CHAPTER 18 D.C. TRANSIENTS
Exercise 104, Page 292
1. An uncharged capacitor of 0.2 F is connected to a 100 V, d.c. supply through a resistor of
100 k. Determine, either graphically or by calculation the capacitor voltage 10 ms after the
voltage has been applied.
1010
t






6
3
Capacitor voltage, VC  V 1  e CR   100 1  e 0.210 10010   100 1  e 0.5  = 39.35 V






3
2. A circuit consists of an uncharged capacitor connected in series with a 50 k resistor and has a
time constant of 15 ms. Determine (a) the capacitance of the capacitor and (b) the voltage drop
across the resistor 5 ms after connecting the circuit to a 20 V, d.c. supply.
(a) Time constant,  = C  R = 15 ms
 15 10 3

hence, capacitance, C =
= 0.309 μF
R 50 103
(b) Resistor voltage, vR  Ve

t
CR
 20e

5103
0.309106 50103
= 14.47 V
3. A 10F capacitor is charged to 120 V and then discharged through a 1.5 M resistor. Determine
either graphically or by calculation the capacitor voltage 2 s after discharging has commenced.
Also find how long it takes for the voltage to fall to 25 V.
When discharging, capacitor voltage, v C  Ve
When the voltage falls to 25 V, 25 = 120e
Hence,

t
 25 
 ln 

15
 120 
and

t
15

t
CR
 120e

2
10106 1.5106
from which,
 120e  0.13333 = 105.0 V
t

25
 e 15
120
 25 
time, t = - 15 ln 
 = 23.53 s
 120 
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233
4. A capacitor is connected in series with a voltmeter of resistance 750 k and a battery. When the
voltmeter reading is steady the battery is replaced with a shorting link. If it takes 17s for the
voltmeter reading to fall to two-thirds of its original value, determine the capacitance of the
capacitor.
When discharging, capacitor voltage, vC  Ve

t
CR
17

3
2
2
v C  V when t = 17 s hence,
V  Ve C(75010 )
3
3
17
2
ln    
C (750 103
3
from which,
and
17

3
2
 e C(75010 )
i.e.
3
capacitance, C = 
17
= 55.90 μF
 750 10  ln  23 
3
5. When a 3 F charged capacitor is connected to a resistor, the voltage falls by 70% in 3.9 s.
Determine the value of the resistor.
For discharge, vC  Ve
Hence,

t
CR
0.30V  Ve
and
from which,
If the voltage falls by 70%, then vC  0.30V

3.9
3106 R

i.e.
0.30 = e
ln 0.30  
3.9
3106 R
3.9
3  106  R
resistance, R = 
3.9
= 1.08 M
3 10  ln 0.30
6
6. A 50 μF uncharged capacitor is connected in series with a 1 k resistor and the circuit is switched
to a 100 V, d.c. supply. Determine: (a) the initial current flowing in the circuit, (b) the time
constant, (c) the value of current when t is 50 ms and (d) the voltage across the resistor 60 ms after
closing the switch.
© John Bird Published by Taylor and Francis
234
(a) The initial value of the current flowing, I =
V
100
=
= 0.10 A
R
110 3
(b) Time constant, τ = CR = (50 × 10- 6)(1 × 103) = 0.05 s = 50 ms
(c) Current, i = I e
-t/τ
= 0.10 e

-t/τ
(d) Resistor voltage, vR = Ve
50103
50103
 0.10e1 = 0.03678 = 36.78 mA
= 100e

60103
50103
= 30.12 V
7. An uncharged 5 F capacitor is connected in series with a 30 k resistor across a 110 V, d.c.
supply. Determine the time constant of the circuit, the initial charging current, and the current
flowing 120 ms after connecting to the supply.
Time constant,  = C  R = 5 106  30 103 = 0.150 s or 150 ms
Initial charging current, I =
V
110

