Bird J. Electrical Circuit Theory and Technology
Подождите немного. Документ загружается.
266 Electrical Circuit Theory and Technology
(c) Current, i D Ie
t/
Working in mA units, i D 0.4e
1/1
D 0.4 ð 0.368 D 0.147 mA
(d) Capacitor voltage,
v
C
D V1 e
t/
D 201 e
2/1
D 201 0.135 D 20 ð 0.865 D 17.3 V
(e) Resistor voltage,
v
R
D Ve
t/
Thus 15 D 20e
t/1
,
15
20
D e
t
,i.e.e
t
D
20
15
D
4
3
Taking natural logarithms of each side of the equation gives
t D ln
4
3
D ln1.3333
i.e, time, t
= 0.288 s
Problem 4. A circuit consists of a resistor connected in series with
a0.5
µF capacitor and has a time constant of 12 ms. Determine
(a) the value of the resistor, and (b) the capacitor voltage 7 ms
after connecting the circuit to a 10 V supply
(a) The time constant D CR, hence R D
C
i.e. R D
12 ð 10
3
0.5 ð 10
6
D 24 ð10
3
D 24 kZ
(b) The equation for the growth of capacitor voltage is:
v
C
D V1 e
t/
Since D 12 ms D 12 ð 10
3
s, V D 10Vand
t D 7msD 7 ð10
3
s,
then
v
C
D 10
1 e
7ð10
3
12ð10
3
D 101 e
0.583
D 101 0.558 D 4.42 V
Alternatively, the value of
v
C
when t is 7 ms may be determined
using the growth characteristic as shown in Problem 1.
17.5 Discharging a
capacitor
When a capacitor is charged (i.e. with the switch in position A in
Figure 17.7), and the switch is then moved to position B, the electrons
stored in the capacitor keep the current flowing for a short time. Initially,
at the instant of moving from A to B, the current flow is such that the
capacitor voltage
v
C
is balanced by an equal and opposite voltage v
R
D iR.
Since initially
v
C
D v
R
D V, then i D I D V/R. During the transient
D.c. transients 267
Figure 17.7
decay, by applying Kirchhoff’s voltage law to Figure 17.7,
v
C
D v
R
.
Finally the transients decay exponentially to zero, i.e.
v
C
D v
R
D 0. The
transient curves representing the voltages and current are as shown in
Figure 17.8.
The equations representing the transient curves during the discharge
period of a series connected C –R circuit are:
decay of voltage,
v
C
D v
R
D Ve
t/CR
D Ve
t/
decay of current, i D Ie
t/CR
D Ie
t/
When a capacitor has been disconnected from the supply it may still be
charged and it may retain this charge for some considerable time. Thus
precautions must be taken to ensure that the capacitor is automatically
discharged after the supply is switched off. This is done by connecting a
high value resistor across the capacitor terminals.
Problem 5. A capacitor is charged to 100 V and then discharged
through a 50 k resistor. If the time constant of the circuit is 0.8 s,
determine:(a) thevalue of the capacitor,(b) the timefor the capacitor
voltage to fall to 20 V, (c) the current flowing when the capacitor
has been discharging for 0.5 s, and (d) the voltage drop across the
resistor when the capacitor has been discharging for one second.
Figure 17.8
Parts (b), (c) and (d) of this problem may be solved graphically as shown
in Problems 1 and 2 or by calculation as shown below.
V D 100 V, D 0.8s,R D 50 k D 50 ð 10
3
(a) Since time constant, D CR, C D /R
i.e. C D
0.8
50 ð 10
3
D 16 mF
(b)
v
C
D Ve
t/
20 D 100e
t/0.8
, i.e.
