9. Capacitor and Resistor Circuits - Hunter · PDF file · 2018-02-289. Capacitor...

9. Capacitor and Resistor

CircuitsIntroduction

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

(1)V =Q

C+ iR

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 1

(2)i@tD =d

dtQ@tD

You can eliminate one of the unknowns between equations (1) and (2) by taking the derivative of equa-

tion (1) with respect to time obtaining

(3)0 =1

C

d

dtQ@tD + R

d

dti@tD

and using equation (2) to eliminate the derivative of the charge

(4)0 =1

Ci@tD + R

d

dti@tD

It is easy enough to solve equation (4) since by rearrangement

(5)d

dti@tD =

−1

RCi@tD

Further

(6)‡ 1

iâi =

−1

RC‡ ât

Integration yields

(7)LogB i@tDi0

F −t

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)

(8)i@tD = i0 ExpB−t

RCF

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

(9)i@0D R = V or i@0D =V

R

as initially the capacitor C play no role.

Setting t=0 in equation (8) and using equation (9) yields

(10)i@0D = i0 =V

R

so we have determined the constant of integration. Utilization of equation (10) in equation (8) yields

finally the solution as


(11)‡ i@tD =V

RExpB−

t

RCF

The product RC has units of time and usually is called the time constant t

(12)τ = R C

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;

τ = R ∗ Cap;

Print@"Time Constant =", τ, " sec"DTime 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

RExpB−

t

τF

Plot@i@tD, 8t, 0, 4 ∗ τ<,

AxesLabel → 8"time", "i@tD"<D

0.02 0.04 0.06 0.08time

0.0002

0.0004

0.0006

0.0008

0.001

0.0012

i@tD

� 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


So initially, right after the switch is closed, the current i[t] is a maximum and thereafter is decreases

exponentially. The initial current is

[email protected]

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@τD0.000459849

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 ∗ τD0.000169169

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


Vr = Plot@i@tD ∗ R, 8t, 0, 4 ∗ τ<,

AxesLabel → 8"time", "VR@tD"<D

0.02 0.04 0.06 0.08time

2

4

6

8

10

VR@tD

� Graphics �

Graph of the Charge Q[t] Combining equations (2) and (11) yields

(13)d

dtQ@tD =

V

RExpB−

t

RCF

and integration yields

(14)Q@tD = Q@0D +V

R‡

0

t

ExpB−t'

τF ât'

where the initial charge on the capacitor is zero Q[0]=0. The integral is easily obtained

Clear@τD;

‡0

t

ExpB−t'

τF ât'

τ − ã−t

τ τ

Combining this with equation (14) and recalling equation (12) we obtain

(15)Q@tD = VC 1 − ExpB−t

τF

which we can also graph.


V = 10; R = 8000.; Cap = 2.5 ∗ 10−6;

τ = R ∗ Cap;

Q@t_D := V ∗ Cap 1 − ExpB−t

τF

Plot@Q@tD, 8t, 0, 4 ∗ τ<,

AxesLabel → 8"time", "Q@tD"<D

0.02 0.04 0.06 0.08time

5·10-6

0.00001

0.000015

0.00002

0.000025Q@tD

� Graphics �

Initially the charge on the capacitor is zero

[email protected].

and this agrees with the above graph. After a time t=t the charge on the capacitor is

Q@τD0.000015803

and this can also be obtained approximately from the graph. After a very long time the charge has its

maximum value

Q@5 ∗ τD0.0000248316

which is almost the same as


V ∗ Cap

0.000025

The voltage across the capacitor is Q

C and we may graph this using

VC = PlotB Q@tDCap

, 8t, 0, 4 ∗ τ<,

AxesLabel → 8"time", "VC@tD"<F

0.02 0.04 0.06 0.08time

2

4

6

8

10

VC@tD

� Graphics �

Replace the Battery and Switch by a Signal Generator

having a Square Wave The circuit diagram now appears


Suppose the square wave generator has a frequency f given by the square of the signal generator can be

graph using

T = τ;

f =1

T;

Print@"frequency =", f, "Hz"Dfrequency =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


V0 = 10.;

SqWave@t_D := IfBt <T

2, V0, 0F

Plot@SqWave@tD, 8t, 0, T<D

0.005 0.01 0.015 0.02

2

4

6

8

10

� 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 CapacitorWe may graph the voltage across the capacitor together with the signal generator voltage and

obtain


T = τ;