= 3.67 mA
R 30 103
Current flowing after 120 ms, i = I e

t
CR
3
 3.67 10 e

120103
15010 3
 3.67 10 3 e  0.8 = 1.65 mA
8. An uncharged 80 F capacitor is connected in series with a 1 k resistor across a 110 V supply.
Determine the time constant of the circuit and the initial value of current flowing. Determine also
the value of current flowing after (a) 40 ms and (b) 80 ms.
Time constant,  = C  R = 80 106 1103 = 0.080 s or 80 ms
Initial charging current, I =
V
110

= 110 mA or 0.11 A
R 1103
(a) Current flowing after 40 ms, i = I e

t
CR

t
CR
(b) Current flowing after 80 ms, i = I e
 0.11e
 0.11e

4010 3
0.080

8010 3
0.080
 0.11e  0.5 = 66.7 mA
 0.11e  1 = 40.5 mA
© John Bird Published by Taylor and Francis
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9. A resistor of 0.5 M is connected in series with a 20 μF capacitor and the capacitor is charged to
200 V. The battery is replaced instantaneously by a conducting link. Draw a graph showing the
variation of capacitor voltage with time over a period of at least 6 time constants. Determine
from the graph the approximate time for the capacitor voltage to fall to 50 V.
Time constant,  = C  R = 20 106  0.5 106 = 10 s
When discharging, capacitor voltage, vC  Ve

t
CR
 200 e

t
10
A table of values is shown below.
Time, t (s)
vC  200e

t
10
10
20
30
73.6 27.1 10.0
40
50
60
3.7 1.3
0.5
A graph of vC against time is shown below.
From the graph, when the capacitor voltage is 50 V, the time is 14 s
By calculation, vC  Ve
and

t
CR
i.e. 50  200e

t
10
t

50
t
 50 
10
e
from which,
and ln 

10
200
 200 
 50 
time, t = - 10 ln 
 = 13.86 s
 200 
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236
10. A 60 F capacitor is connected in series with a 10 k resistor and connected to a 120 V d.c.
supply. Calculate (a) the time constant, (b) the initial rate of voltage rise, (c) the initial charging
current, (d) the time for the capacitor voltage to reach 50 V.
(a) Time constant,  = C  R = 60 106 10 103 = 0.60 s
(b) Initial rate of voltage rise =
Vm 120

= 200 V/s (see diagram below, where

0.60
gradient = AB/0B)
(c) Initial charging current, I =
V
120

= 12 mA
R 10 103
t



CR
(d) Capacitor voltage, vC  V 1  e




When the capacitor voltage is 50 V, then
i.e.
Hence,
t

50
0.60
 1 e
120

from which,
t
50 

 ln 1 

0.60
 120 
and
t



50 = 120 1  e 0.60 


e

t
0.60
 1
50
120
50 

time, t = - 0.60 ln  1 
 = 0.323 s
 120 
11. If a 200 V dc supply is connected to a 2.5 M resistor and a 2 F capacitor in series. Calculate
(a) the current flowing 4 s after connecting, (b) the voltage across the resistor after 4 s, and
(c) the energy stored in the capacitor after 4 s.
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237
(a) Time constant,  = C  R = 2 106  2.5 106 = 5 s
Initial charging current, I =
V
200

= 80 A
R 2.5  106
Current flowing after 4 s, i = I e

t
CR
 80 106 e
(b) Voltage across the resistor after 4 s, v = V e
(c) Energy stored in capacitor, W =

t
CR

4
5
 80 106 e 0.8 = 35.95 A
 200e

4
5
 200e 0.8 = 89.87 V
1
CV 2 where V is the capacitor voltage.
2
Since, after 4s, the supply voltage is 200 V and the resistor voltage is 89.87 V from part (b), then
the capacitor voltage must be 200 – 89.87 = 110.13 V
Hence, energy stored after 4s, W =
1
1
2
CV 2   2 106  110.13 = 12.13 mJ
2
2
12. (a) In the circuit below, with the switch in position 1, the capacitor is uncharged. If the
switch is moved to position 2 at time t = 0 s, calculate the (i) initial current through the
0.5 M, (ii) the voltage across the capacitor when t = 1.5 s, and (iii) the time taken for the
voltage across the capacitor to reach 12 V.
(b) If at the time t = 1.5 s, the switch is moved to position 3, calculate (i) the initial current
through the 1 M resistor, (ii) the energy stored in the capacitor 3.5 s later (i.e. when t = 5 s).
(c) Sketch a graph of the voltage across the capacitor against time from t = 0 to t = 5 s, showing
the main points.
© John Bird Published by Taylor and Francis
238
(a)(i) Initial current, I 
V
40

