Skip to main content

Interview Question On Circuit Breaker And Relay

Q. Explain different types of faults?

A. Shunt Faults: Basically shunt faults are short circuits. When the path of the load current is cut short because of breakdown of insulation, we say that a 'short circuit' has occurred which is nothing but a shunt fault.
Series Faults: Series faults are nothing but a break in the path of current. It is observed in practice that most of the open conductor faults sooner or later develop into some or the other short-circuit fault.

Q. What are the various states of operation of a power system?

A. Normal, Alert, Emergency, In extremis and Restorative.

Q. Explain the working of a Current transformer?

A. The current transformer has two jobs to do. Firstly, it steps down the current to such a level that it can be easily handled by the relay current coil. Secondly, it isolates the relay circuitry from the high voltage of the EHV system. A CT whose primary is in series with the line in which current is to be measured.

Q. Why the secondary of a CT should not be open?

A. If the secondary of a CT is open current through it is zero and hence the ampere turns produced by secondary which generally oppose the primary ampere turns becomes zero. As there is no counter m.m.f, primary m.m.f produces high flux in the core which leads to excessive core losses and overheating.

Q. Explain the working of a Potential transformer?

A. The voltage transformer or potential transformer steps down the high voltage of the line to a level safe enough for the relaying system (pressure coil of relay) and personnel to handle. A PT primary is connected in parallel at the point where a measurement is desired.

Q. Explain the working principal of the circuit breaker?

A. The circuit breaker is an electrically operated switch, which is capable of safely making, as well as breaking, short-circuit currents. It has two contacts namely fixed contact & moving contact. Under normal conditions the moving contact comes in contact with fixed contact thereby forming the closed contact for the flow of current. During abnormal & faulty conditions (when current exceeds the rated value) an arc is produced between the fixed & moving contacts & thereby it forms the open circuit arc is extinguished by the Arc Quenching media like air, oil, vacuum etc.

Q. What are the different types of circuit breakers?

A.     1. Air Break Circuit Breaker
    2. Oil Circuit Breaker
    3. Minimum Oil Circuit Breaker
    4. Air Blast Circuit Breaker
    5. Vacuum Circuit Breaker
    6. SF6 Circuit Breaker

Q. What is SF6 Circuit Breaker? Its advantages?

A. If SF6 (Sulpher Hexa Fluoride) gas is used as arc quenching medium in a Circuit breaker, it is termed as SF6 CB. SF6 gas has superior arc quenching capacity and high dielectric strength. So, they have very short arcing time and can interrupt high currents. 

Q. What is meant by reach point of the relay?

A. The farthest point from the relay location, which is still inside the zone of protection, is called the reach point.

Q. What is difference between fuse and breaker?

A. Fuses are burnt when over current flows in the circuit but circuit breakers will just open during over current. Thus fuses are used for only once but breakers can used by multiple number of times.

Q. What is the difference between isolators and electrical circuit breakers?

A. Isolators are mainly for switching purpose under normal conditions but they cannot operate in fault conditions. Actually they used for isolating the CBs for maintenance. Whereas CB gets activated under fault conditions according to the fault detected.

Q. Explain the working of an Over current relay and mention the types of it?

A. An over-current (OC) relay has a single input in the form of ac current. The output of the relay is a normally-open contact, which changes over to closed state when the relay trips. The relay has two settings. These are the time setting and the plug setting. The time setting decides the operating time of the relay while the plug setting decides the current required for the relay to pick up.
Types:
1. Instantaneous Over-current Relay
2. Definite Time Over-current Relay
3. Inverse Time Over-current Relay

Q. Explain the working of an IDMT relay?

A. In an IDMT (inverse definite minimum time) relay, the operating principle is based on the idea that with the more severe a fault is, the faster it should be cleared to avoid damage to the apparatus. It is inverse in the sense, the tripping time will decrease as the magnitude of fault current increase.

