicsLab9.nb
2
d
i@tD =
dt
(2)
Q@tD
You can eliminate one of the unknowns between equations (1) and (2) by taking the derivative of equation (1) with respect to time obtaining
0=
1
d
d
Q@tD + R
C dt
dt
(3)
i@tD
and using equation (2) to eliminate the derivative of the charge
0=
1
C
d
i@tD + R
dt
(4)
i@tD
It is easy enough to solve equation (4) since by rearrangement
d
dt
i@tD =
-1
RC
(5)
i@tD
Further
à
1
i
âi =
-1
RC
à ât
(6)
Integration yields
LogB
i@tD
i0
F-
t
(7)
RC
where i0 is a constant of integration which we will determine shortly. Using a property of the exponential
function, we obtain from equation (7)
i@tD = i0 ExpB-
t
RC
F
(8)
Initially at t=0, when the switch is closed, the capacitor has zero charge and therefore there is zero
potential across it. The current in the circuit is determined entirely by the battery potential V and the
resistance R through Ohm's law
i@0D R = V
or
i@0D =
V
(9)
R
as initially the capacitor C play no role.
Setting t=0 in equation (8) and using equation (9) yields
i@0D = i0 =
V
R
so we have determined the constant of integration.
finally the solution as
(10)
Utilization of equation (10) in equation (8) yields
ElectronicsLab9.nb
3
à i@tD =
V
R
ExpB-
t
RC
F
(11)
The product RC has units of time and usually is called the time constant t
t=RC
(12)
Graph of the Solution for the current
Suppose the numerical values V=10 volts, R=8,000 ohms, and C=2.5 microfarads as indicated
then
V = 10; R = 8000.; Cap = 2.5 * 10-6 ;
t = R * Cap;
Print@"Time Constant =", t, " sec"D
Time Constant =0.02 sec
IMPORTANT: C is a protect variable assigned to something specific in Mathematica so instead we use
Cap as the symbol for capacitance.
i@t_D :=
V
R
ExpB-
t
t
F
Plot@i@tD, 8t, 0, 4 * t<,
AxesLabel ® 8"time", "i@tD"<D
i@tD
0.0012
0.001
0.0008
0.0006
0.0004
0.0002
0.02
0.04
0.06
0.08
time
… Graphics …
So initially, right after the switch is closed, the current i[t] is a maximum and thereafter is decreases
exponentially. The initial current is
ElectronicsLab9.nb
4
So initially, right after the switch is closed, the current i[t] is a maximum and thereafter is decreases
exponentially. The initial current is
i@0D
0.00125
which agrees with the graph above. The time constant (t=0.02 seconds in this case) determines the rate
of decay. After a time t the current has decreased to
i@tD
0.000459849
and this is equal to equation (11) with t=t namely
i@0D * ã-1
0.000459849
Note that
ã-1.
0.367879
so after one time t=t the current has dropped 37% in its value. After t=2t, the current is
i@2 * tD
0.000169169
which is 14% of the original value of the current since
ã-2.
0.135335
and so on.
The voltage across the resistor is i*R so we may graph this voltage as
ElectronicsLab9.nb
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Vr = Plot@i@tD * R, 8t, 0, 4 * t<,
AxesLabel ® 8"time", "VR @tD"<D
VR @tD
10
8
6
4
2
0.02
0.04
0.06
0.08
time
… Graphics …
Graph of the Charge Q[t]
Combining equations (2) and (11) yields
d
dt
Q@tD =
V
R
ExpB-
t
RC
F
(13)
and integration yields
Q@tD = Q@0D +
V
R
à ExpBt
0
t'
t
F ât'
(14)
where the initial charge on the capacitor is zero Q[0]=0. The integral is easily obtained
Clear@tD;
t
t'
F ât'
à ExpBt
0
t
t - ã- t t
Combining this with equation (14) and recalling equation (12) we obtain
Q@tD = VC 1 - ExpBwhich we can also graph.
t
t
F
(15)
ElectronicsLab9.nb
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V = 10; R = 8000.; Cap = 2.5 * 10-6 ;
t = R * Cap;
t
Q@t_D := V * Cap 1 - ExpB- F
t
Plot@Q@tD, 8t, 0, 4 * t<,
AxesLabel ® 8"time", "Q@tD"<D
Q@tD
0.000025
0.00002
0.000015
0.00001
-6
5´10
0.02
0.04
0.06
0.08
time
… Graphics …
Initially the charge on the capacitor is zero
Q@0.D
0.