1
5
D e
t/0.8
Thus e
t/0.8
D 5 and taking natural logarithms of each side, gives
t
0.8
D ln5, i.e., t D 0.8ln5
Hence t
= 1.29 s
(c) i D Ie
t/
The initial current flowing, I D
V
R
D
100
50 ð 10
3
D 2mA
Working in mA units, i D Ie
t/
D 2e
0.5/0.8
D 2e
0.625
D 2 ð 0.535 D 1.07 mA
268 Electrical Circuit Theory and Technology
(d) v
R
D v
C
D Ve
t/
D 100e
1/0.8
D 100e
1.25
D 100 ð 0.287 D 28.7 V
Problem 6. A 0.1 µF capacitor is charged to 200 V before being
connected across a 4 k resistor. Determine (a) the initial discharge
current, (b) the time constant of the circuit, and (c) the minimum
time required for the voltage across the capacitor to fall to less
than 2 V
(a) Initial discharge current, i D
V
R
D
200
4 ð 10
3
D 0.05 A or 50 mA
(b) Time constant D CR D 0.1 ð10
6
ð 4 ð 10
3
D 0.0004 s or 0.4 ms
(c) The minimum time for the capacitor voltage to fall to less than 2 V,
i.e., less than or
2
200
or 1% of the initial value is given by 5.
5 D 5 ð 0.4 D 2ms
In a d.c. circuit, a capacitor blocks the current except during the
times that there are changes in the supply voltage.
Further problems on transients in series connected C-R circuits may be
found in Section 17.12, problems 1 to 8, page 276.
17.6 Current growth in
an L–R circuit
(a) The circuit diagram for a series connected L –R circuit is shown
in Figure 17.9. When switch S is closed, then by Kirchhoff’s
voltage law:
V D
v
L
C v
R
17.3
(b) The battery voltage V is constant. The voltage across the inductance
is the induced voltage, i.e.
v
L
D L ð
change of current
change of time
D L
di
dt
The voltage drop across R,
v
R
is given by iR. Hence, at all times:
V D Ldi/dt C iR 17.4
Figure 17.9
D.c. transients 269
Figure 17.10
(c) At the instant of closing the switch, the rate of change of current is
such that it induces an e.m.f. in the inductance which is equal and
opposite to V, hence V D
v
L
C 0, i.e. v
L
D V. From equation (17.3),
because
v
L
D V, then v
R
D 0andi D 0
(d) A short time later at time t
1
seconds after closing S, current i
1
is
flowing, since there is a rate of change of current initially, resulting
in a voltage drop of i
1
R across the resistor. Since V (constant) D
v
L
C v
R
the induced e.m.f. is reduced, and equation (17.4) becomes:
V D L
di
1
dt
1
C i
1
R
(e) A short time later still, say at time t
2
seconds after closing the
switch, the current flowing is i
2
, and the voltage drop across the
resistor increases to i
2
R. Since v
R
increases, v
L
decreases.
(f) Ultimately, a few seconds after closing S, the current flow is entirely
limited by R, the rate of change of current is zero and hence
v
L
is zero. Thus V D iR. Under these conditions, steady state current
flows, usually signified by I. Thus, I D V/R,
v
R
D IR and v
L
D 0
at steady state conditions.
(g) Curves showing the changes in
v
L
, v
R
and i with time are shown
in Figure 17.10 and indicate that
v
L
is a maximum value initially
(i.e equal to V), decaying exponentially to zero, whereas
v
R
and i
grow exponentially from zero to their steady state values of V and
I D V/R respectively.
17.7 Time constant for
an L–R circuit
With reference to Section 17.3, the time constant of a series connected
L –R circuit is defined in the same way as the time constant for a series
connected C –R circuit. Its value is given by:
time constant, t = L=R seconds
17.8 Transient curves for
an L–R circuit
Transient curves representing the induced voltage/time, resistor
voltage/time and current/time characteristics may be drawn graphically, as
outlined in Section 17.4. A method of construction is shown in Problem 7.
Each of the transient curves shown in Figure 17.10 have mathematical
equations, and these are:
decay of induced voltage,
v
L
D Ve
Rt/L
D Ve
t/
growth of resistor voltage, v
R
D V1 e
Rt/L
D V1 e
t/
270 Electrical Circuit Theory and Technology
growth of current flow, i D I1 e
Rt/L
D I1 e
t/
These equations are derived analytically in Chapter 45.
