LENDI INSTITUTE ENGINEERING TECHNOLOGY CIRCUIT BREAKERS

LENDI INSTITUTE OF ENGINEERING AND TECHNOLOGY
Jonnada, Andhra Pradesh- 535005
UNIT -I
CIRCUIT BREAKERS (PART I)
Presented by,
Dr. Rohit Babu, Associate Professor
Department of Electrical and Electronics Engineering

SYLLABUS
Miniature Circuit Breaker (MCB)– Elementary principles of arc interruption–
Restriking Voltage and Recovery voltages– Restriking phenomenon - RRRV–
Average and Max. RRRV– Current chopping and Resistance switching–
Introduction to oil circuit breakers– Description and operation of Air Blast–
Vacuum and SF6 circuit breakers– CB ratings and specifications– Concept of
Auto reclosing.

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MINIATURE CIRCUIT BREAKER (MCB)
A Miniature Circuit Breaker (MCB) is an
automatically operated electrical switch
used to protect low voltage electrical
circuits from damage caused by excess
current from an overload or short circuit.
MCBs are typically rated up to a current
up to 125 A, do not have adjustable trip
characteristics, and can be thermal or
thermal-magnetic in operation.
Fig. 1 MCB

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MCB: Working Principle
Fig. 2. MCB working
framework

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ELEMENTARY PRINCIPLES OF ARC
INTERRUPTION
Arc Interruption Theory
The insulating material (may be fluid or air) used in circuit breaker should serve two
important functions.
They are written as follows:
1. It should provide sufficient insulation between the contacts when circuit breaker
opens.
2. It should extinguish the arc occurring between the contacts when circuit breaker
opens.

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ELEMENTARY PRINCIPLES OF ARC INTERRUPTION:
METHODS OF ARC INTERRUPTION
There are two methods by which interruption is done.
1. High resistance method
• In this method, its resistance is increased so as to reduce the current to a value
insufficient to maintain the arc.
• The arc resistance can be increased by cooling, lengthening, constraining and splitting
the arc.
• When current is interrupted the energy associated with its magnetic field appears in
the form of electrostatic energy.
• A high voltage appears across the contacts of the circuit breaker.
• If this voltage is very high and more than the withstanding capacity of the gap between
the contacts, the arc will strike again.

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METHODS OF ARC INTERRUPTION Contd.
• This method is applicable only in case of ac circuit breakers.
• In case of ac supply, the current wave passes through a zero point, 100 times per second
at the supply frequency of 50 Hz. This feature of ac is utilized for arc interruption.
• The current is not interrupted at any point other than the zero current instant,
otherwise a high transient voltage will occur across the contact gap.
• The current is not allowed to rise again after a zero current occurs.
• There are two theories to explain the zero current interruption of the arc.
(i) Recovery rate theory (Slepain’s Theory)
(ii) Energy balance theory (Cassie’s Theory)
2. Current Zero Interruption

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• Restriking Voltage
• Recovery Voltage
• Active Recovery Voltage
• Arc Voltage
Before going in details about these theories, we should know the following terms.

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(i) Recovery rate theory (Slepain’s Theory)
• The arc is a column of ionised gases.
• To extinguish the arc, the electrons and ions are to be removed from the gap immediately
after the current reaches a natural zero.
• Ions and electrons can be removed either by recombining them into neutral molecules or by
sweeping them away by inserting insulating medium (gas or liquid) into the gap.
• The arc is interrupted if ions are removed from the gap at a rate faster than the rate of
ionisation.
• In this method, the rate at which the gap recovers its dielectric strength is compared with
the rate at which the gap recovers its dielectric strength is compared with the rate at which
the restriking voltage (transient voltage) across the gap rises.

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• If the dielectric strength increases more rapidly than the restriking voltage, the arc is
extinguished.
• If the restriking voltage rises more rapidly than the dielectric strength, the ionisation persists and
breakdown of the gap occurs, resulting in an arc for another half cycle.
• Fig. 3 (a & b) explains the principle of recovery rate theory.
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(ii) Energy balance theory (Cassie’s Theory)
• The space between the contacts contains some ionised gas immediately after current zero and
hence, it has a finite post-zero resistance.
• At the current zero moment, power is zero because restricking voltage is zero.
• When the arc is finally extinguished, the power again becomes zero, the gap is fully de-ionised
and its resistance is infinitely high.
• In between these two limits, first the power increases, reaches a maximum value, then decreases
and finally reaches zero value as shown in Fig. 14.6.
• Due to the rise of restriking voltage and associated current, energy is generated in the space
between the contacts.
• The energy appears in the form of heat.