PlotB: Q@tDCap

, SqWave@tD>, :t, 0,T

2>F

0.002 0.004 0.006 0.008 0.01

2

4

6

8

10

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 τ;

PlotB: Q@tDCap


2>F

0.005 0.01 0.015 0.02

2

4

6

8

10

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


T = 3 τ;

PlotB: Q@tDCap


2>F

0.005 0.01 0.015 0.02 0.025 0.03

2

4

6

8

10

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 ResistorWe may graph the voltage across the resistor together with the signal generator voltage and obtain

T = τ;

PlotB8i@tD R, SqWave@tD<, :t, 0,T

2>, PlotRange → 80, 10<F

0.002 0.004 0.006 0.008 0.01

2

4

6

8

10

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 τ;



0.005 0.01 0.015 0.02

2

4

6

8

10



T = 3 τ;



0.005 0.01 0.015 0.02 0.025 0.03

2

4

6

8

10

About 2 volts now appears across the resistor.

The Second Part of the Square Wave of the Signal

Generator.


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 V0during the second part

of the cycle

T = τ;

V0 = 10.;


2, 0, V0F

SigGen = Plot@SqWave@tD, 8t, 0, T<D

0.005 0.01 0.015 0.02

2

4

6

8

10

� 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


Kirchoff circuit law after the switch is closed is

(16)0 =Q

C+ iR

which is the same as equation (1) without the battery. Taking the time derivative of equation (16) and

using equation (2) yields

(17)d

dti = −

i

R C

Equation (17) can be solved using the same techniques as before and we obtain again equation (11)

(18)d

dtQ@tD = i@tD = i0 ExpB−

t

τF

However, the initial condition i0is different this time as we shall see. Equation (18) can be integrated for

the charge Q[t] obtaining

(19)Q@tD = Q@0D + i0 τ 1 − ExpB−t

τF

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

(20)Q@0D = C V

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−∞

τE=0 and equation (19) reduces to

(21)Q@∞D = C V + i0 τ = 0

and it follows that

(22)i0 = −C V

τ= −

V

R

Combining equations (20) and (22) with equation (19) yields

(23)Q@tD = C V + VC ExpB−t

τF − 1 = V C ExpB−

t

τF

Equation (23) should make intuitive sence, since during the second half of the square wave cycle, the

capacitor is discharging. The voltage across the capacitor is V=Q

C initially so


T = τ;

V = 10.;


2, 0, VF

PlotB:V ∗ ExpB−t

τF, SqWave@tD>, :t, 0,

T

2>F

0.002 0.004 0.006 0.008 0.01

2

4

6

8

10

� 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


T = 3 ∗ τ;

V = 10.;


2, V0, 0F


τF, SqWave@tD>, :t, 0,

T

2>F

0.005 0.01 0.015 0.02 0.025 0.03

4

6

8

10

� Graphics �

The Voltage Across the ResistorThe current in the circuit is obtained by taking the derivative of the charge equation (23) obtaining

(24)i@tD = −V C

τExpB−

t

τF = −

V

RExpB−

t

τF

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


T = τ;

V = 10.;


τF, −V ∗ ExpB−

t

τF>, :t, 0,

T

2>F

0.002 0.004 0.006 0.008 0.01

-10

-5

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 ∗ τ;

V = 10.;



t

τF>, :t, 0,

T

2>F

0.005 0.01 0.015 0.02

-10

-5

5

10

� Graphics �


If the signal generator period T is three the time constant t then we obtain

T = 3 ∗ τ;

V = 10.;



t

τF>, :t, 0,

T

2>F

0.005 0.01 0.015 0.02 0.025 0.03

-10

-5

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 prop-

erly make a note of the maximum voltage in your lab notebook. Choose a square wave and make the

frequency f of the signal generator such that f=1

T with T=t=RC at first. With channel 1 of the oscillo-

scope, 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:


R = 6.8 × 103;

c = 0.1 × 10−6;

τ = 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 exam-

ple above,

T = τ;

f = 1 ê 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 measure-

ments of PART A above.



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 measure-

ments 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 capaci-

tance is given by

Ceff = C1 + C2


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.


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9. Capacitor and Resistor Circuits - Hunter · PDF file · 2018-02-289. Capacitor...

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