The application of these equations is shown in Problem 9.
Problem 7. A relay has an inductance of 100 mH and a resistance
of 20 . It is connected to a 60 V, d.c. supply. Use the ‘initial slope
and three point’ method to draw the current/time characteristic and
hence determine the value of current flowing at a time equal to two
time constants and the time for the current to grow to 1.5 A
Before the current/time characteristic can be drawn, the time constant and
steady-state value of the current have to be calculated.
Time constant, D
L
R
D
100 ð 10
3
20
D 5ms
Final value of current, I D
V
R
D
60
20
D 3A
The method used to construct the characteristic is the same as that used
in Problem 2.
(a) The scales should span at least five time constants (horizontally),
i.e. 25 ms, and 3 A (vertically).
(b) With reference to Figure 17.11, the initial slope is obtained by
making AB equal to 1 time constant, (5 ms), and joining OB.
Figure 17.11
(c) At a time of 1 time constant, CD is 0.632 ðI D 0.632 ð 3 D
1.896 A
At a time of 2.5 time constants, EF is 0.918 ð I D 0.918 ð 3 D
2.754 A
At a time of 5 time constants, GH is I D 3A
D.c. transients 271
(d) A smooth curve is drawn through points O, D, F and H and this
curve is the current/time characteristic.
From the characteristic, when t D 2, i ³ 2.6A. [This may be checked
by calculation using i D I1 e
t/
, where I D 3andt D 2, giving
i D 2.59 A]. Also, when the current is 1.5 A, the corresponding time is
about 3.6 ms. [This may be checked by calculation, using i D I1 e
t/
where i D 1.5, I D 3and D 5 ms, giving t D 3.466 ms.]
Problem 8. A coil of inductance 0.04 H and resistance 10 is
connected to a 120 V, d.c. supply. Determine (a) the final value of
current, (b) the time constant of the circuit, (c) the value of current
after a time equal to the time constant from the instant the supply
voltage is connected, (d) the expected time for the current to rise
to within 1% of its final value.
(a) Final steady current, I D
V
R
D
120
10
D 12 A
(b) Time constant of the circuit, D
L
R
D
0.04
10
D 0.004 s or 4 ms
(c) In the time s the current rises to 63.2% of its final value of 12 A,
i.e. in 4 ms the current rises to 0.632 ð 12 D 7.58 A
(d) The expected time for the current to rise to within 1% of its final
value is given by 5 s, i.e. 5 ð4 D 20 ms
Problem 9. The winding of an electromagnet has an inductance
of 3 H and a resistance of 15 . When it is connected to a 120 V,
d.c. supply, calculate:
(a) the steady state value of current flowing in the winding,
(b) the time constant of the circuit,
(c) the value of the induced e.m.f. after 0.1 s,
(d) the time for the current to rise to 85% of its final value, and
(e) the value of the current after 0.3 s
(a) The steady state value of current is I D V/R,i.e.I D 120/15 D 8A
(b) The time constant of the circuit, D L/R D 3/15 D 0.2 s
Parts (c), (d) and (e) of this problem may be determined by drawing
the transients graphically, as shown in Problem 7 or by calculation
as shown below.
272 Electrical Circuit Theory and Technology
(c) The induced e.m.f., v
L
is given by v
L
D Ve
t/
. The d.c. voltage V
is 120 V, t is 0.1 s and t is 0.2 s, hence
v
L
D 120e
0.1/0.2
D 120e
0.5
D 120 ð 0.6065
i.e.
v
L
D 72.78 V
(d) When the current is 85% of its final value, i D 0.85I.
Also, i D I1 e
t/
, thus 0.85I D I1 e
t/
0.85 D 1 e
t/
and D 0.2, hence
0.85 D 1 e
t/0.2
e
t/0.2
D 1 0.85 D 0.15
e
t/0.2
D
1
1.15
D 6.
