Question:
What is the VA output required for a CT of \(5\text{ A}\) rated secondary current when burden consists of a relay requiring \(7.5\text{ VA}\) at \(5\text{ A}\) and connecting lead resistance of \(0.08\text{ }\Omega\)?
What is the VA output required for a CT of \(5\text{ A}\) rated secondary current when burden consists of a relay requiring \(7.5\text{ VA}\) at \(5\text{ A}\) and connecting lead resistance of \(0.08\text{ }\Omega\)?
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Always ensure that the lead resistance given is the total loop resistance. If the problem specifies single-way lead resistance, you must double it ($2R$) to account for the complete return path of the current loop.
Updated On: Jun 25, 2026
- \(7.5\text{ VA}\)
- \(15.0\text{ VA}\)
- \(9.5\text{ VA}\)
- \(1.3\text{ VA}\)
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The Correct Option is C
Solution and Explanation
Concept:
The total burden or Volt-Ampere (VA) output capacity required for a Current Transformer (CT) must accommodate both the power consumption of the connected protective equipment (such as a relay) and the power losses incurred within the connecting secondary loop leads. The total burden is mathematically represented as the sum of the individual component burdens:
\[
\text{Total VA Required} = \text{VA}_{\text{relay}} + \text{VA}_{\text{leads}}
\]
The power loss across the leads can be calculated using the classical Joule heating expression $P = I^2 R$, where $I$ represents the rated secondary current, and $R$ represents the total loop resistance of the connecting leads.
Step 1: Identify the given data from the problem statement.
* Rated secondary current of the CT (\(I\)) = \(5\text{ A}\) * Volt-Ampere burden requirement of the relay (\(\text{VA}_{\text{relay}}\)) = \(7.5\text{ VA}\) * Resistance of the connecting leads (\(R\)) = \(0.08\text{ }\Omega\)
Step 2: Calculate the power loss (VA burden) contributed by the connecting leads.
The power dissipated in the wire leads due to their internal electrical resistance when carrying the rated current is: \[ \text{VA}_{\text{leads}} = I^2 \times R \] Substituting the provided physical parameters: \[ \text{VA}_{\text{leads}} = (5\text{ A})^2 \times 0.08\text{ }\Omega \] \[ \text{VA}_{\text{leads}} = 25 \times 0.08 = 2.0\text{ VA} \]
Step 3: Evaluate the cumulative VA output demand for the Current Transformer.
Summing the individual burdens together: \[ \text{Total VA Required} = \text{VA}_{\text{relay}} + \text{VA}_{\text{leads}} \] \[ \text{Total VA Required} = 7.5\text{ VA} + 2.0\text{ VA} = 9.5\text{ VA} \] Hence, the total mandatory VA capacity of the CT is exactly \(9.5\text{ VA}\), which matches option (C).
Step 1: Identify the given data from the problem statement.
* Rated secondary current of the CT (\(I\)) = \(5\text{ A}\) * Volt-Ampere burden requirement of the relay (\(\text{VA}_{\text{relay}}\)) = \(7.5\text{ VA}\) * Resistance of the connecting leads (\(R\)) = \(0.08\text{ }\Omega\)
Step 2: Calculate the power loss (VA burden) contributed by the connecting leads.
The power dissipated in the wire leads due to their internal electrical resistance when carrying the rated current is: \[ \text{VA}_{\text{leads}} = I^2 \times R \] Substituting the provided physical parameters: \[ \text{VA}_{\text{leads}} = (5\text{ A})^2 \times 0.08\text{ }\Omega \] \[ \text{VA}_{\text{leads}} = 25 \times 0.08 = 2.0\text{ VA} \]
Step 3: Evaluate the cumulative VA output demand for the Current Transformer.
Summing the individual burdens together: \[ \text{Total VA Required} = \text{VA}_{\text{relay}} + \text{VA}_{\text{leads}} \] \[ \text{Total VA Required} = 7.5\text{ VA} + 2.0\text{ VA} = 9.5\text{ VA} \] Hence, the total mandatory VA capacity of the CT is exactly \(9.5\text{ VA}\), which matches option (C).
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