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PGCIL DT Electrical 13 Aug 2021 Official Paper (NR I)

Option 1 : greater than that at unity power factor

Voltage regulation: The voltage regulation of an AC alternator is,

Percentage voltage regulation \(= \frac{{{E_{0}} - {V}}}{{{V}}} × 100\)

E0 is the internally generated voltage per phase at no load and it is given by,

E_{0} = V + I_{a}Z_{s}

V is the terminal voltage per phase at full load

Voltage regulation indicates the drop in voltage from no load to the full load.

There are three causes of voltage drop in the alternator.

- Armature circuit voltage drop
- Armature reactance
- Armature reaction

The first two factors always tend to reduce the generated voltage, the third factor may tend to increase or decrease the generated voltage. The nature of the load affects the voltage regulation of the alternator.

**Voltage Regulation Curve at different Power Factor:**

**From The curve:**

**Voltage regulation at the lagging power factor will be more than the voltage regulation at the unity power factor.**- Negative voltage regulation and zero voltage regulation occur at the leading power factor.
- Positive voltage regulation occurs at both unities as well as lagging power factors.

**Illustration:**

**Phasor Diagram at Unity Power Factor:**

At unity power factor the value of phase angle will be zero and hence, phasor can be drawn as,

**∴ E _{0} = V + I_{a}R_{a} .... (1)**

**Phasor Diagram at 0.8 lagging Power Factor:**

Given, cos ϕ = 0

∴ ϕ = 36.86°

Now, R_{a} = Z cos ϕ & X_{s} = sin ϕ

Phasor diagram can be drawn as,

From Phasor, the value of E_{0} given by

(E_{0})2 = OA^{2} + AB^{2} = (OD + DA)^{2} + (AC + CD)^{2}

**(E _{0})^{2} = (V cosϕ + I_{a}R_{a})2 + (V sin ϕ + I_{a}X_{s})^{2} .... (2)**

**Proof:**

Let consider,

V = 200 volts (single phase)

R_{a} = 0.06 Ω

X_{s} = 0.8 Ω

Since. X_{s} >> R_{a}

I_{a} = 50 A

**At unity Power,**

E_{0} = V + I_{a}R_{a} = 200 + (50 × 0.06) = 200 + 3 = 203 volts

**∴ Change in voltage = 203 - 200 = 3 volts**

**At 0.8 Lagging Power factor,**

cos ϕ = 0.8 → sin ϕ = 0.6

Now, V cosϕ = 200 × 0.8 = 160 volts

V sin ϕ = 200 × 0.6 = 120 volts

I_{a}R_{a} = 3 volts

I_{a}X_{s} = 50 × 0.8 = 40 volts

From equation (2),

(E_{0})2 = (V cosϕ + I_{a}R_{a})^{2} + (V sin ϕ + I_{a}X_{s})^{2}

(E_{0})2 = (160 + 3)^{2} + (120 + 40)^{2} = 163^{2} + 160^{2}

∴ E_{0} = 228 volts

**∴ Change in voltage = 228 - 200 = 28 volts**

**Hence, Voltage regulation at 0.8 lagging power factor will be more than the voltage regulation at unity power factor.**