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```CURRENT AND VOLTAGE IN A C-R CIRCUIT WHEN THE CAPACITOR IS
CHARGING
Figure 1 shows a circuit consisting of a capacitor C and resistor R in series connected to a
d.c. source of e.m.f. V volts via a switch S.
R
S
Current i
V
Volts
C
Figure 1
At the instant of closing the switch, the capacitor is uncharged, that is the voltage across it is
zero; so all the voltage applied to the circuit is developed across the resistor R, and the
current flowing in the circuit is:
I =V/R
This current charges up the capacitor. As the capacitor begins to charge up, the potential
difference across the capacitor rises, causing the voltage across the resistor to fall, which in
turn causes the charging current to fall.
Finally, a point is reached when the potential difference across the capacitor is equal to the
e.m.f. of the source, and as a result the current ceases. The capacitor is now fully charged, the
p.d. across the capacitor has reached its final value and the circuit is in its steady state
condition.
Figure 2 shows the variations of Vc, the voltage across C, and VR, the voltage across R, from
the time when switch S is closed until the time when the capacitor becomes fully charged.
The current in the circuit varies in a similar way to VR.
Voltage
V
A
Vc
VR
T
Switch closed
Figure 2
1
You should note that the current, and the rate of change of current, are greatest at the instant
the switch is closed.
INITIAL RATE OF INCREASE OF VOLTAGE
In the circuit of Figure 1, the charge q on the capacitor is given by q = C VC
Also, current i = rate of change of q = C x (rate of change of VC). At the start, when the
switch closes,
Note q = CVc and q = it it = CVc and hence i = CVc (1)
t
As i = V (2)
R
Equating (1) and (2) V = CVc  Vc = V where Vc is the
R
t
t CR
t
i=V
R
(rate of change of VC ) = V
CR
rate of change of Vc
If the initial rate of change was maintained, the voltage VC would rise as shown by line A in
Figure 2. This is a straight line graph whose slope
equals the rate of change V
CR
Applying the equation for a straight line graph, y = mx + c
v =V t + 0
CR
Note: V is the slope of line A and the c the
CR intercept passes through 0.
Therefore, when t = CR, v = V.
So if this initial rate of change was maintained, V volts would be reached in a time of CR
seconds.
CURRENT AND VOLTAGE AFTER THE SWITCH CLOSES
The initial rate of change we worked out above is not maintained; as the voltage across the
capacitor increases, so the current decreases, and with it the rate of change of capacitor
voltage.
It can be shown by mathematics that this situation, where the rate of change of current
depends on the voltage, is represented by an exponential function, so we often refer to an
exponential rise or fall. The voltage across the capacitor is given by the expression:
Vc = V (1- e-t/CR)
If you have forgotten your mathematics course, or haven’t got to this point yet, you'll need to
know that e = 2.718.The usual log tables or a scientific calculator will give values of e raised
to various powers, or the power to which e must be raised to obtain a given value x that is,
logex. You can also use the formula logex = 2.303log10x. In a later section you will see how
to work out the answers step-by-step without using tables.
The voltage across the resistor can be found as follows. Vc + VR = V
2
VR = V- VC = V- V(1- e-t/CR)
VR = Ve-t/CR
The rate of change of voltage, dVc , is given by
dt:
dVc = Ve-t/CR
dt CR
At the instant of closing the switch, that is when t = 0:
dVc = Ve-t/CR which agrees with our calculations of the initial rate of rise above.
dt CR
Time constant
The variation of current and voltage with time depend only on the product CR, not the actual
values of C and R individually. So we usually work with T = CR, which is known as the time
constant of the circuit, and is defined as the time which would be taken for steady rate
conditions to be reached if the initial rate of change was maintained.
The value of T = CR ; T is measured in seconds, with C in farads and R in ohms. Some books
write the equation the other way round as T = RC, which of course means exactly the same.
Example 1
Calculate the time constant of a circuit consisting of a 12 kohm resistor in series with a
200 F capacitor.
T = CR
= 200 x10-6 x 12 x 103
= 2.4 s
PROGRESS QUESTION C.1
Calculate the time constant of the following combinations of components:
a)
Capacitor 1 F, resistor 22 k
