DC SWITCHING TRANSIENTS

Both inductors L and capacitors C have the ability to store energy.

Inductance stores magnetic field energy when current is flowing through it.

Energy stored W = ½ LI2 where W = energy stored (joules)

L = inductance (henry)
I = current flowing (amp)

This energy storage means that the current through the inductor cannot be changed instantly.

Capacitance stores electric field energy in its dielectric when a voltage exists across the
capacitor.

Energy stored W = ½ C υ 2 where W = energy stored (joules)

C = capacitance (Farads)
υ = voltage across the capacitor

This energy storage means that the voltage across the capacitor cannot be changed instantly.

Example 1

An inductance of 10mH takes a dc current of 5A. What is the value of the energy stored in
the magnetic field of the inductor?

W = ½ L I2 L = 10 mH
I = 5A

∴ W = ½ x 10 x 52

∴ W = 125 mW

GROWTH AND DECAY OF CURRENT IN R-L CIRCUIT

R L

V
+• •-

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A circuit consisting of a resistance R in series with an inductance L is connected through a
switch S to a dc supply of V volts. When the switch is closed the current starts to build up
but it cannot rise instantly because an induced EMF in the inductance acts against V which
allows the inductance to absorb energy from the supply.

The current will slowly increase until it reaches its maximum possible value which is
determined by the supply voltage V and the circuit resistance R.

Final value of current I = V

R

Current FINAL VALUE OF CURRENT

Time
t

switch closed
t=0

The time when the current is building up is called the TRANSIENT. During the transient the
changing current is given the symbol i

The growth of current during the transient is EXPONENTIAL and is given by the equation

i = I (l - e-t/τ)

where i = current during the transient

I = final value of current
=V
R

t = time after the switch is closed

τ = TIME CONSTANT of the circuit

τ = L
R

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Example 2

A coil has a resistance of 5 Ω and an inductance of 150 mH and is switched across a 30 v dc

supply.

Calculate EACH of the following:

(a) the current time constant

(b) the final steady current in the coil

(c) the value of the current 18 ms after switch on

(d) energy stored in the magnetic field after 18 ms

(e) the time taken for the current to reach 4.5A.

(a) Time Constant τ = L τ - seconds

R L – Henrys
R – ohms

= 150
5

= 30 ms

(b) Final Value of Current I = V

R

= 30
5

= 6A

(c) i = I (l-e-t/τ) I = 6A

= 6 (l-e-18/30) t = 18 ms

= 6 (l-e-0.6)

= 6 (l – 0.549)

= 6 x 0.451

= 2.706A

(d) Energy stored = ½ Li2 L - inductance

i - current after 18 ms

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 = ½ x 150 x 2.7062

= 549.2 mW

(e) i = I (l-e-t/τ)

4.5 = 6 (l-e-t/30)

4.5 = l-e-t/30
6

0.75 = l-e-t/30

e-t/30 = 1 – 0.75 = 0.25

e-t/30 = 0.25

Take natural log of both sides: - t = -1.386

30

∴ t = 30 x 1.386
t = 41.59 ms

When the switch S is opened the current through the circuit will fall but it cannot fall to zero
instantly because the energy stored in the inductance must be dissipated. As the current falls
it induces an EMF in the inductance which maintains current flow until the stored energy is
dissipated as heat in the resistance and in the arc which occurs at the switch contacts.

Current I=V
R

t=0 time
switch opened t

Just as the growth of current after switch-on was exponential, so the decay of current when
the switch is opened is exponential and given by the equation:

i = Ie-t/τ where i = current during the transient

I = initial value of current
t = time after switch is opened
τ = Time Constant of the circuit = L
R

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Example 3

A coil has a resistance of 12 Ω and an inductance of 0.3 H and carries a steady current of 2 A.

Calculate each of the following:

(a) The circuit time constant.

(b) The coil current 20 ms after the supply is switched off.

