Short Circuit Capacity MCQ Quiz - Objective Question with Answer for Short Circuit Capacity - Download Free PDF

Ring distribution system

• A ring distribution system is an electrical power distribution network where feeders connect in a closed loop, providing multiple paths for power delivery.
• This design ensures that if one feeder fails, power can continue to flow through an alternative path, maintaining reliability and preventing power outages.
• There are fewer voltage fluctuations at the consumer's terminal.

Circuit breaker

• A circuit breaker is an electrical safety device designed to protect an electrical circuit from damage caused by current more than that which the equipment can safely carry (overcurrent).
• Its basic function is to interrupt current flow to protect equipment and to prevent fire.
• In power systems, protection schemes follow the principle of selective coordination. The circuit breaker closest to the fault (downstream) must trip first to isolate the fault while keeping the rest of the system operational.
• This prevents unnecessary disconnection of large portions of the electrical network.

Statement 1 is True.

Relay

• Relays are electrically operated switches that open and close the circuits by receiving electrical signals from outside sources.
• Relays should operate as quickly as possible to minimize damage to equipment and ensure safety.

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Statement 2 is False.

Commutation

Commutation refers to the process of turning off a conducting thyristor. There are six major types of commutation methods, classified into two categories:

Category 1: Forced Commutation

• It is used in DC circuits where the current does not naturally reach zero. It requires external components (capacitors, inductors) to force the current to zero.
• Forced commutation is classified into five commutation techniques.
   

Class A (Resonant Commutation)

• Uses an LC circuit to create oscillations that naturally bring the current to zero, turning off the thyristor.
• Completely independent of external circuit components.

Class B (Resonant-Pulse Commutation)

• The thyristor is turned off using an LC resonant circuit discharging when the current reaches zero.

Both Class A and Class B rely on the circuit's natural behavior to turn off the thyristor, making them self-commutation methods.

Class C: Complementary Commutation

• Uses another thyristor (auxiliary SCR) and a capacitor to turn off the main thyristor.
• Used in DC choppers and inverters.

Class D: Auxiliary Commutation

• Similar to Class C but uses an external voltage source to force commutation. Used in high-power inverters and choppers.​

Class E: External Pulse Commutation

• Uses an external commutation pulse from a circuit to turn off the thyristor.
• Used in Cycloconverters and high-power circuits.

Category 2: Class F (Natural Commutation)

• It occurs in AC circuits where the current naturally becomes zero. In AC applications, the supply voltage reverses polarity every half-cycle, automatically bringing the current to zero.
• Example: Line commutation in rectifiers.

Overload in Power System

• An overload occurs when electrical equipment or a circuit carries more current than its rated capacity, but there is no direct fault or connection between phases or ground.
• This condition can cause the wiring to overheat, potentially leading to a fire. Common signs include tripped circuit breakers, flickering lights, and unusual buzzing or sizzling sounds from electrical outlets.
• In an overload condition, the electrical load exceeds the rated capacity, causing a voltage drop but not a complete loss of voltage. Unlike a short circuit, where voltage at the fault point is nearly zero, an overload only reduces the voltage.

Hence, Statement 1 is True.

Short Circuit in the Power System

• A short circuit is an unintentional, low-resistance connection between two points in a circuit, allowing excessive current to flow, bypassing the intended load path. This can be caused by damaged insulation, faulty wiring, or accidental contact, leading to potential hazards like overheating, fires, and equipment damage. 
• In an overload, the current is higher than normal but still within a limited range (e.g., 2-5 times the rated current).
• In a short circuit, the current can be hundreds or thousands of times the normal current, depending on system impedance.
• Therefore, overload current is substantially lower compared to short-circuit current.

Hence, Statement 2 is True.

Explanation: 

Connecting a lead from the negative to the positive of a battery creates a direct connection between the two terminals of the battery without any significant resistance or load in between.

This condition is known as a short circuit.

In a short circuit, the current flow is very high because the resistance of the connection is very low. This can lead to overheating and potential damage to the battery and the circuit.

