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

Last updated on Apr 22, 2025

Latest Current Setting MCQ Objective Questions

Current Setting Question 1:

When the plug setting multiplier is 50, the fault current is 2500 A and CT ratio is 250/5, then the current setting will be

  1. 50%
  2. 100%
  3. 150%
  4. 200%

Answer (Detailed Solution Below)

Option 2 : 100%

Current Setting Question 1 Detailed Solution

Concept:

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

Plug setting multiplier =faultcurrentpick up current=faultcurrentcurrentsetting×CT secondary current

Current setting = Relay setting × Secondary CT current

Relay current (Irelay) = Fault current / CT ratio

Calculation:

Given that,

Plug setting multiplier = 50,

Fault current = 2500 A,

CT ratio = 250/5 = 50

∴ Current setting can be calculated as 

50=2500currentsetting×50

Current setting = 50×502500

Current setting = 1

∴ Current setting = 100%

Current Setting Question 2:

A 3 phase, 15 MVA, 13 kV, star connected alternator is protected by current balancing system of protection. If the ratio of CT is 1500/4, the minimum operating current of the relay is 0.8 A and the neutral point resistance is 4 Ω. The percentage of each phaser of stator winding which is unprotected against faults when the machine is operating at normal voltage is _______ (in %)

Answer (Detailed Solution Below) 15.5 - 16.5

Current Setting Question 2 Detailed Solution

Let %x be the stator winding which is unprotected.

Emf induced in this unprotected winding =13×1033×x100=1303xV

Fault current, If=1303x4=18.76xA

The minimum current which will operate the relay

=15004×0.8=300 

Now, 18.76x = 300

⇒ x = 15.98%

Current Setting Question 3:

A 3 – Phase transformer rated for 33 kV/ 11kV is connected in delta/star as shown in figure. The current transformers (CTs) on low land high voltage sides have a ratio of 500/5. Find the current i1 and i2, of the fault current is 300 A as shown in figure

Gate EE 2015 paper 2 Images-Q43

  1. i1=13A,i2=0A
  2. i1=0Ai2=0A
  3. i1=0A,i2=13A
  4. i1=13A,i2=13A

Answer (Detailed Solution Below)

Option 1 : i1=13A,i2=0A

Current Setting Question 3 Detailed Solution

i2=0

Since entire current flows through fault,    Primary kVA = Secondary kVA

3×33000×IL=3×11000×(300×5500),     IL=1A

IL=3IPhIph=i1=13A

Top Current Setting MCQ Objective Questions

Determine operating current of the relay, for 5 A, 2.2-sec IDMT rating, and 125% relay setting?

  1. 11 A
  2. 4 A
  3. 12.5 A
  4. 6.25 A

Answer (Detailed Solution Below)

Option 4 : 6.25 A

Current Setting Question 4 Detailed Solution

Download Solution PDF

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

A 3 – Phase transformer rated for 33 kV/ 11kV is connected in delta/star as shown in figure. The current transformers (CTs) on low land high voltage sides have a ratio of 500/5. Find the current i1 and i2, of the fault current is 300 A as shown in figure

Gate EE 2015 paper 2 Images-Q43

  1. i1=13A,i2=0A
  2. i1=0Ai2=0A
  3. i1=0A,i2=13A
  4. i1=13A,i2=13A

Answer (Detailed Solution Below)

Option 1 : i1=13A,i2=0A

Current Setting Question 5 Detailed Solution

Download Solution PDF

i2=0

Since entire current flows through fault,    Primary kVA = Secondary kVA

3×33000×IL=3×11000×(300×5500),     IL=1A

IL=3IPhIph=i1=13A

Current Setting Question 6:

When the plug setting multiplier is 50, the fault current is 2500 A and CT ratio is 250/5, then the current setting will be

  1. 50%
  2. 100%
  3. 150%
  4. 200%

Answer (Detailed Solution Below)

Option 2 : 100%

Current Setting Question 6 Detailed Solution

Concept:

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

Plug setting multiplier =faultcurrentpick up current=faultcurrentcurrentsetting×CT secondary current

Current setting = Relay setting × Secondary CT current

Relay current (Irelay) = Fault current / CT ratio

Calculation:

Given that,

Plug setting multiplier = 50,

Fault current = 2500 A,

CT ratio = 250/5 = 50

∴ Current setting can be calculated as 

50=2500currentsetting×50

Current setting = 50×502500

Current setting = 1

∴ Current setting = 100%

Current Setting Question 7:

Determine operating current of the relay, for 5 A, 2.2-sec IDMT rating, and 125% relay setting?

  1. 11 A
  2. 4 A
  3. 12.5 A
  4. 6.25 A

Answer (Detailed Solution Below)

Option 4 : 6.25 A

Current Setting Question 7 Detailed Solution

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

Current Setting Question 8:

A 3 phase, 15 MVA, 13 kV, star connected alternator is protected by current balancing system of protection. If the ratio of CT is 1500/4, the minimum operating current of the relay is 0.8 A and the neutral point resistance is 4 Ω. The percentage of each phaser of stator winding which is unprotected against faults when the machine is operating at normal voltage is _______ (in %)

Answer (Detailed Solution Below) 15.5 - 16.5

Current Setting Question 8 Detailed Solution

Let %x be the stator winding which is unprotected.

Emf induced in this unprotected winding =13×1033×x100=1303xV

Fault current, If=1303x4=18.76xA

The minimum current which will operate the relay

=15004×0.8=300 

Now, 18.76x = 300

⇒ x = 15.98%

Current Setting Question 9:

A 3 – Phase transformer rated for 33 kV/ 11kV is connected in delta/star as shown in figure. The current transformers (CTs) on low land high voltage sides have a ratio of 500/5. Find the current i1 and i2, of the fault current is 300 A as shown in figure

Gate EE 2015 paper 2 Images-Q43

  1. i1=13A,i2=0A
  2. i1=0Ai2=0A
  3. i1=0A,i2=13A
  4. i1=13A,i2=13A

Answer (Detailed Solution Below)

Option 1 : i1=13A,i2=0A

Current Setting Question 9 Detailed Solution

i2=0

Since entire current flows through fault,    Primary kVA = Secondary kVA

3×33000×IL=3×11000×(300×5500),     IL=1A

IL=3IPhIph=i1=13A

Current Setting Question 10:

An over – current relay having a current setting of 10% is connected to a supply circuit through a current transformer of ratio 400/5. The pickup value is – (in A)

Answer (Detailed Solution Below) 0.5

Current Setting Question 10 Detailed Solution

Pick up value = operating current × current setting

= 5 × 0.10 = 0.5 A

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