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• The circuit breaker is designed to remove this
generated heat as early as possible by cooling
the gap, giving a blast of air or flow of oil at
high velocity and pressure.
• If the rate of removal of heat is faster than the
rate of heat generation the arc is extinguished.
• If the rate of heat generation is more than the
rate of heat dissipation, the space breaks down
again resulting in an arc for another half cycle,
as shown in Fig. 4

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RESTRIKING VOLTAGE AND RECOVERY
VOLTAGE
• The transient voltage which appears
across the breaker contacts at the instant
of arc being extinguished is known as
restriking voltage.
• The power frequency rms voltage, which
appears across the breaker contacts after
the arc is finally extinguished and
transient oscillations die out is called
recovery voltage.
• Fig. 5 shows the restriking and recovery
voltage.

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Arc formation: Video 1

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Expression for Restriking Voltage and RRRV
Considered a simple circuit, having a circuit breaker CB, as shown in the fig. 6.
Let
L be the inductance per phase of the system up to the fault point;
R be the resistance per phase of the system up to the fault point, and
C be the capacitance of the circuit.
When the breaker contacts are opened, and the arc
certainly quenches at some current zero, a voltage v
is suddenly applied across the capacitor and
therefore across the circuit breaker contacts. The
current i which would flow to the fault is not
injected in the capacitor and inductor. Thus

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Expression for Restriking Voltage and RRRV Contd.
Assuming Zero time at zero currents when t = 0 and the value of current and voltage before opening
of circuit breaker is expressed as

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On substituting the above values in equation (1), we get
The solution of the standard equation is

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From the equation,
The above expression is for restriking voltage where Vmax is the peak value of recovery voltage
(phase -to-neutral) t is time is seconds. L is inductance in Henrys, C is the capacitance in farads and
v is the restriking voltage in volts. The maximum value of restriking voltage is 2Vmax and occurs at

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Expression for Restriking Voltage and RRRV:
Characteristic of Restriking Voltage
The important characteristic of restriking voltage which affects the performance of the circuit
breaker is as follows –
Amplitude Factor – It is defined as the ratio of the peak of transient voltage to the peak system
frequency voltage.
The rate of Rising of Restriking Voltage – It is defined as the slope of the steepness tangent of the
restriking voltage curve. It is expressed in kV/µs. RRRV is directly proportional to the natural
frequency. The expression for the restriking voltage is expressed as

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Current Chopping
What is Current Chopping?
• Current Chopping in circuit breaker is defined as a phenomena in which current is forcibly
interrupted before the natural current zero.
• Current Chopping is mainly observed in Vacuum Circuit Breaker and Air Blast Circuit Breaker.
• There is no such phenomena in Oil Circuit Breaker.
• Current chopping is predominant while switching Shunt Reactor or unloaded Transformer.

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Current Chopping contd.
Theory of Current Chopping
• Generally the arc extinction in a circuit breaker take place at natural current zero.
• But this is true if the capacity of the breaker to extinguish the arc is varies with the level of fault
current.
Consider a shunt reactor as
shown in figure.

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The stored energy in the inductance of reactor is basically transferred to the capacitor. Therefore
mathematically we can write as
LI2 / 2 = CV2 / 2
Here V = Voltage across the capacitor
Thus, V = I √(L/C)
This is the prospective voltage across the capacitor during current chopping.
Let us consider a simple example to have an idea of magnitude of prospective voltage. Let the value
of L = 64 mH and C = 0.001 uF then the induced voltage for a chopping current of 10 A will be
V = 10x√(64×10-3 / 0.001×10-6 )= 80 kV

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Carefully observe the figure. In the figure
you can see, 4 current chopping. In each
current chopping the magnitude of current
reduces. This is because of damping effect of
losses in the equipment like eddy current
loss and hysteresis loss.
Why there is no Current Chopping in
Oil Circuit Breaker?

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Resistance switching
A fixed connection of resistance in parallel with the
contact space or arc is called the resistance switching.
Severe voltage occurs in the system because of
two reasons, firstly because of the breaking of
low voltage current, and secondly because of the
breaking of capacitive current.
The severe voltage may endanger the operation
of the system.
It can be avoided by using resistance switching
(by connecting the resistor across the contacts of
the circuit breaker).

