A 6600 v 50 hz transformer operates at flux density of 1.5 T. Each lin...
Problem Statement:
A 6600 V 50 Hz transformer operates at a flux density of 1.5 T. Each linear dimension of the core is doubled while the primary and secondary turns are halved. If the transformer now operates at 13200 V 50 Hz, what will be the core flux density?
Solution:
Given:
- Voltage (V1) = 6600 V
- Frequency (f1) = 50 Hz
- Flux density (B1) = 1.5 T
- Primary and Secondary turns (N1 = N2) are halved
- Linear dimensions of the core are doubled
- Voltage (V2) = 13200 V
- Frequency (f2) = 50 Hz
Step 1: Calculate the voltage ratio
The voltage ratio is given by:
```
V2 / V1 = N2 / N1
```
Substituting the given values, we get:
```
13200 / 6600 = N2 / (N1 / 2)
N2 = N1
```
Therefore, the number of turns in the primary and secondary remain the same.
Step 2: Calculate the flux density
The flux density is given by:
```
B2 / B1 = (V2 / V1) * (f1 / f2) * (N1 / N2) * (l2 / l1)
```
where l is the mean length of the magnetic path.
Since the linear dimensions of the core are doubled, the mean length of the magnetic path is also doubled. Therefore,
```
l2 / l1 = 2
```
Substituting the given values, we get:
```
B2 / 1.5 = (13200 / 6600) * (50 / 50) * (1 / 1) * (2 / 1)
B2 = 3.0 T
```
Step 3: Conclusion
Therefore, the core flux density of the transformer operating at 13200 V 50 Hz is 3.0 T.