P
6
Taking natural logarithms of each side of this equation gives:
lne
t/0.2
D ln6.
P
6, and by the laws of logarithms
t
0.2
lne D ln6.
P
6. But lne D 1, hence
t D 0.2ln6.
P
6i.e.t D 0.379 s
(e) The current at any instant is given by i D I1 e
t/
When I D 8,t D 0.3and D 0.2, then
i D 81 e
0.3/0.2
D 81 e
1.5
D 81 0.2231 D 8 ð 0.7769 i.e., i D 6.215 A
17.9 Current decay in an
L–R circuit
When a series connected L –R circuit is connected to a d.c. supply as
shown with S in position A of Figure 17.12, a current I D V/R flows
after a short time, creating a magnetic field / I associated with the
inductor. When S is moved to position B, the current value decreases,
causing a decrease in the strength of the magnetic field. Flux linkages
occur, generating a voltage
v
L
, equal to Ldi/dt. By Lenz’s law, this
voltage keeps current i flowing in the circuit, its value being limited by
R. Thus
v
L
D v
R
. The current decays exponentially to zero and since v
R
is
proportional to the current flowing,
v
R
decays exponentially to zero. Since
v
L
D v
R
, v
L
also decays exponentially to zero. The curves representing
these transients are similar to those shown in Figure 17.8.
The equations representing the decay transient curves are:
decay of voltages,
v
L
D v
R
D Ve
Rt/L
D Ve
t/
decay of current, i D Ie
Rt/L
D Ie
t/
Figure 17.12
D.c. transients 273
Problem 10. The field winding of a 110 V, d.c. motor has a resis-
tance of 15 and a time constant of 2 s. Determine the inductance
and use the tangential method to draw the current/time characteristic
when the supply is removed and replaced by a shorting link. From
the characteristic determine (a) the current flowing in the winding
3 s after being shorted-out and (b) the time for the current to decay
to 5 A.
Since the time constant, D
L
R
,LD R
i.e. inductance L D 15 ð2 D 30 H
The current/time characteristic is constructed in a similar way to that used
in Problem 1.
(i) The scales should span at least five time constants horizontally, i.e.
10 s, and I D V/R D 110/15 D 7.
P
3 A vertically.
(ii) With reference to Figure 17.13, the initial slope is obtained by
making OB equal to 1 time constant, (2 s), and joining AB.
(iii) At, say, i D 6 A, let C be the point on AB corresponding to a
current of 6 A. Make DE equal to 1 time constant, (2 s), and join
CE.
(iv) Repeat the procedure given in (iii) for current values of, say, 4 A,
2 A and 1 A, giving points F, G and H.
(v) Point J is at five time constants, when the value of current is zero.
Figure 17.13
(vi) Join points A, C, F, G, H and J with a smooth curve. This curve is
the current/time characteristic.
(a) From the current/time characteristic, when t D 3s,i D 1.5 A. [This
may be checked by calculation using i D Ie
t/
, where I D 7.
P
3,
t D 3and D 2, giving i D 1.64 A.] The discrepancy between the
two results is due to relatively few values, such as C, F, G and H,
being taken.
(b) From the characteristic, when i D 5A,t
= 0.70 s. [This may be
checked by calculation using i D Ie
t/
, where i D 5, I D 7.
P
3, D
2, giving t D 0.766 s.] Again, the discrepancy between the graph-
ical and calculated values is due to relatively few values such as C,
F, G and H being taken.
Problem 11. A coil having an inductance of 6 H and a resistance
of Ris connected in series with a resistor of 10 to a 120 V,
d.c. supply. The time constant of the circuit is 300 ms. When
steady-state conditions have been reached, the supply is replaced
instantaneously by a short-circuit. Determine: (a) the resistance of
the coil,(b) the current flowing in the circuit one second after the
shorting link has been placed in the circuit, and (c) the time taken
for the current to fall to 10% of its initial value.