and this agrees with the above graph. After a time t=t the charge on the capacitor is
Q@tD
0.000015803
and this can also be obtained approximately from the graph. After a very long time the charge has its
maximum value
Q@5 * tD
0.0000248316
which is almost the same as
ElectronicsLab9.nb
7
V * Cap
0.000025
The voltage across the capacitor is
VC = PlotB
Q@tD
Cap
Q
C
and we may graph this using
, 8t, 0, 4 * t<,
AxesLabel ® 8"time", "VC @tD"<F
VC @tD
10
8
6
4
2
0.02
0.04
0.06
0.08
time
… Graphics …
Replace the Battery and Switch by a Signal Generator
having a Square Wave
The circuit diagram now appears
ElectronicsLab9.nb
8
Suppose the square wave generator has a frequency f given by the square of the signal generator can be
graph using
T = t;
1
f= ;
T
Print@"frequency =", f, "Hz"D
frequency =50.Hz
Notice that the period of the signal generator is chosen to be the same as the time constant in the RC
circuit. We will discuss this more later. Suppose the voltage amplitude of the signal generator is 10 volts
(the same as the battery voltage) The square wave of the signal generator is graphed using
ElectronicsLab9.nb
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V0 = 10.;
SqWave@t_D := IfBt <
T
, V0 , 0F
2
Plot@SqWave@tD, 8t, 0, T<D
10
8
6
4
2
0.005
0.01
0.015
0.02
… Graphics …
Effectively having the signal generator in the circuit is the same as having the battery in the circuit for
time 0<t<0.01 seconds so the equation (11) for the current and equation (15) for the charge hold for these
times.
The Voltage Across the Capacitor
We may graph the voltage across the capacitor together with the signal generator voltage and
obtain
ElectronicsLab9.nb
10
T = t;
PlotB:
Q@tD
Cap
, SqWave@tD>, :t, 0,
T
2
>F
10
8
6
4
2
0.002
0.004
0.006
0.008
0.01
If the period T of the signal generator is longer, for example T=2*t, then the capacitor has more time to
charge
T = 2 t;
Q@tD
T
PlotB:
, SqWave@tD>, :t, 0, >F
Cap
2
10
8
6
4
2
0.005
0.01
0.015
0.02
Further if the period T of the signal generator is longer still, for example T=3*t, then the capacitor has
more time to charge
ElectronicsLab9.nb
11
T = 3 t;
Q@tD
T
PlotB:
, SqWave@tD>, :t, 0, >F
Cap
2
10
8
6
4
2
0.005
0.01
0.015
0.02
0.025
0.03
and the capacitor almost has time to fully charge and have all the 10 volts appear across it. About 8 volts
now appears across the capacitor.
The Voltage Across the Resistor
We may graph the voltage across the resistor together with the signal generator voltage and obtain
T = t;
PlotB8i@tD R, SqWave@tD<, :t, 0,
T
2
>, PlotRange ® 80, 10<F
10
8
6
4
2
0.002
0.004
0.006
0.008
0.01
If the period T of the signal generator is longer, for example T=2*t, then the current gets smaller still and
the voltage across the resistor is reduced further
ElectronicsLab9.nb
12
If the period T of the signal generator is longer, for example T=2*t, then the current gets smaller still and
the voltage across the resistor is reduced further
T = 2 t;
PlotB8i@tD R, SqWave@tD<, :t, 0,
T
2
>, PlotRange ® 80, 10<F
10
8
6
4
2
0.005
0.01
0.015
0.02
If the period T of the signal generator is longer, for example T=3*t, then the current gets smaller still and
the voltage across the resistor is reduced further
T = 3 t;
PlotB8i@tD R, SqWave@tD<, :t, 0,
T
2
>, PlotRange ® 80, 10<F
10
8
6
4
2
0.005
0.01
0.015
0.02
0.025
0.03
About 2 volts now appears across the resistor.
The Second Part of the Square Wave of the Signal
Generator.
ElectronicsLab9.nb
13
The Second Part of the Square Wave of the Signal
Generator.
