Transformer Voltage Drop: What, Why, How To Find And Detailed Facts


This article highlights the transformer voltage drop and the related frequently asked questions. Transformer voltage drop is a significant factor that affects the efficiency and performance of a transformer.

Many reasons can bring about the transformer voltage drop. The two most significant factors are load and internal resistance of the supply. The voltage drop measure moderately differs in single-phase transformers to three-phase transformers. Both transformer voltage drops are functions of current, reactance, and resistance.

Read more on….How Do Transformers Increase Voltage To Decrease Current: Exhaustive FAQs

What is transformer voltage drop?

Load resistance and cumulative series resistance in the transformer’s primary winding and secondary winding result in transformer voltage drop. These are caused by improper mutual inductance.

The transformer voltage drop is also known as “voltage regulation” as the voltage drops due to an increase in load resistance. Voltage regulation shows the amount of voltage drop that occurs in the secondary winding/ load of the transformer. Transformer voltage drop is also influenced by the I2R losses.

Equivalent circuit of a real transformer
Equivalent circuit of a real transformer; Image credit: Wikipedia

Causes of voltage drop in transformer?

The internal resistance of the source is the primary reason for voltage drop in a circuit. The more we draw current from the supply, the more the voltage drops across internal resistance and less the overall source voltage.

If there is a small load connected across the secondary winding of the transformer, the load impedance induces current to flow through the internal winding. Because of the impedance of the internal coils of the transformer, voltage drops. Also, the leakage reactance accounts for the change in output terminal voltage.

Read more on…Mutual Inductance Transformer : Mutual Inductance Equivalent Circuit and 10+ Critical FAQs

Voltage drop in transformer formula?

Transformer voltage drop is a significant factor that affects the efficiency of an electrical system. Excessive voltage drop in the transformer may lead to low voltage at the part of the system where the load is present.

The formula for calculating the transformer voltage drop-

Single Phase Transformer: Voltage Drop [Latex]  V_{d}= I\left ( R\cos \theta + X\sin \theta  \right ) [/Latex]

Three Phase Transformer: Voltage Drop [Latex] V_{d}= \sqrt{3} I\left ( R\cos \theta + X\sin \theta  \right ) [/Latex]

where: 

Vd = voltage drop

R = Resistance 

X = Reactance

Θ = power factor angle

How to calculate voltage drop in transformer?

We can calculate the voltage drop in a transformer in either approximate or exact form. We need to know the resistance and reactance both to find out any kind of transformer voltage drop.

The approximate transformer voltage drop referred to the primary side [Latex] = I_{1} R_{01} \cos \theta \pm  I_{1} X_{01} \sin \theta [/Latex] and to the secondary side [Latex] = I_{2} R_{02} \cos \theta \pm  I_{2} X_{02} \sin \theta [/Latex]

The exact transformer voltage drop [Latex] =\left ( I_{2} R_{02} \cos \theta \pm  I_{2} X_{02} \sin \theta  \right ) + \frac{ \left ( I_{2} X_{02} \cos \theta \mp  I_{2} R_{02} \sin \theta  \right )^{2} } {2\: _{0}^{}\textrm{} V_{2}} [/Latex]

Approximate voltage drop in a transformer?

At no-load, the induced voltage at the primary side is the same as the applied voltage, and the induced voltage at the secondary side is the same as the secondary terminal voltage. Suppose, at no-load, 0V2 is the secondary terminal voltage. So, we can say E2 = 0V2. Let us say V2 is the on-load secondary voltage. Figure 1 depicts the phasor diagram of a transformer referred to as secondary.

In Figure 1, R02 and X02 are respectively the net equivalent resistance and reactance of the transformer, referred to the secondary side. Keeping the center at O, we draw an arc that intersects the extended OA at H. From C, we draw a perpendicular on OH that intersects it at G. Now AC represents the exact drop, and AG represents the approximate drop.

The approximate transformer voltage drop

= AG = AF+ FG = AF+ BE

[Latex] =  \left ( I_{2} R_{02} \cos \theta +  I_{2} X_{02} \sin \theta  \right ) [/Latex]

This is the approximate voltage drop for a lagging power factor.