274 Electrical Circuit Theory and Technology
(a) The time constant, D
circuit inductance
total circuit resistance
D
L
R C 10
Thus R D
L
10 D
6
0.3
10 D 10 Z
Parts (b) and (c) may be determined graphically as shown in Prob-
lems 7 and 10 or by calculation as shown below.
(b) The steady-state current, I D
V
R
D
120
10 C 10
D 6A
The transient current after 1 second, i D Ie
t/
D 6e
1/0.3
Thus i D 6e
3.
P
3
D 6 ð 0.03567 D 0.214 A
(c) 10% of the initial value of the current is 10/100 ð 6, i.e. 0.6 A
Using the equation i D Ie
t/
gives
0.6 D 6e
t/0.3
i.e.
0.6
6
D e
t/0.3
or e
t/0.3
D
6
0.6
D 10
Taking natural logarithms of each side of this equation gives:
t
0.3
D ln10
t D 0.3ln10D 0.691 s
Problem 12. An inductor has a negligible resistance and an induc-
tance of 200 mH and is connected in series with a 1 k resistor
to a 24 V, d.c. supply. Determine the time constant of the circuit
and the steady-state value of the current flowing in the circuit. Find
(a) the current flowing in the circuit at a time equal to one time
constant, (b) the voltage drop across the inductor at a time equal
to two time constants and (c) the voltage drop across the resistor
after a time equal to three time constants.
The time constant, D
L
R
D
0.2
1000
D 0.2 ms
The steady-state current I D
V
R
D
24
1000
D 24 mA
(a) The transient current, i D I1 e
t/
and t D 1
Working in mA units gives, i D 241 e
1t/
D 241 e
1
D 241 0.368 D 15.17 mA
D.c. transients 275
(b) The voltage drop across the inductor, v
L
D Ve
t/
When t D 2, v
L
D 24e
2/
D 24e
2
D 3.248 V
(c) The voltage drop across the resistor,
v
R
D V1 e
t/
When t D 3,
v
R
D 241 e
3/
D 241 e
3
D 22.81 V
Further problems on transients in series L –R circuits may be found in
Section 17.12, problems 9 to 12, page 277.
17.10 Switching
inductive circuits
Energy stored in the magnetic field of an inductor exists because a current
provides the magnetic field. When the d.c. supply is switched off the
current falls rapidly, the magnetic field collapses causing a large induced
e.m.f. which will either cause an arc across the switch contacts or will
break down the insulation between adjacent turns of the coil. The high
induced e.m.f. acts in a direction which tends to keep the current flowing,
i.e. in the same direction as the applied voltage. The energy from the
magnetic field will thus be aided by the supply voltage in maintaining
an arc, which could cause severe damage to the switch. To reduce the
induced e.m.f. when the supply switch is opened, a discharge resistor R
D
is connected in parallel with the inductor as shown in Figure 17.14. The
magnetic field energy is dissipated as heat in R
D
and R and arcing at the
switch contacts is avoided.
Figure 17.14
17.11 The effects of time
constant on a rectangular
waveform
Integrator circuit
By varying the value of either C or R in a series connected C–R circuit,
the time constant D CR, of a circuit can be varied. If a rectan-
gular waveform varying from CE to E is applied to a C –R circuit as
shown in Figure 17.15, output waveforms of the capacitor voltage have
various shapes, depending on the value of R. When R is small, t D CR
is small and an output waveform such as that shown in Figure 17.16(a)
is obtained. As the value of R is increased, the waveform changes to that
shown in Figure 17.16(b). When R is large, the waveform is as shown in
Figure 17.16(c), the circuit then being described as an integrator circuit.
Differentiator circuit
If a rectangular waveform varying from CE to E is applied to a series
connected C–R circuit and the waveform of the voltage drop across the
resistor is observed, as shown in Figure 17.17, the output waveform alters
as R is varied due to the time constant, D CR, altering. When R is
small, the waveform is as shown in Figure 17.18(a), the voltage being
generated across R by the capacitor discharging fairly quickly. Since the
Figure 17.15