= 80 A
R 0.5 106
t
 


(ii) Capacitor voltage, vC  V 1  e 


where time constant,  = C  R = 5 106  0.5 106 = 2.5 s
1.5



Hence, when t = 1.5 s, capacitor voltage, vC  40 1  e 2.5  = 18.05 V


t



(iii) When v C = 12 V, then 12 = 40 1  e 2.5 


and
e

from which,
t
2.5
 1
12
 0.7
40
thus,
from which,
t

12
 1  e 2.5
40
t
 ln 0.7
2.5

time, t = -2.5 ln 0.7 = 0.892 s
(b)(i) Initial current through 1 M resistor, I =
V
40

= 40 A
R 1 106
(ii) For discharge, time constant, dis  C  R  2 106 1106 = 2 s
and
capacitor voltage, vC  V e
Energy stored 3.5 s into discharge, W =

t
dis
 40e

3.5
2
= 6.95 V
1
1
2
CV 2   2 106   6.95  = 48.30 J
2
2
(c) A sketch of capacitor voltage against time for the 5 s period is shown below.
© John Bird Published by Taylor and Francis
239
Exercise 105, Page 297
1. A coil has an inductance of 1.2 H and a resistance of 40  and is connected to a 200 V, d.c.
supply. Either by drawing the current/time characteristic or by calculation determine the value of
the current flowing 60 ms after connecting the coil to the supply.
The graph of current/time is shown on page 294, Figure 18.11(c) of textbook.
t
 


By calculation, current flowing, i  I 1  e 


and final current, I =
and time constant,  =
L 1.2

= 0.03 s
R 40
V 200

=5A
R 40
t
0.06
 




Hence, when t = 60 ms = 0.06 s, current, i = I 1  e    5 1  e 0.03   5 1  e  2  = 4.32 A




2. A 25 V d.c. supply is connected to a coil of inductance 1 H and resistance 5 . Use a graphical
method to draw the exponential growth curve of current and hence determine the approximate
value of the current flowing 100 ms after being connected to the supply.
Before the current/time characteristic can be drawn, the time constant and steady-state value of the
current have to be calculated.
τ=
L
1
= = 0.2 s
5
R
Final value of current, I =
V
25
=
=5A
R
5
Time constant,
(a) The scales should span at least five time constants (horizontally), i.e. 1 s, and 5 A (vertically)
(b) With reference to the diagram, the initial slope is obtained by making AB equal to 1 time
constant,
(i.e. 0.2 s), and joining 0B
(c) At a time of 1 time constant, CD is 0.632  I = 0.632  5 = 3.16 A
© John Bird Published by Taylor and Francis
240
At a time of 2.5 time constants, EF is 0.918  I = 0.918  5 = 4.59 A
At a time of 5 time constants, GH is I = 5 A
(d) A smooth curve is drawn through points 0, D, F and H and this curve is the current/time
characteristic.
From the characteristic, when t = 100 ms, i.e. 0.10 s, i ≈ 1.95 A
[This may be checked by calculation using i = I(1 - e-t/τ), where I = 5 and t = 0.1 s, giving i = 1.97 A ]
3. An inductor has a resistance of 20  and an inductance of 4 H. It is connected to a 50 V d.c.
supply. Calculate (a) the value of current flowing after 0.1 s, and (b) the time for the current to
grow to 1.5 A
t
 