(a) τ = L
R

= 0.3
12

= 0.025 s

(b) i = Ie-t/τ

= 2e-20/25

= 2e-0.8

= 2 x 0.449

= 0.899 A

TIME CONSTANT

i = I (1- et/τ) if the time after switch-on t = τ the value of the time constant then

= I (1- et/τ)

= I (1- e-1)

= I (1 – 0.3678)

= 0.632 I

i.e. = 63.2% of I

Time constant can be defined as:

The time taken to reach 63.2% of the final value.

The transient period can be considered finished after a time equal to 5 x time constant i.e.
after 5τ.

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GROWTH AND DECAY OF CURRENT IN R-C CIRCUIT

R C

V
+• •-

A circuit consisting of a resistance R in series with a capacitance C is connected through a

switch S to a d.c. supply of V volts. When the switch is closed the voltage across the
capacitor at this instant is zero so the circuit current immediately reaches an initial value of:

I = V
R

The capacitor, at the instant of switch-on acts as a short circuit. But now the capacitor begins
to charge and a voltage υc develops across it, this reduces the circuit current. So, the current
decays from I = V/R and the capacitor voltage υc grows.

When υc = V, the supply voltage, then the circuit current = 0 and the capacitor is fully
charged.
i
I=V
R
-V
υc

time

The time when the current is decaying and the voltage across the capacitor building up is
called the TRANSIENT.

The transient current is given the symbol i and the transient capacitor voltage is given the
symbol υc.

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The transients are exponential and given by the equations:

υc = V (1- et/τ)

where υc = transient capacitor voltage

V = supply voltage and final capacitor voltage.

t = time after switch on

τ = circuit TIME CONSTANT

τ = CR

and i = I e-t/τ where I = initial charging current I = V

R

If the switch S is opened then the capacitor will hold its charge and the voltage across its
plates until it is discharged through a resistor.

R1
S1
•

R C

V
+• •-

If the capacitor C has been fully charged from the supply through switch S then the voltage
on the capacitor will be V.

The circuit time constant during the charging transient will be:

τ = CR

If switch S is now opened and S1 closed, the capacitor C will discharge through resistor R1
with a time constant of τ1 = CR1.

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V INITIAL VOLTAGE V

Voltage υc

current

INITIAL DISCHARGE CURRENT I1 = V

R1

Again, the curves are exponential, note that the discharge current is in the opposite direction
to the charging current.

The equations for current and voltage during the discharge are:

υc = Ve-t/τ1 V = initial capacitor voltage

υc = transient capacitor voltage
t = time after switch-on
τ1 = circuit time constant
τ = CR1

and i = Ie-t/τ1 I = initial discharge current

=V
R1

As with the R-L circuit, the time constant is the time taken to reach 63.2% of the final value
and the transient period can be considered complete after 5 time constants.

The circuit diagram shows the overall circuit conditions during the charging transient.

R C

υR υc
o
S
o

i V
+• •-

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υc = V (Ie-t/τ)

i = Ie-t/τ

υR = V - υc

or υR = iR

Example 4

A 40 µF capacitor in series with a 1 k Ω resistor is switched on to a 50 V dc supply.

Calculate EACH of the following:

(a) The time constant of the circuit.

(b) The voltage across the capacitor after 60 ms.

(c) The charging current after 60 ms.

(d) The time taken for the capacitor voltage to reach 20 V.

(e) Estimate the time for the charging current to fall to zero.

1 kΩ 40 µ F

υR υc
o
S
o

+• •-
50 V

I = 50 mA V = 50 V υc
1 I = 50mA
i
= 50 mA

time

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(a) τ = CR

= 40 x 10-6 x 1 x 103

= 40 ms

(b) υc = V (l - e-t/τ)

= 50 (1 - e-60/40)

υc = 50 (1 - e-1.5)

= 50 (1 – 0.2231)

=38.845 V

(c) There are two ways to calculate this. We know that υc = 38.845 V after 60 ms so:

υR = V - υc

= 50 – 38.845

= 11.155 V

i = υR
R

= 11.155
1K

= 11.155 mA

Alternatively:

= Ie-t/τ

= 50 x 0.2231
1K

= 50 x 0.2231

= 11.155 mA

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FURTHER WORKED EXAMPLES

Example 5

A relay coil has an operating current of 20 mA i.e. the relay contacts will not operate until the
current in the relay coil reaches 20 mA.

