Over Current Relays MCQ Quiz - Objective Question with Answer for Over Current Relays - Download Free PDF

Induction Type Directional Overcurrent Relay

Definition: An induction type directional overcurrent relay is a protective device used in electrical power systems to detect overcurrent conditions while ensuring the current flows in a specific direction. This relay combines the principles of overcurrent protection and direction discrimination, making it suitable for applications where fault directionality is critical, such as in ring main systems or parallel feeder systems.

Net Torque at Balance Point:

When an induction type directional overcurrent relay is on the verge of operating, the net torque acting on the relay must be analyzed. In this condition, the relay is balanced, and the mechanical torque generated within the relay coil system reaches equilibrium. This balance point is crucial for determining the relay’s operating threshold.

Correct Option: The net torque at the balance point is equal to zero.

Explanation:

The torque in an induction type directional overcurrent relay is developed by two fluxes: the flux produced by the current coil and the flux produced by the voltage coil. These fluxes interact with each other, inducing currents in the relay’s disc or rotor, which ultimately produce a resultant torque. This torque is responsible for moving the relay’s disc or rotor to operate the relay contacts.

At the balance point, the relay is on the verge of operation, which means that the torque produced by the interaction of the fluxes is in equilibrium. In this state:

• The driving torque, which tends to move the disc or rotor, is exactly balanced by the restraining torque, which opposes the motion.
• As a result, the net torque acting on the relay is zero.

This balance ensures that the relay does not operate unnecessarily during normal conditions or minor disturbances but operates reliably when the fault current exceeds the threshold and flows in the specified direction.

Correct Option Analysis:

The correct option is:

Option 4: Equal to zero.

This is because, at the balance point, the driving torque and restraining torque cancel each other out, resulting in no net torque on the relay’s disc or rotor. This equilibrium condition ensures that the relay is stable and does not operate until the system conditions necessitate it.

Additional Information

To further understand the analysis, let’s evaluate the other options:

Option 1: Equal to K2.

This option is incorrect because the net torque at the balance point is not a constant value such as K2. The balance point is defined by the condition of zero net torque, where the driving and restraining torques are equal and opposite.

Option 2: Equal to VI cos (θ - α).

This option represents the torque equation in terms of the product of voltage (V) and current (I) with a phase angle difference (θ - α). While this expression is related to the torque produced in the relay, it does not define the net torque at the balance point, which is zero.

Option 3: Equal to the pick-up torque of the relay.

The pick-up torque is the minimum torque required to initiate the motion of the relay’s disc or rotor. However, at the balance point, the net torque is zero, as the driving and restraining torques are in equilibrium. Therefore, this option is incorrect.

Conclusion:

The net torque at the balance point of an induction type directional overcurrent relay is equal to zero. This equilibrium condition ensures the relay remains stable under normal operating conditions and operates reliably when the fault current exceeds the threshold and flows in the specified direction. Understanding the torque balance in such relays is essential for designing and applying protective systems in electrical power networks.

Explanation:

The Time Setting Multiplier (TSM) of a Relay

The Time Setting Multiplier (TSM) is a critical parameter in the operation of protective relays, especially in the context of overcurrent and differential protection schemes. The TSM adjusts the time delay characteristic of the relay, which influences how quickly the relay responds to fault conditions. This adjustment is crucial for coordinating the protection devices within an electrical system to ensure selective tripping and minimize disruption to the power supply.

Analysis of Other Options:

Option 1: Adjust the Operating Current Threshold of the Relay

This option is incorrect because the operating current threshold, also known as the pickup current, is typically adjusted using the current setting parameter of the relay. The TSM is concerned with the time delay characteristics rather than the current threshold.

Option 2: Increase the Current Setting of the Relay

This option is incorrect as well. Increasing the current setting of the relay would mean changing the current level at which the relay starts to operate (pickup level), which is not the function of the TSM. The TSM specifically adjusts the time delay after the relay has picked up a fault current.

Option 4: Decrease the Time Delay of the Relay

While this option is partially correct, it is incomplete. The TSM can indeed decrease the time delay of the relay, but its primary function is to modify the time delay based on fault severity, meaning it can both increase and decrease the time delay as required for proper coordination. Therefore, option 3 is more comprehensive and accurate.