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The shunt resistor also helps in limiting the oscillatory growth of restriking voltage transients.
It can be proved mathematically that the natural frequency (fn) of oscillations of the circuit shown is
given as
To sum up, resistor across the circuit breaker contacts may be used to perform any one or more of
the following functions.
•It reduces the RRRV ( Rate of Rising of Restriking Voltage ) burden on the circuit breaker.
•It reduces the high-frequency restriking voltage transients during switching out inductive or
capacitive loads.
•In a multi-break circuit breaker, it helps in distributing the transient recovery voltage more
uniformly across the contact gaps.

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• To reduce the transient recovery voltage requires a considerably lower value of resistor whereas
for voltage equalisation a resistor of relatively high ohmic value will be required.
• In this case it is required that its resistance be low compared with the reactance of the
capacitance, shunting the breaks at the frequency of the recovery transient.
• It is often necessary to compromise and make one resistor do more than one of these jobs Critical
restriking voltage damping is obtained if

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• To reduce the transient recovery voltage requires a considerably lower value of resistor whereas
for voltage equalisation a resistor of relatively high ohmic value will be required.
• In this case it is required that its resistance be low compared with the reactance of the
capacitance, shunting the breaks at the frequency of the recovery transient.
• It is often necessary to compromise and make one resistor do more than one of these jobs Critical
restriking voltage damping is obtained if

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Expressions of Resistance switching
The analysis of resistance switching can be made to find out the critical value of the shunt resistance
to obtain complete damping of transient oscillations.
Fig. shows the equivalent
electrical circuit for such an
analysis.
As the period of transient oscillations is very small, the change in the power frequency term
during this short period is very little and hence negligible, because cos wt = 1 (nearly).

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Hence, the sinusoidally varying voltage Vm cos w t can be assumed to remain constant at Vm during
the transient periods, i.e., Vm cos w t = Vm.
Hence, the voltage equation is given by
Therefore, the above equation becomes
(1)

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(2)
(3)
Substituting these values in Eq. (1), we get
(4)
Taking Laplace Transform, of both sides of Eq. (4), we get
Other terms are zero, as vc = 0 at t = 0

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(5)
For no transient oscillation, all the roots of the equation should be real. One root is zero, i.e. S = 0
which is real. For the other two roots to be real, the roots of the quadratic equation in the
denominator should be real. For this, the following condition should be satisfied.
(6)

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Therefore, if the value of the resistance connected across the contacts of the circuit breaker is equal
to or less than ½ *root (L/C) there will be no transient oscillation.
If R > ½*root (L/C) , there will be oscillation. R = ½*root(L/C) is known as critical resistance.
The frequency of damped oscillation is given by

Numerical Examples
Circuit Breaker (PART-I)

Numerical Examples- 1
Q 1. In a system of 132 kV, the line to ground capacitance is 0.01 microF and the inductance is 5
henries. Determine the voltage appearing across the pole of a C.B. if a magnetising current of 5 amps
(instantaneous value) is interrupted. Determine also the value of resistance to be used across the
contacts to eliminate the restriking voltage.
Ans 1. This is a case of conversion of electromagnetic energy into electrostatic energy and hence the
voltage appearing across breaker contacts is nothing but the voltage across the capacitor which is
given by
In order to eliminate the transient critically the value of resistance across the breaker contacts
required is

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Q 2. In a short circuit test on a 132 kV 3-phase system, the breaker gave the following results: p.f. of
the fault 0.4, recovery voltage 0.95 of full line value; the breaking current is symmetrical and the
restriking transient had a natural frequency of 16 kHZ. Determine the rate of rise of restriking
voltage. Assume that the fault is grounded.
Ans 2. The peak value of line to neutral voltage
*****If the neutral line is present in a given wiring configuration, then we can measure three-phase
voltages as line-neutral voltages. The equations for these calculations are as follows:
•VLINE-NEUTRAL (VL-N) = VL-L/√3
•VPEAK = √2 * VL-N (in this case, 392 V)
•VPK-PK = 2 * VPEAK

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Numerical Examples- 2 contd.
Since the recovery voltage is 0.95 times the full line value, the recovery voltage = 107.75 × 0.95 = 102.4
kV.
Since the power factor of fault is 0.4, the value of the voltage when the current is zero will be Vm sin
θ, where θ = cos–1 0.4 = 66.42° or sin θ = 0.916.
The active recovery voltage = 102.4 × 0.916 = 93.85 kV
The maximum restriking voltage = 2 × 93.85 = 187.7 kV

