# Phasor Diagram for Synchronous Generator

**phasor diagram for synchronous generator**. Now, let us write the various notations for each quantity at one place, this will help us to understand the phasor diagram more clearly. In this phasor diagram we are going to use:E

_{f}which denotes excitation voltage V

_{t}which denotes terminal voltage I

_{a}which denotes the armature current θ which denotes the phase angle between V

_{t}and I

_{a}ᴪ which denotes the angle between the E

_{f}and I

_{a}δ which denotes the angle between the E

_{f}and V

_{t}r

_{a}which denotes the armature per phase resistance In order to draw the phasor diagram we will use V

_{t}as reference. Consider these two important points which are written below:

- We already know that if a machine is working as a synchronous generator then direction of Ia will be in phase to that of the E
_{f}. - Phasor E
_{f}is always ahead of V_{t}.

**phasor diagram of synchronous generator**. Given below is the phasor diagram of synchronous generator:

In this phasor diagram we have drawn the direction of the Ia is in phase with that of the Ef as per the point number 1 mentioned above. Now let us derive expression for the excitation emf in each case. We have three cases that are written below:

- Generating operation at lagging power factor.
- Generating operation at unity power factor.
- Generating operation at leading power factor.

_{f}by first taking the component of the V

_{t}in the direction of I

_{a}. Component of V

_{t}in the direction of Ia is V

_{t}cosΘ, hence the total voltage drop is (V

_{t}cosΘ+I

_{a}r

_{a}) along the I

_{a}. Similarly we can calculate the voltage drop along the direction perpendicular to I

_{a}. The total voltage drop perpendicular to I

_{a}is (V

_{t}sinθ+I

_{a}X

_{s}). With the help of triangle BOD in the first phasor diagram we can write the expression for E

_{f}as (b) Generating operation at unity power factor: Here also we can derive the expression for the E

_{f}by first taking the component of the V

_{t}in the direction of I

_{a}. But in this case the value of theta is zero and hence we have ᴪ=δ.

With the help of triangle BOD in the second phasor diagram we can directly write the expression for E_{f} as
(c) Generating operation at leading power factor: Component in the direction of Ia is V_{t}cosΘ. As the direction of Ia is same to that of the V_{t} thus the total voltage drop is (V_{t}cosΘ+I_{a}r_{a}). Similarly we can write expression for the voltage drop along the direction perpendicular to I_{a}. The total voltage drop comes out to be (V_{t}sinθ-I_{a}X_{s}). With the help of triangle BOD in the first phasor diagram we can write the expression for E_{f} as

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