Explanation:

Overcurrent Relay

Definition: An overcurrent relay is a type of protective relay that operates when the current exceeds a predetermined value. It is widely used in power systems to protect electrical equipment and lines from damage caused by excessive current due to overloads or short circuits.

Working Principle: The overcurrent relay works based on the principle that the relay will trip the circuit breaker when the current flowing through it exceeds a set threshold. The time of operation of the relay is inversely proportional to the magnitude of the current. This means that the higher the current, the faster the relay will operate to disconnect the faulty section of the power system.

Correct Option Analysis:

The correct option is:

Option 2: The square of the current

This option correctly states that the time of operation of an overcurrent relay is inversely proportional to the square of the current. This relationship can be described by the formula:

T ∝ 1/I²

Where:

• T is the operating time of the relay
• I is the current flowing through the relay

This inverse square relationship means that as the current increases, the time required for the relay to operate decreases rapidly. This characteristic is crucial for the protection of electrical systems, as it ensures that in the event of a high current fault, the relay will operate almost instantaneously to prevent damage.

Applications: Overcurrent relays are used in various applications within power systems, including:

• Protection of transformers, generators, and motors from overcurrent conditions.
• Protection of transmission and distribution lines from faults and overloads.
• Ensuring the selective tripping of circuit breakers to isolate only the faulty section of the system.

Additional Information

To further understand the analysis, let’s evaluate the other options:

Option 1: The cube of the current

This option is incorrect because the time of operation of an overcurrent relay is not inversely proportional to the cube of the current. The correct relationship is the inverse square of the current, as explained earlier.

Option 3: The frequency

This option is also incorrect. The time of operation of an overcurrent relay is not dependent on the frequency of the electrical system. Overcurrent relays are designed to respond to the magnitude of the current, not its frequency.

Option 4: The voltage

This option is incorrect as well. The time of operation of an overcurrent relay is not inversely proportional to the voltage. Overcurrent relays are specifically designed to respond to current levels, irrespective of the voltage.

Conclusion:

Understanding the relationship between the operating time of an overcurrent relay and the current is crucial for effective protection of electrical systems. The correct relationship is that the time of operation is inversely proportional to the square of the current. This ensures rapid response to high current faults, thereby protecting electrical equipment and maintaining system stability. Evaluating other options highlights the importance of distinguishing between current-dependent and other potential factors, such as frequency and voltage, which do not influence the operating time of overcurrent relays in the same manner.

Concept:

Plug setting multiplier is the ratio of fault current and the product of plug setting and CT ratio.

Plug setting multiplier \(= \frac{{fault\;current}}{{plug\;setting \times CT\;ratio}}\)

Calculation:

Plug setting = relay setting × secondary CT current

= 1.5 × 5 = 7.5

CT ratio \(= \frac{{400}}{5} = 80\)

Concept:

The pick-up current of a relay is given by:

Pick-up current = Rated secondary current of CT × Current setting

Calculation:

Given, the CT ratio = 400/5

Secondary current = 5 A

Relay setting = 12.5%

Pick-up current = 5 × 0.125

Pick-up current = 0.625 A

Concept:

The pick-up current of a relay is given by:

Pick-up current = Rated secondary current of CT × Current setting

Calculation:

Given, the CT ratio = 400/5

Secondary current = 5 A

Relay setting = 12.5%

Pick-up current = 5 × 0.125

Pick-up current = 0.625 A

Concept:

Plug setting multiplier:

Plug setting multiplier is the ratio of fault current and the product of plug setting and CT ratio.

Plug setting multiplier \(= \frac{{fault\;current}}{{plug\;setting \times CT\;ratio}}\)

Calculation:

Plug setting = relay setting × secondary CT current

= 0.5 × 5 = 2.5

CT ratio \(= \frac{{400}}{5} = 80\)

Concept:

Plug setting multiplier is the ratio of fault current and the product of plug setting and CT ratio.

Plug setting multiplier \(= \frac{{fault\;current}}{{plug\;setting \times CT\;ratio}}\)

Calculation:

Plug setting = relay setting × secondary CT current

= 1.5 × 5 = 7.5

CT ratio \(= \frac{{400}}{5} = 80\)

Concept:

Plug setting multiplier is the ratio of fault current and the product of plug setting and CT ratio.

