Question

An overcurrent relay operates at 1.2pu for 2 s. Fault current = 3pu. What is the expected operation time if relay is inverse-time type?

• A. $<$2s
• B. 2s
• C. $>$2s
• D. Depends on system voltage

Hint:

"Inverse-time" means "More Current = Less Time".
This is a core design feature of protective relays to clear severe, high-current faults as quickly as possible to protect power system components from thermal damage.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
The question asks to compare the operating time of an inverse-time overcurrent relay under two different current conditions.
An inverse-time relay has a characteristic where the operating time is inversely proportional to the magnitude of the operating current.

Step 2: Key Formula or Approach:
The fundamental characteristic of an inverse-time relay is:
\[ T \propto \frac{1}{I^n - 1} \quad \text{or} \quad T \propto \frac{1}{\text{PSM}^n - 1} \] where:
\(T\) is the operating time,
\(I\) (or \(\text{PSM}\)) is the current in per-unit (or Plug Setting Multiplier), and
\(n\) is a positive constant depending on the relay type (e.g., \(n=0.02\) for standard inverse, \(n=1\) or \(n=2\) for very/extremely inverse).
Crucially, as the current increases, the operating time decreases.

Step 3: Detailed Explanation:

• The relay operates at a current of \(I_1 = 1.2\text{ pu}\) with an operating time of \(T_1 = 2\text{ s}\).

• The new fault current is \(I_2 = 3\text{ pu}\).

• Since \(I_2 = 3\text{ pu} > I_1 = 1.2\text{ pu}\), the magnitude of the fault current has increased significantly.

• By the definition of an "inverse-time" characteristic, a higher current will result in a faster relay operation.

• Therefore, the operating time \(T_2\) at \(3\text{ pu}\) must be less than the operating time \(T_1 = 2\text{ s}\) at \(1.2\text{ pu}\).

• Thus, \(T_2 < 2\text{ s}\).

Step 4: Final Answer:
The expected operation time is $<$2s.