(a) Current flowing, i  I 1  e  


and final current, I =
and time constant,  =
L 4

= 0.20 s
R 20
V 50

= 2.5 A
R 20
© John Bird Published by Taylor and Francis
241
t
0.1
 




Hence, when t = 0.1 s, current, i = I 1  e    2.5 1  e 0.2   2.5 1  e 0.5  = 0.984 A




t



(b) When current, i = 1.5 A, then: 1.5 = 2.5 1  e 0.2 


Rearranging gives:
e

t
0.2
 1
1.5
 0.4
2.5
and
from which,

t

1.5
 1  e 0.2
2.5
t
 ln 0.4
0.2
Thus, the time to reach 1.5 A, t = - 0.2 ln 0.4 = 0.183 s
4. The field winding of a 200 V d.c. machine has a resistance of 20  and an inductance of 500 mH.
Calculate: (a) the time constant of the field winding, (b) the value of current flow one time constant
after being connected to the supply, and (c) the current flowing 50 ms after the supply has been
switched on.
(a) Time constant, τ =
L 500 10 3

= 25 ms
R
20
t
 

V 200


(b) Current flowing, i  I 1  e  and final current, I =
= 10 A
R 20


2510
t


 

3

Hence, when t = τ, current, i = I 1  e   10 1  e 2510




3
5010
t


 

3

(c) When t = 50 ms, current, i = I 1  e   10 1  e 2510




3

  10 1  e1  = 6.32 A



  10 1  e 2  = 8.65 A


5. A circuit comprises an inductor of 9 H of negligible resistance connected in series with a 60 
resistor and a 240 V d.c. source. Calculate (a) the time constant, (b) the current after 1 time
constant, (c) the time to develop maximum current, (d) the time for the current to reach 2.5 A,
and (e) the initial rate of change of current.
(a) Time constant,  
L 9

= 0.15 s
R 60
t

 
 

V
240
(b) Current after 1 time constant, i  Im 1  e    1  e   
1  e1  = 2.528 A
R
60




© John Bird Published by Taylor and Francis
242
Alternatively, after 1 time constant the current will rise to 63.2% of its final value of 4 A, i.e.
i = 0.632 × 4 = 2.528 A
(c) Time to develop maximum current = 5  = 5 × 0.15 = 0.75 s
t
t
 




(d) When i = 2.5 A, then: 2.5  4 1  e    4 1  e 0.15 




from which,
t

2.5
 1  e 0.15
4
and
  t 
Taking logarithms gives: ln  e 0.15   ln 0.375


e

t
0.15
i.e.
 1

2.5
 0.375
4
t
 ln 0.375
0.15
and the time to reach 2.5 A, t = - 0.15 ln 0.375 = 0.147 s
(e) Initial rate of increase of current means the gradient of the tangent at t = 0,
i.e. initial rate of increase =
Im
4

= 26.67 A/s
 0.15
6. In the inductive circuit shown below, the switch is moved from position A to position B until
maximum current is flowing. Calculate (a) the time taken for the voltage across the resistance to
reach 8 volts, (b) the time taken for maximum current to flow in the circuit, (c) the energy stored
in the inductor when maximum current is flowing, and (d) the time for the current to drop to
750 mA after switching to position C.
t


(a) Resistor voltage, v R  V 1  e ch






where ch arg e 
L 400 103

= 40 ms = 0.04 s
R
10
© John Bird Published by Taylor and Francis
243
t



When the resistor voltage is 8 V, then: 8  10 1  e 0.04 


e
Rearranging gives:

t
0.04
 1
8
 0.2
10
and 
and
t

8
 1  e 0.04
10
t
 ln 0.2
0.04
from which, time to reach 8 V, t = - 0.04 ln 0.2 = 0.06438 s = 64.38 ms
(b) The time taken for maximum current to flow = 5 ch arg e = 5  0.04 = 0.20 s
(c) Maximum current, I m 
Energy stored, W =
V 10

=1A
R 10
1
1
2
LI m 2   400 103  1 = 0.20 J
2
2

(d) On discharge, current, i = I e
t
dis
where dis 
When i = 750 mA = 0.75 A, then:
from which,
ln e

t
0.02666..
0.75 = 1 e
= ln 0.75
L 400 103

= 0.02666…
R
10  5

t
0.02666..
i.e.

t
 ln 0.75
0.02666..
and time for the current to fall to 750 mA, t = - 0.02666…ln 0.75 = 7.67 ms
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244