Chapter 18 D.C. Transients

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.

Capacitor voltage, = 39.35 V

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

hence, capacitance, C = = 0.309 μF

(b) Resistor voltage, = 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, = 105.0 V

When the voltage falls to 25 V, 25 = from which,

Hence, and time, t = - 15 = 23.53 s

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,

when t = 17 s hence, i.e.

from which,

and capacitance, C = = 55.90 μF

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, If the voltage falls by 70%, then

Hence, i.e. 0.30 =

and

from which, resistance, R = = 1.08 M

6. A 50 μF uncharged capacitor is connected in series with a 1 k resistorand 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.

(a) The initial value of the current flowing,I = = = 0.10 A

(b) Time constant,τ= CR = (50 × 10-6)(1 × 103) = 0.05 s = 50ms

(d) Resistor voltage, vR = Ve-t/τ == 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 = = 0.150 s or 150 ms

Initial charging current, I = = 3.67 mA

Current flowing after 120 ms, i= = 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 = = 0.080 s or 80 ms

Initial charging current, I = = 110 mA or 0.11 A

(a) Current flowing after 40 ms, i= = 66.7 mA

(b) Current flowing after 80 ms, i= = 40.5 mA

9. A resistor of 0.5 M is connected in series with a 20 μF capacitor andthe 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 = = 10 s

When discharging, capacitor voltage,

A table of values is shown below.

Time, t (s) 10 20 30 40 50 60

73.6 27.1 10.0 3.7 1.3 0.5

A graph of against time is shown below.

From the graph, when the capacitor voltage is 50 V, the time is 14 s

By calculation, i.e. from which, and

and time, t = - 10 = 13.86 s

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 = = 0.60 s

(b) Initial rate of voltage rise = = 200 V/s (see diagram below, where

gradient = AB/0B)

(d) Capacitor voltage,

When the capacitor voltage is 50 V, then 50 = 120

i.e. from which,

Hence, and time, t = - 0.60 = 0.323 s

11. If a 200 V dc supply is connected to a 2.5 M resistor and a 2 F capacitor in series. Calculate

(a) thecurrent flowing 4 s after connecting, (b) the voltage across the resistor after 4 s, and

(a) Time constant,  = C  R = = 5 s

Initial charging current, I = = 80 A

Current flowing after 4 s, i = = 35.95 A

(b) Voltage across the resistor after 4 s, v = = 89.87 V

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 = = 12.13 mJ

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).

the main points.

(a)(i) Initial current, = 80 A

(ii) Capacitor voltage,

where time constant,  = C  R = = 2.5 s

Hence, when t = 1.5 s, capacitor voltage, = 18.05 V

(iii) When = 12 V, then 12 = 40 from which,

and thus,

from which, time, t = -2.5 ln 0.7 = 0.892 s

(b)(i) Initial current through 1 M resistor, I = = 40 A

(ii) For discharge, time constant, = 2 s

and capacitor voltage, = 6.95 V

Energy stored 3.5 s into discharge, W = = 48.30 J

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.

By calculation, current flowing, and time constant,  = = 0.03 s

and final current, I = = 5 A

Hence, when t = 60 ms = 0.06 s,current, i = = 4.32 A

2. A 25 V d.c. supply is connected to a coil of inductance 1 H and resistance5 . 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.

Time constant,τ = = = 0.2 s

Final value of current,I = = = 5 A

(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

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 isthe 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 = 1.97A ]

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

(a) Current flowing, and time constant,  = = 0.20 s

and final current, I = = 2.5 A

Hence, when t = 0.1 s,current, i = = 0.984 A

(b) When current, i = 1.5 A, then: 1.5 = 2.5 from which,

Rearranging gives: and

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 aninductance of 500 mH.

Calculate:(a) the time constant of the field winding,(b) the value of current flow one time constant

after being connected tothe supply, and(c) the current flowing 50 ms after the supply has been

switched on.

(a) Time constant, τ = = 25 ms

(b) Current flowing, and final current, I = = 10 A

Hence, when t = τ,current, i == 6.32 A

(c)When t = 50 ms,current, i == 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, = 0.15 s

(b) Current after 1 time constant, = 2.528 A

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

(d) When i = 2.5 A, then:

from which, and

Taking logarithms gives: i.e.

andthe time to reach 2.5 A, t = -0.15ln 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 = = 26.67 A/s

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.

(a) Resistor voltage, where = 40 ms = 0.04 s

When the resistor voltage is 8 V, then: and

Rearranging gives: and

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= 5  0.04 = 0.20 s

Energy stored, W = = 0.20 J

(d) On discharge, current, i = where = 0.02666…

When i = 750 mA = 0.75 A, then: 0.75 = 1

from which, ln = ln 0.75 i.e.

andtime for the current to fall to 750 mA, t = - 0.02666…ln 0.75 = 7.67 ms