Therefore, the correct answer is:

Option 2: a short circuit

Concept:

Short circuit MVA = \( \frac{{\;\ MVA}_{base}}{{{X_{pu}}}}\)

Where,

MVA_base = Full load or base MVA

X_pu = Per unit reactance

Calculation:

Given that, MVA_base = 500 kVA = 0.5 MVA, X_pu= 5% = 0.05

Since the transformer is connected to an infinite bus, the p.u. the reactance of the circuit will be 0.05 i.e., the p.u. reactance offered by the transformer.

∴ short circuit MVA can be calculated as

Short circuit MVA = \( \frac{{\ 0.5}}{{{0.05}}} =10 ~MVA\)

The correct answer is option 2): (200 A )

Concept:

The short circuit current is given by

I_sc = I × \(100 \over percentage\: of \: reactance\)

Where 

I is the full load current 

Calculation:

Isc = I × \(100 \over percentage\: of \: reactance\)

= 40 × \(100 \over 20\)

 = 200 A 

Concept:

Accelerating power can be calculated as

P_a = P_m – P_e

Accelerating torque is given as

\({T_a} = \frac{{{P_a}}}{{{\omega _s}\;}}\)

Where,

P_m = Mechanical input power

P_e = Electrical output power

ω_s = Synchronous speed in radian/sec

P = Number of poles

F = Supply frequency

Synchronous speed in rpm can be calculated as

\({N_s} = \frac{{120f}}{P}\)

Also, \({\omega _s} = \frac{{2\pi {N_s}}}{{60}}\) 

Calculation:

Given-

f = 50 Hz, P = 4

Before fault

P_m = P_e = 500 cosϕ = 500 x 0.8

P_m = P_e = 400 MW

After fault P_m remains same, while P_e reduces by 40%

So electrical output at the time of fault is

P_e’ = (1 – 0.4) = 0.6 x 400

P_e’= 240 MW

Now Accelerating power at the time of fault is

P_a’ = 400 – 240 = 160 MW

Now synchronous speed is

\({N_s} = \frac{{120 \times 50}}{4} = 1500\;rpm\)

\({\omega _s} = \frac{{2\pi \times 1500}}{{60}} = 157\;radian/sec\)

Hence accelerating torque is

\({T_a} = \frac{{160 \times {{10}^6}}}{{157}} = 1.018\; \times {10^6}\;N - m\)

T_a = 1.018 MN-m

Concept:

Short circuit MVA or Fault MVA = (Base MVA) x 100 / (% X)

Where,

X = Thevenin equivalent reactance between fault point and zero power bus (ZPB)

Calculation:

Base MVA = 10 MVA

According to the given data, the circuit diagram & per unit diagram can be drawn as given below.

Line diagram:

Per unit diagram:

Thevenin equivalent circuit:

\(X = \frac{{0.25}}{5} = 0.05\;{\rm{\Omega }}\)

% X = 5 %

Now short circuit MVA can be calculated as

Short circuit MVA or Fault MVA = (10)/0.05

Short circuit MVA or Fault MVA = 200 MVA 

The correct answer is option 2):(short circuit breaking current)

Concept:

• The short circuit breaking current  is defined as the maximum current that can flow through the breaker from the time occurring short circuit to the time of clearing the short circuit without any permanent damage to the Circuit breaker
• The rated short-circuit-breaking-current of a circuit-breaker is the highest value of short circuit current which a circuit-breaker is capable of breaking under specified conditions of transient recovery voltage and power frequency voltage. It is expressed in kA r.m.s. at contact separation.

Additional Information

• The rated short-circuit-making current is the maximum peak current that the switch shall be capable of making at its rated voltage.
• The rated normal current of a circuit breaker is the r.m.s. value of the current which the circuit breaker can carry continuously and with temperature rise of the various parts within specified limits

Reactance of each alternator = 0.25

As all the five alternators are connected in parallel, the equivalent reactance

\({X_{eq}} = \frac{{0.25}}{5} = 0.05\;pu\)

The correct answer is option 2):(Short circuit faults)

Concept:

• A short circuit can be defined as an abnormal connection of very low impedance between two points of different potential, whether made intentionally or accidentally.
• These are the most common and severe kinds of faults, resulting in the flow of abnormally high currents through the equipment or transmission lines
• These faults can be caused by a lightning strike, weather conditions or equipment failure. During the fault, the system can see a direct connection to the earth and current flows from all sources into it.