Plug setting multiplier \(= \frac{{fault\;current}}{{plug\;setting\; \times\; CT\;ratio}}\)

Calculation:

Plug setting = relay setting × secondary CT current

= 0.5 × 5 = 2.5

CT ratio \(= \frac{{200}}{5} = 40\)

Concept:

Plug setting multiplier is the ratio of fault current and the product of plug setting and CT ratio.

Plug setting multiplier \(= \frac{{fault\;current}}{{plug\;setting \times CT\;ratio}}\)

Calculation:

Plug setting = relay setting × secondary CT current

= 1.5 × 5 = 7.5

CT ratio \(= \frac{{400}}{5} = 80\)

Concept:

Plug setting multiplier is the ratio of fault current and the product of plug setting and CT ratio.

Plug setting multiplier \(= \frac{{fault\;current}}{{plug\;setting \times CT\;ratio}}\)

Calculation:

Given: Fault current is 4000 A, CT ratio is  400/5 A, Relay setting is 50%.

Plug setting = relay setting × secondary CT current

= 0.5 × 5 = 2.5

CT ratio \(= \frac{{400}}{5} = 80\)

Concept:

Plug Setting Multiplier (PSM): It is the ratio between the actual fault current in the relay operating coil to pick up current or the relay current setting.

\(PSM=\frac{Fault~Current}{Current~Setting× CT~Ratio}\)

• Plug setting multiplier (PSM) Indicates the severity of the fault.
• The plug setting multiplier is used only in Electromagnetic relays, not in numerical relays.
• The operating Times of the Relay operation are depending upon the PSM. The high value of PSM indicates low operating time.

Calculation:

Given that, PSM = 8

Relay setting = 125 % = 1.25

CT ratio = 400/1

\(PSM=\frac{I_f}{1.25× 1× \left( \frac{400}{1} \right)}=8\)

\(\frac{I_f}{1.25× 1× \left( \frac{400}{1} \right)}=8\)

∴ Fault current (I_f) = 1.25 × 400 × 8

Fault current (If) = 4000 A

Concept:

The plug setting multiplier of a relay is defined as the ratio of secondary fault current to the pick-up current.

PSM = Secondary fault current / Relay current setting

Pick up current or operating current = (Rated secondary current in CT) x (Current setting)

Calculation:

Given that,

Current setting = 125% = 1.25, Rated secondary current = 5 A

Therefore, operating current of relay = 5 x 1.25 = 6.25 A

Concept:

The plug setting multiplier of a relay is defined as the ratio of the secondary fault current to the pickup current.

\(PSM = \frac{Fault\hspace{0.1cm}current}{Relay\hspace{0.1cm}setting×C.T. Ratio×C.T.Secondary\hspace{0.1cm}current}\)

Significance of PSM:

• In the electromagnetic relay, the current setting can be done by adding a resistance value. This action is performed by inserting plugs. The resistances are connected in series with the relay operating coil.

• The plug is a small short link that connects the relay operating coils and resistance. The high number of plug positions increases the resistance value, therefore to operate the relay coils they need a high value of current to energize.

• At the same time, the lower number of plug positions would be of less resistance, hence to operate relay coil, which needs low current value.

• Therefore, The plug position ensures the current setting value of the relay. Plug setting multiplier (PSM) Indicates the severity of the fault.

Calculation:                                   

\(PSM = \frac{1500}{1.5×\frac{150}{5}×5} = 6.67\)

Therefore; the Plug setting Multiplier(PSM) of the relay is 6.67

Concept:

The plug setting multiplier of a relay is defined as the ratio of the secondary fault current to the pick-up current.

PSM = Secondary fault current/Relay current setting

Pick up current = (Rated secondary current in CT) x (Current setting)

Calculation:

Given that,

Current setting = 150% = 1.5

Relay CT ratio = 200/5

Rated secondary current = 5 A

Therefore, pickup value current of relay = 5 x 1.5 = 7.5 A

For 7.5 A, the current in the primary of the CT is \( = \frac{{200}}{5} \times 7.5 = 300\;A\)