During the second part of the period of the signal generator for times
T
2
< t < T, the voltage is
zero in the original circuit. It helps make the analysis simpler to change the wave form a little and have
the signal generator voltage zero during the first part of the cycle and a constant V0 during the second part
of the cycle
T = t;
V0 = 10.;
, 0, V0 F
2
SigGen = Plot@SqWave@tD, 8t, 0, T<D
SqWave@t_D := IfBt <
T
10
8
6
4
2
0.005
0.01
0.015
0.02
… Graphics …
This corresponds to the times t in the range 0.01 sec < t < 0.02 sec in the previous diagram. Effectively
for the first part of the cycle the battery is removed from the circuit and replaced by a shorting wire and
the circuit looks like
ElectronicsLab9.nb
14
Kirchoff circuit law after the switch is closed is
0=
Q
C
+ iR
(16)
which is the same as equation (1) without the battery. Taking the time derivative of equation (16) and
using equation (2) yields
d
dt
i=-
i
(17)
RC
Equation (17) can be solved using the same techniques as before and we obtain again equation (11)
d
dt
Q@tD = i@tD = i0 ExpB-
t
t
F
(18)
However, the initial condition i0 is different this time as we shall see. Equation (18) can be integrated for
the charge Q[t] obtaining
Q@tD = Q@0D + i0 t 1 - ExpB-
t
t
F
(19)
The capacitor is assumed fully charged initially (which can happen if the time constant t is short
compared with the period T of the square wave) so initially
Q@0D = C V
(20)
and when t=0 the part of equation (19) involving the exponential function vanishes. For long times there
is no charge on the capacitor Q[¥]=0. Since ExpA- t E=0 and equation (19) reduces to
¥
Q@¥D = C V + i0 t = 0
(21)
and it follows that
i0 = -
CV
t
=-
V
(22)
R
Combining equations (20) and (22) with equation (19) yields
Q@tD = C V + VC ExpB-
t
t
F - 1 = V C ExpB-
t
t
F
(23)
Equation (23) should make intuitive sence, since during the second half of the square wave cycle, the
Q
capacitor is discharging. The voltage across the capacitor is V= C initially so
ElectronicsLab9.nb
15
T = t;
V = 10.;
SqWave@t_D := IfBt <
PlotB:V * ExpB-
t
t
T
2
, 0, VF
F, SqWave@tD>, :t, 0,
T
2
>F
10
8
6
4
2
0.002
0.004
0.006
0.008
0.01
… Graphics …
Further if the signal generator is longer say three times the time constant, T=3t then the capacitor has
even more time to discharge
ElectronicsLab9.nb
16
T = 3 * t;
V = 10.;
SqWave@t_D := IfBt <
PlotB:V * ExpB-
t
t
T
2
, V0 , 0F
F, SqWave@tD>, :t, 0,
T
2
>F
10
8
6
4
0.005
0.01
0.015
0.02
0.025
0.03
… Graphics …
The Voltage Across the Resistor
The current in the circuit is obtained by taking the derivative of the charge equation (23) obtaining
i@tD = -
VC
t
ExpB-
t
t
F=-
V
R
ExpB-
t
t
F
(24)
and the voltage across the resistor is just R*i[t]. Graphing the voltage across the capacitor and the
voltage across the resistor for the second half the cycle yields
ElectronicsLab9.nb
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T = t;
V = 10.;
PlotB:V * ExpB-
t
t
F, - V * ExpB-
t
t
F>, :t, 0,
T
2
>F
10
5
0.002
0.004
0.006
0.008
0.01
-5
-10
… Graphics …
Notice the sum of the voltage of the capacitor and the voltage of the resistor is just zero as required by
Kirchoff's law. If the signal generator period T is twice the time constant t then we obtain
T = 2 * t;
V = 10.;
PlotB:V * ExpB-
t
t
F, - V * ExpB-
t
t
F>, :t, 0,
T
2
>F
10
5
0.005
-5
-10
… Graphics …
0.01
0.015
0.02
ElectronicsLab9.nb
18
If the signal generator period T is three the time constant t then we obtain
T = 3 * t;
V = 10.;
PlotB:V * ExpB-
t
t
F, - V * ExpB-
t
t
F>, :t, 0,
T
2
>F
10
5
0.005
0.01
0.015
0.02
0.025
0.03
-5
-10
… Graphics …
Laboratory Exercises
PART A: Place a signal generator in series with a resistor and capacitor. Pretty much any output level
(the output voltage) of the signal generator will do OK but after you get the oscilloscope working properly make a note of the maximum voltage in your lab notebook. Choose a square wave and make the
1
frequency f of the signal generator such that f= T with T=t=RC at first. With channel 1 of the oscilloscope, measure the voltage across the signal generator and with channel 2 measure the voltage across the
capacitor. Compare with the graphs of the first example above. Make the frequency f of the signal
generator smaller (T larger) so the capacitor has more time to charge. Keep decreasing f. Sketch the
oscilloscope figures you get and indicate the values of the voltage on the vertical scale and the time on
the horizontal scales.