Additional Information

•  An unsymmetrical or asymmetrical fault is defined as a fault that affects one or two phases of a three-phase system in contrast with the previously studied balanced or symmetrical faults which equally affect each of the three phases
• A symmetrical fault is a fault where all phases are affected so that the system remains balanced
• An open-circuit fault occurs if a circuit is interrupted by a failure of a current-carrying wire (phase or neutral) or a blown fuse or circuit breaker.

The correct answer is option 2): (625)

Concept:

Short-circuit MVA  is a measure of the electrical strength of the bus. It determines the dimension of the bus bar and the interrupting capacity of a circuit breaker.  The short circuit MVA of each component is defined as the ratio of  MVA rating to its own per unit reactance.

It is also called a Fault MVA.

Short circuit = \(\frac{Base MVA }{Reactance}\)

Calculation: 

Short Circuit MVA = \(\frac {Number of alternators×MVA ~rating }{ Sub-transient reactance (in per unit)}\) 

Given

No of alternators = 5

Base MVA = 25 MVA

Rated Voltage = 11 KV

Sub transient reactance of each alternator = 20 %= 0.2  per unit

X = \(\frac{0.2}{5}\) = 0.04 per unit (as 5 alternator connected in parallel)

Short circuit = \(\frac{Base MVA }{Reactance}\)

= \(\frac{25}{0.04}\)= 625 MVA

Short Circuit Capacity:

• It is the measure of the electrical strength of the bus
• It is also called short-circuit MVA stated in MVA.
• It determines the dimension of the bus bar and the interrupting capacity of a circuit breaker.

Finding the fault level:

\({I_{SC}} = \frac{{{V_b}}}{Z} = \frac{{{V_b}/{Z_b}}}{{Z/{Z_b}}} = \frac{{{I_b}}}{{{Z_{p.u.}}}}\)

Where,

I_SC is the short circuit current

V_b is the base voltage

Z impedance of the alternator

Z_b is the base impedance of the alternator

Z_p.u. per unit impedance of the alternator

\(\sqrt 3 {V_b}{I_{SC}} = \frac{{\sqrt 3 {V_b}{I_b}}}{{{Z_{p.u.}}}}\)

\(MV{A_{SC}} = \frac{{MV{A_b}}}{{{Z_{p.u.}}}}\)

The percent value is the per-unit value multiplied by 100 (Z% = Z_p.u. × 100)

Calculation:

Given,

MVA_b = 50 MVA

As the alternators having only reactance (Z_p.u. = X_p.u.)

Z_p.u. = 15 / 100

= 0.15

As the alternators are connected in parallel

X_p.u. = 0.15 / 2

= 0.075

The fault level for a fault on one of the feeders near the bus bar

MVA_SC = 50 / 0.075

= 666.67 MVA

Therefore, the fault level for a fault on one of the feeders near the bus bar is 666.67 MVA

\(\begin{array}{l} {X_{{G_1}}} = 0.25 \times \frac{{100}}{{250}}{\left( {\frac{{15}}{{15}}} \right)^2} = 0.1\\ {X_{{G_2}}} = 0.1 \times \frac{{100}}{{100}}{\left( {\frac{{15}}{{15}}} \right)^2} = 0.1\\ {X_{{L_1}}} = 0.225 \times 10 \times \frac{{100}}{{{{\left( {15} \right)}^2}}} = j1.0\\ {X_{{L_2}}} = 0.225 \times 10 \times \frac{{100}}{{{{\left( {15} \right)}^2}}} = j1.0 \end{array}\)

Reactance diagram:

We can see that at bus 3, equivalent Thevenin's impedance is given by 

X_th = (j 0.1 + j 1.0)||(j 0.1 + j 1.0) = j .55 pu

Fault MVA = (Base MVA / X_th) = 100 / .55 = 181.82 MVA