A three-phase transformer has 600 primary turns and 200 secondary turns. If the supply voltage is 1000 V, find the secondary line voltage on no-load when the windings are connected in star–delta.
192.45 V
Three-Phase Transformer Star-Delta Voltage Calculation
Let's calculate the secondary line voltage of the three-phase transformer based on the given information.
We are given:
- Primary turns, \(N_p = 600\)
- Secondary turns, \(N_s = 200\)
- Primary supply line voltage, \(V_{p,line} = 1000\) V
- Winding connection: Primary is Star (Y), Secondary is Delta (\(\Delta\))
We need to find the secondary line voltage, \(V_{s,line}\), on no-load.
First, let's consider the primary winding which is connected in Star (Y). In a star connection, the relationship between line voltage and phase voltage is:
\[V_{line} = \sqrt{3} \times V_{phase}\]So, the primary phase voltage \(V_{p,phase}\) is:
\[V_{p,phase} = \frac{V_{p,line}}{\sqrt{3}} = \frac{1000}{\sqrt{3}}\]Next, the ratio of phase voltages in a transformer is equal to the ratio of turns:
\[\frac{V_{p,phase}}{V_{s,phase}} = \frac{N_p}{N_s}\]We can use this to find the secondary phase voltage \(V_{s,phase}\):
\[V_{s,phase} = V_{p,phase} \times \frac{N_s}{N_p}\] \[V_{s,phase} = \left(\frac{1000}{\sqrt{3}}\right) \times \left(\frac{200}{600}\right)\] \[V_{s,phase} = \left(\frac{1000}{\sqrt{3}}\right) \times \left(\frac{1}{3}\right)\] \[V_{s,phase} = \frac{1000}{3\sqrt{3}}\]Finally, let's consider the secondary winding which is connected in Delta (\(\Delta\)). In a delta connection, the relationship between line voltage and phase voltage is:
\[V_{line} = V_{phase}\]So, the secondary line voltage \(V_{s,line}\) is equal to the secondary phase voltage \(V_{s,phase}\).
\[V_{s,line} = V_{s,phase} = \frac{1000}{3\sqrt{3}}\]Now, let's calculate the numerical value:
\[V_{s,line} = \frac{1000}{3 \times \sqrt{3}} \approx \frac{1000}{3 \times 1.73205} \approx \frac{1000}{5.19615} \approx 192.45 \text{ V}\]Therefore, the secondary line voltage on no-load is approximately 192.45 V.
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