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Q 3. In a short circuit test on a 3-pole, 132 kV C.B. the following observations are made: p.f. of fault
0.4, the recovery voltage 0.90 times full line value, the breaking current symmetrical, the frequency
of oscillations of restriking voltage 16 kHz. Assume that the neutral is grounded and the fault does
not involve ground, determine the average rate of rise of restriking voltage.
Ans 3. Peak value of L-G voltage
Instantaneous value of recovery voltage is
where
K = K1K2
and K1 = multiplying factor due to system voltage
K2 = 1.5 here as fault does not involve ground

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Vr = 0.90 × 1.5 × 107.77 × 0.92 = 133.85 kV
Now

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Q 4. For a 132 kV system, the reactance and capacitance up to the location of the circuit breaker is 3
ohms and 0.015 m F, respectively. Calculate the following:
(a) The frequency of transient oscillation
(b) The maximum value of restriking voltage across the contacts of the circuit
breaker
(c) The maximum value of RRRV
Ans 4. (a) The frequency of transient oscillation

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(b) The restriking voltage
The maximum value of the restriking voltage = 2Vm
(c) The maximum value of RRRV = wnVm

Department of Electrical and Electronics Engineering 42Department of Electrical and Electronics Engineering 42Department of Electrical and Electronics Engineering 42Department of Electrical and Electronics Engineering 42Department of Electrical and Electronics Engineering 42Department of Electrical and Electronics Engineering 42Department of Electrical and Electronics Engineering 42Department of Electrical and Electronics Engineering 42Department of Electrical and Electronics Engineering 42Department of Electrical and Electronics Engineering 42Department of Electrical and Electronics Engineering 42Department of Electrical and Electronics Engineering 42Department of Electrical and Electronics Engineering 42Department of Electrical and Electronics Engineering 42Department of Electrical and Electronics Engineering 42Department of Electrical and Electronics Engineering 42Department of Electrical and Electronics Engineering 42Department of Electrical and Electronics Engineering 42Department of Electrical and Electronics Engineering 42Department of Electrical and Electronics Engineering 42Department of Electrical and Electronics Engineering 42Department of Electrical and Electronics EngineeringDepartment of Electrical and Electronics Engineering 42Department of Electrical and Electronics Engineering 42Department of Electrical and Electronics Engineering 42Department of Electrical and Electronics Engineering 42Department of Electrical and Electronics Engineering 42Department of Electrical and Electronics Engineering 42Department of Electrical and Electronics EngineeringDepartment of Electrical and Electronics Engineering 42Department of Electrical and Electronics Engineering 42
Q 5. In a 220 kV system, the reactance and capacitance up to the location of circuit breaker is 8 W
and 0.025 m F, respectively. A resistance of 600 ohms is connected across the contacts of the circuit
breaker. Determine the following:
(a) Natural frequency of oscillation
(b) Damped frequency of oscillation
(c) Critical value of resistance which will give no transient oscillation
(d) The value of resistance which will give damped frequency of oscillation, one-fourth of the
natural frequency of oscillation
Ans 5.
(i) Natural frequency of oscillation

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(ii) Frequency of damped oscillation is given by
(iii) The value of critical resistance

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(iv) The damped frequency of oscillation is

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Q 6. A circuit breaker interrupts the magnetising current of a 100 MVA transformer at 220 kV. The
magnetising current of the transformer is 5% of the full load current. Determine the maximum
voltage which may appear across the gap of the breaker when the magnetising current is
interrupted at 53% of its peak value. The stray capacitance is 2500 microF. The inductance is 30 H.
Ans 6. The full load current of the transformer Let
M = Transformer rating in MVA
V = Transformer winding voltage in kV
I = Transformer corresponding winding current
in Amp
Then
I = (M x 1000) / {sqrt(3) × V}
If the transformer winding voltage is 220kV,
then the 220kV winding current, according to
the above formula, will be 262.44A.
100MVA transformer is a 3 phase transformer,
hence sqrt(3) appears in the denominator of the
formula. For a single phase transformer,
I = M × 1000 / V

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LENDI INSTITUTE ENGINEERING TECHNOLOGY CIRCUIT BREAKERS

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