When a 24 V dc supply is switched onto the relay coil it does not operate until 50 ms after the
supply has been switched on. The relay coil resistance is measured as 500 Ω .

(a) Calculate EACH of the following:

(i) The time constant of the relay coil

(ii) The inductance of the relay coil

(b) Calculate the new operating time of the relay if a 100 Ω resistor is connected in series
with the relay coil.

RELAY COIL

R L

24 V
+• •-

i = I (l - e-t/τ) I = V
R

20 = 48 (l - e-50/τ) = 24
0.5

20 = l - e-50/τ = 48 mA
48

0.41667 = l - e-50/τ

e-50/τ = 1 – 0.41667

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e-50/τ = 0.5833

take natural logs - 50 = 1n 0.5833

τ

- 50 = -0.539
τ

∴ τ = 50
0.539

τ = 92.764 ms

(b) τ = L
R

∴ L = τR

= 92.764 x 10-3 x 500

L = 46.38 H

RELAY COIL

100 Ω 500 Ω 46.38 H

24 V
• + •-

(b) Total circuit resistance is now 100 + 500 = 600 Ω

New time constant = L

R

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 = 46.38
0.6

= 77.3 ms

∴ 20 = 24 (l - e-t/77.3)
0.6

20 = 40 (l - e-t/77.3)

20 = l - e-t/77.3
40

∴ e-t/77.3 = 0.5

Take natural logs -t = -0.693

77.3

t = 0.693 x 77.3

t = 53.58 ms

Example 6

A resistor of 2 k Ω is connected in series with 180 µ F capacitor. The circuit is switched on

to a 110 v dc supply.

(a) Estimate the time taken for the capacitor to be charged up to the supply voltage.

(b) Calculate EACH of the following:

(i) the initial charging current

(ii) the voltage across the resistor 300 ms after closing the switch.

(b) The charged capacitor is now discharged through a 50 k Ω resistor. Calculate the
current in the circuit 4.5 seconds after switch-on.

2 kΩ 180 µ F

110 v
+• •-

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(a) The capacitor will be fully charged up to the supply voltage after 5 time constants.

τ = CR
= 180 x 10-6 x 2 x 103
= 360 ms

∴ Time to full charge = 5 x 360

= 1800 ms
= 1.8 s

(b) (i) I = V
R

= 110
2k

= 55 mA

(ii) υc = V (l - e-t/τ)

= 110 (l - e-300/360)

= 110 (l - e-0.833)

υc = 110 (l – 0.4346)

= 110 x 0.565

= 62.194 V

υR = V - υc

υR = 110 – 62.194

= 47.81 V

(c)

50 k Ω

S 180 µ F

+ -
V =110V

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 New time constant = CR

= 180 x 10-6 x 50 x 103

= 9000 ms

= 9s

Initial discharge current I = V

R

I = 110
50k

= 2.2 mA

= 2.2 (l - e-4.5/9)

= 2.2 x 0.393

= 0.865 mA

Example 7

A contactor coil has a resistance of 260 Ω and an inductance of 5 H. The coil is switched on
to a 110V dc supply.

Calculate EACH of the following:

(a) The time constant of the coil(19.2 ms)

(b) The steady-state value of the coil current (0.423A)
(c) The value of the coil current 19.2 ms after switch-on (0.267A)
(d) Estimate the time taken for the coil current to reach its steady-state value (96 mS)

Example 8

A capacitor of 100 µ F is charged from a 24 V dc supply and is fully charged in 1.4 seconds.
The capacitor is then disconnected and discharged through a 10 k Ω resistor.

(a) Estimate the value of the charging resistor (2.8 k Ω )

(b) Calculate EACH of the following:

(i) The initial discharge current (2.4 mA)

(ii) The time taken for the discharge current to fall to 1.5 mA (0.47 s)

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