b)
Capacitor 2.2 nF, resistor 15 k
c)
Capacitor 47 pF, resistor 82 
d)
Capacitor 68 F, resistor 1 k
Check on page 11
3
PROGRESS QUESTION C.2
Calculate the value of the other components needed to make up C-R circuits with the
specified time constants:
a)
Time constant 2.2 s, resistor 1 k
b)
Time constant 3 ms, resistor 15 k
c)
Time constant 10 s, resistor 2.2 M
d)
Time constant 22 ns, capacitor 22 pF.
Check on page 11
4
If we substitute t = T = CR in the expression for v:
Vc =V(1- e-CR/CR)
Vc =V(1-e-1)
Substituting e = 2.718 gives:
Vc = 0.632V
that is, the voltage rises to 63.2% of its final value in a time t equal to the time constant T.
Similarly, the current can be shown to be:
i = V e-t/T,whereT = CR
R
Substituting for t = T and e = 2.718 shows that i = 0.368 V = 0.368 I the current, which is
R
decreasing in value, falls to 36.8% of its initial value at a time t equal to the time constant.
The voltage across the resistor falls in a similar way to the current.
The current falls more slowly as time goes by, so theoretically it never reaches zero. But for
most practical purposes, when t = 5T, the value of v is 0.993 V and the value of i is 0.007 I,
and steady state conditions are assumed to have been reached.
Example 2
Calculate the voltage across a 500 F capacitor 2 seconds after it is connected to a 100 volt
battery via a 10 kohm resistor.
Vc = V(1- e-t/CR)
But:
t/CR=2/(500x 1-6x l0 x 103) = 0.4
Therefore:
Vc = 100(1- e-0.4) = 100(1-0.67) = 33 Volts
5
PROGRESS QUESTION C.3
Calculate the voltage across a 10,000 F capacitor, connected in series with a 10 kohm
resistor one minute after the circuit is connected across a 12 volt supply.
Check on page 11
PROGRESS QUESTION C.4
Calculate the voltage across a 100 F, capacitor 0.2 seconds after it is connected to a 100
volt battery in series with a 10 kohm resistor
Check on page 12
PROGRESS QUESTION C.5
Calculate the current in a 500 F capacitor 2 seconds after it is connected to a 100 volt
battery via a 10 kohm resistor
Check on page 12
6
CURRENT AND VOLTAGE IN A C-R CIRCUIT WHEN THE CAPACITOR IS
DISCHARGING
R
S Charge
Current i
Direction shown is during
charging
Discharge
V
Volts
C
Figure3
Assuming the capacitor has been charged to a potential V, then at the instant the switch is
moved to the ‘discharge’ position this potential will also appear across the resistor, and
maximum current will flow. The capacitor will therefore discharge its energy into the
resistor. Figure 4 shows the discharge curves of current and voltage.
Voltage
Current
V
VC
i
I
Switch moved
to discharge position
Figure 4
Some time later, when all the energy has been converted to heat in the resistor, the e.m.f. will
have fallen to zero and no current will flow. At this point the circuit is said to be in a steady
state.
The initial current when the switch is moved to the discharge position, which flows in the
opposite direction to the charging current, is given by:
I =- V
R
7
The equations representing the instantaneous voltage and current are given by:
Vc = V e-t/T
i = I e-t/T = V e-t/T
R
T is the time constant, that is, the time which would be taken for steady state conditions to be
achieved if initial rates of change were maintained. It is the same time constant as we saw
above, that is T=CR. Note that this is because we are looking at the same C-R circuit as
above; practical circuits often charge and discharge a capacitor through different resistances,
making the charging and discharging time constants different.
The values of voltage and current fall to 0.368 x initial value when t = T and steady rate
conditions are assumed to have been reached when t = 5T.
PROGRESS QUESTION C.6
Calculate the current in a 0.1 ohm resistor 2 s after it is connected to a 10 F, capacitor
charged to 100 V
Check on page 12
Figures 6 and 7 summarise in graphical form the voltages and currents in C-R circuits like
that shown in Figure 3, repeated as Figure 5. -.
All the graphs are exponential in form.
R
S Charge
Current i
Direction shown is during
charging
Discharge
V
Volts
C
Figure 5
8
Voltage
V
Vc
0.63 V
0.5 V
0.36 V
VR
Time t
O.7T 1T
2T
3T
4T
5T
Figure 6
Figure 6 shows the voltages in a C-R circuit during charging for the circuit shown in Figure
5.
Voltage
V
0.5 V
0.36 V
Time t
O.7T 1T
2T
3T
4T
5T
Figure 7
Figure 7 shows the voltage during the discharge of the capacitor shown in Figure 5.
The current varies in the same way, that is exponentially.
SUMMARY
Charging