For a leading power factor, the approximate voltage drop is [Latex]  \left ( I_{2} R_{02} \cos \theta –  I_{2} X_{02} \sin \theta  \right ) [/Latex]

( ‘+’ sign represents lagging power factor and ‘-’ sign represents leading power factor)

Similarly, we can find the voltage drop referred to the primary as [Latex]  \left ( I_{1} R_{01} \cos \theta \pm  I_{1} X_{01} \sin \theta  \right ) [/Latex] 

Exact and approximate transformer voltage drop - phasor diagram
Phasor Diagram of the Transformer Referred to the Secondary side

Exact voltage drop in transformer?

According to Figure 1, the exact voltage drop is AH. We can find AH by adding GH to AG which has already been obtained.

By the right-angled triangle OCG. We have

OC2 = OG2 + GC2

i.e. OC2 – OG2 = GC2

i.e. (OC – OG)(OC + OG) = GC2

i.e. (OH –OG)(OC + OG) = GC2

i.e. GH.2.OC= GC2 [Considering. OC = OG]

i.e.   [Latex] GH = \frac{GC^{2}} {2OC}= \frac{\left ( CE-GE \right )^{2}} {2OC}= \frac{\left ( CE-BF \right )^{2}} {2OC}= \frac{ \left ( I_{2} X_{02} \cos \theta –  I_{2} R_{02} \sin \theta  \right )^{2} } {2\: _{0}^{}\textrm{} V_{2}} [/Latex]

For lagging power factor, the exact voltage drop is = AG+ GH = [Latex] =\left ( I_{2} R_{02} \cos \theta +  I_{2} X_{02} \sin \theta  \right ) + \frac{ \left ( I_{2} X_{02} \cos \theta –  I_{2} R_{02} \sin \theta  \right )^{2} } {2\: _{0}^{}\textrm{} V_{2}} [/Latex]

For leading power factor, the exact voltage drop is 

[Latex] =\left ( I_{2} R_{02} \cos \theta –  I_{2} X_{02} \sin \theta  \right ) + \frac{ \left ( I_{2} X_{02} \cos \theta +  I_{2} R_{02} \sin \theta  \right )^{2} } {2\: _{0}^{}\textrm{} V_{2}} [/Latex]

Generally, the exact voltage drop is [Latex] =\left ( I_{2} R_{02} \cos \theta \pm  I_{2} X_{02} \sin \theta  \right ) + \frac{ \left ( I_{2} X_{02} \cos \theta \mp  I_{2} R_{02} \sin \theta  \right )^{2} } {2\: _{0}^{}\textrm{} V_{2}} [/Latex].

FAQs

Transformer voltage drop under load?

Generally, we calculate the primary voltage of a step-up transformer at the primary winding. The load is joined to the secondary. We join a long wire that connects the primary and the AC voltage source.

For this, the resistance of the wire reduces the primary voltage. The AC voltage source sometimes fails to handle the load applied to the secondary terminal of the transformer. Transformer overload will cause a very high primary current to flow. For all these reasons, transformer voltage drops under the load.

Read more on…Transformer Example: Exhaustive List of Examples

Transformer voltage drop during motor starting?

When an induction motor starts at full voltage, it can even draw five to ten times or more of the entire load current of the motor and have a negative effect. This phenomenon is also known as the line starting.

This line starting current of the motor lasts until the motor nearly approaches the synchronous or rated speed. At these starting conditions, the motors have very low power factors( around 10-30 %). The combined effect of high starting current and low power factor results in the voltage drop across the motors.

Induction motor - Wikipedia
Induction Motor Equivalent Circuit; Image credit: Wikipedia

Transformer voltage drop current?

Transformer voltage drop is the measure of the voltage lost through all or part of the transformer due to resistance/ impedance. Voltage in a transformer drops when the current increases because of the source impedance. 

Current is the driving force for the voltage drop in a transformer. When the current passes through the transformer windings, voltage drops. When current flows through the primary winding, it creates magnetic flux. This flux, being passed through the secondary winding, lets current flow through the load.

Kaushikee Banerjee

I am an electronics enthusiast and currently devoted towards the field of Electronics and Communications . My interest lies in exploring the cutting edge technologies. I'm an enthusiastic learner and I tinker around with open-source electronics. LinkedIn ID- https://www.linkedin.com/in/kaushikee-banerjee-538321175

Recent Posts