Q. How to select the pickup value of a relay?

A. We can set the pick-up value of the relay, keeping in mind that the relay should allow normal load as well as a certain degree of overload to be supplied. Thus the pick-up value of the relay should be more than the allowable maximum load. At the same time, the relay should be sensitive enough to respond to the smallest fault. Thus, the pick-up value should be less than the smallest fault current.

Q. How to set the operating time of a relay?

A. A relay must get an adequate chance to protect the zone under its primary protection. Only if the primary protection does not clear the fault, the back-up protection should initiate tripping. Thus as soon as the fault takes place, it is sensed by both the primary and the back-up protection.
Naturally the primary protection is the first to operate, it’s operating time being less than that of the back-up relay. We have to allow for the overshoot of the primary relay, so that there is proper coordination between the primary and the back-up.

Q. When do we use IDMT relays and DTOC relays?

A. IDMT relays have better fault clearing times over DTOC relays. But, DTOC relays are used for lines which are short in length as Zs>>Zl and If remains almost same.

Comments

Popular posts from this blog

Transformer multiple choice questions part 1

Hello Engineer's Q.[1] A transformer transforms (a) frequency (b) voltage (c) current (d) voltage and current Ans : D Q.[2] Which of the following is not a basic element of a transformer ? (a) core (b) primary winding (c) secondary winding (d) mutual flux. Ans : D Q.[3] In an ideal transformer, (a) windings have no resistance (b) core has no losses (c) core has infinite permeability (d) all of the above. Ans : D Q.[4] The main purpose of using core in a transformer is to (a) decrease iron losses (b) prevent eddy current loss (c) eliminate magnetic hysteresis (d) decrease reluctance of the common magnetic circuit. Ans :D Q.[5] Transformer cores are laminated in order to (a) simplify its construction (b) minimize eddy current loss (c) reduce cost (d) reduce hysteresis loss. Ans : B Q.[6] A transformer having 1000 primary turns is connected to a 250-V a.c. supply. For a secondary voltage of 400 V, the number of secondary turns should be (a) 1600 (b) 250 (c) 400 (d) 1250 A

Condition for Maximum Power Developed In Synchronous Motor

The value of δ for which the mechanical power developed is maximum can be obtained as, Note : Thus when R a is negligible, θ = 90 o for maximum power developed. The corresponding torque is called pull out torque. 1.1 The Value of Maximum Power Developed        The value of maximum power developed can be obtained by substituting θ = δ in the equation of P m .        When R a is negligible,     θ = 90 o  and cos (θ) = 0 hence, . . .               R a = Z s cosθ   and X s = Z s sinθ        Substituting   cosθ = R a /Z s in equation (6b) we get,         Solving the above quadratic in E b we get,        As E b is completely dependent on excitation, the equation (8) gives the excitation limits for any load for a synchronous motor. If the excitation exceeds this limit, the motor falls out of step. 1.2 Condition for Excitation When Motor Develops ( P m ) R max        Let us find excitation condition for maximum power developed. The excitation

Effect of Slip on Rotor Parameters : Part2

Effect of Slip on Rotor Parameters 2. Effect of Slip on Magnitude of Rotor Induced E.M.F        We have seen that when rotor is standstill, s  = 1, relative speed is maximum and maximum e.m.f. gets induced in the rotor. Let this e.m.f. be,                 E 2 = Rotor induced e.m.f. per phase on standstill condition         As rotor gains speed, the relative speed between rotor and rotating magnetic field decreases and hence induced e.m.f. in rotor also decreases as it is proportional to the relative speed N s - N. Let this e.m.f. be,                E 2r = Rotor induced e.m.f. per phase in running condition  Now        E 2r α N s while E 2r α N s - N        Dividing the two proportionality equations,               E 2r /E 2 = ( N s - N)/N s    but (N s - N)/N = slip s               E 2r /E 2 = s               E 2r = s E 2        The magnitude of the induced e.m.f in the rotor also reduces by slip times the