The voltage across a capacitor, initially uncharged, in series with a resistor rises during
charge from zero to equal the potential of the source of e.m.f.

The current in a C-R circuit is a maximum at the start of the charge falling to zero when
the capacitor is charged up.

The voltage across the resistor at the start of the charge is a maximum falling to zero at
the end of the charge.
9


The changes of current and voltage in charging a capacitor are exponential.
The growth of voltage across a capacitor in a series C-R circuit during the charging
period is given by: Vc = V(1- e-t/CR)

The decay of current in a series C-R circuit during charge is given by:
i = I e-t/T = V e-t/T
R
Discharging

The discharge of a capacitor through a resistor produces similar falls in both current
and voltage.

At the start of discharge the current and voltage have maximum values.

During the discharge period, current and voltage fall most rapidly at the start.

when all the energy initially stored in the capacitor has been converted to heat, the
current and voltage cease.

The fall of voltage in a series (%R circuit during the discharging period is given by:
Vc = V e-t/CR

The decay of current in a series ~R circuit during discharge is given by:
i = -I e-t/CR = -V e-t/CR
R
Time constant

The time constant of a series C-R circuit is defined as the time which would be taken
for steady state conditions to be reached if the initial rates of change of voltage and
current were maintained.

Its numerical value is the product of C and R.

If C is measured in farads and R in ohms, T is in seconds.
10
CHECK ON PROGRESS QUESTION C.1
The time constant T is given by T = CR, where T is in seconds if C is in farads and R is in
ohms.
a)
T = CR = 1 x 10-6 x 22 x 103 = 22 x103 seconds T = 22 ms
b)
T = CR = 2.2 x 10-9x 15 x 103 = 33 x 10-6seconds T =33 s
c)
T = CR = 47 x 10-l2 x 82 = 3850 x 10-12 seconds T = 3.85ns
d)
T = CR = 68 x 10-6 x 103 = 68 x 103 seconds T = 68 ms
CHECK ON PROGRESS QUESTION C.2
a)
T = CR,therefore C = T
R
C = 22 x10-6 = 2.2 x 10-9 F
103
C = 2.2nF
b)
C = 3 x 10-3 = 0.2 x10-6 F C = 0.2F
15 x103
c)
C = 10 = 4.54 x 10-6 F C = 4.54F
2.2 x 106
d)
R = T = 22x109 = 103  R = 1k
C
22 x l0-12
CHECK ON PROGRESS QUESTION C.3 Use the formula Vc = V(1- e-t/CR)
First work out:
t/CR = 60/(1. 102x1x 104)=0.6
Therefore Vc = 12 (1- e-0.6)
= 12 (1 - 0.5488) = 12 x 0.4511 = 5.41 volts
11
CHECK ON PROGRESS QUESTION C.4 Use the formula Vc = V(1- e-t/CR)
Time constant T = CR = 100 x 10-6 x 10 x 103 = 1 second
t/T = 0.2/1 = 0.2
Therefore VC = 100 (1 – e-0.2)
= 100 (1 - 0.819) = 18.1 volts
CHECK ON PROGRESS QUESTION C.5
Initial current I = V/R= 100/10000 = 0.01A
=
I e-t/CR
But:
t/CR = 2/(500 x 10-6 x 10 x 103) = 0.4
Therefore:
ic =
0.01e-0.4 = 6.7 mA
CHECK ON PROGRESS QUESTION C.6
Time constant T = CR = 10 x 10-6 x 0 1 = 10-6 seconds
t/T = 2 x 10-6/10-6
=2
Initial current I = 100/0.1 = 1000 A
Using Ic = -I e-t/CR
= -1000 e-2
= - 135 A
12
```
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