Example: Suppose C=0.1 mF and R=6.8 kW then the time constant t=RC=0.00068 sec. as indicated
below:
ElectronicsLab9.nb
19
R = 6.8 ´ 103 ;
c = 0.1 ´ 10-6 ;
t = R´c
0.00068
NOTE: The value of R and C you use need not be the values given above. Use your digital ohmmeter to
measure the value of the resistor and make sure it is the same as given by the color code. Use your digital
capacitor meter to measured the value of the capacitor and it should agree with the capacitor code (which
is not that standardized so check with the maker of the capacitor and use your meter For example, a
capacitor labeled 250 B means 1 is the first digit and 5 is the second digit for the capacitance. 2 is the
multiplier
in
powers
of
10
so
this
capacitor
is
C=25×100 mF = 25 mF
where
mF=10-6 F. Capacitors can be much smaller and pF = 10-12 F is often used
t is the time it take the capacitor to charge to 67% of the maximum voltage (which is the maximum
voltage of the signal generator). The signal generator frequency should be set to have a period T=t at
first but what you actually control is the frequency f of the signal generator where f=1/T. For the example above,
T = t;
f = 1•T
1470.59
So the frequency f=1,470 Hz= 1.5 kHz corresponds to one time constant t. The horizontal time scale of
the oscilloscope had better be something like this frequency f. If the oscilloscope is set at too high a
frequency, the time will be too short to see the voltage rise. On the other hand, if the oscilloscope is set
at too low a frequency, there will not be enough time to see the voltage rise across the capacitor. You
also must make sure to set the voltage scale at roughly the output voltage of the oscilloscope which you
should have measured first before connecting the capacitor and resistor in the circuit.
PART B: With channel 1 of the oscilloscope, measure the voltage across the signal generator and with
channel 2 measure the voltage across the resistor. Compare with the graphs of the second example
above. Make the frequency f of the signal generator smaller (T larger) so the capacitor has more time to
charge. Keep decreasing f the frequency of the signal generator. Sketch the oscilloscope figures you get.
PART C: Call the capacitor used above C1 . Take a second capacitor and call it C2 . Combine the two
capacitors in SERIES without the signal generator and oscilloscope attached. The effective capacitance is
given by
1
Ceff
=
1
C1
+
1
C2
Compute the numerical value of the effective capacitance and check it with the digital capacitance meter.
Note the effective capacitance of two capacitors in SERIES is less than both C1 and C2 . Use the SERIES
combination of C1 and C2 together with the signal generator and oscilloscope and repeat the measurements of PART A above.
ElectronicsLab9.nb
20
Compute the numerical value of the effective capacitance and check it with the digital capacitance meter.
Note the effective capacitance of two capacitors in SERIES is less than both C1 and C2 . Use the SERIES
combination of C1 and C2 together with the signal generator and oscilloscope and repeat the measurements of PART A above.
PART D: Call the capacitor used above C1 . Take a second capacitor and call it C2 . Combine the two
capacitors in PARALLEL without the signal generator and oscilloscope attached. The effective capacitance is given by
Ceff = C1 + C2
Compute the numerical value of the effective capacitance and check it with the digital capacitance meter.
Note the effective capacitance of two capacitors in PARALLEL is less than both C1 and C2 . Use the
PARALLEL combination of C1 and C2 together with the signal generator and oscilloscope and repeat
the measurements of PART A above.