Is it Possible to Operate a 50Hz Transformer on 5Hz or 500Hz Frequency?

What Happens if a 50Hz Transformer is Connected to 5Hz or 500Hz Supply Frequency?

What would happen if a power transformer designed for operation on 50Hz frequency were connected to a 5 Hz frequency or 500Hz source of the same voltage?

Power transformer is made to operate on specific frequency, usually 50Hz or 60Hz. Let’s see what happens if a 60Hz or 50Hz transformer is connected to the 5Hz and 500Hz frequency then.

Transformer Rating & Parameters

Suppose, a transformer rating as follow where the rated frequency is 50Hz.

• V = Voltage = 11kV
• R = Resistance = 100Ω
• L = Inductance = 0.3 Henry
• f = Frequency = 5Hz, 50Hz & 500Hz

Related Post: Which Transformer is More Efficient When Operates on 50Hz or 60Hz?

50 Hz Transformer Operated on rated 50Hz 

We can find the transformer primary current by I = V/Z (Ohm’s Law i.e. I = V/R) where the Z is the impedance (resistance of AC circuits) which further depends on inductive reactance (X_L).

To calculate the circuit impedance, we will have to find the inductive reactance first.

Inductive Reactance = X_L = 2πfL = 2 x 3.1415 x 50 x 0.3

X_L = 94.2Ω

and

Impedance Z = √ (R²+X_L²)

Z = √ (100²+94.2 ²)

Z = 137.4 Ω

The current in the transformer primary

I = 11kV / 137.4 Ω

 I = 80 A 

Now, Power of the circuit

P = V x I x Cos θ …. (i.e. P ∝ I in this case)

Current is directly proportional to the current.

Power factor = Cos θ = R/Z

Cos θ = 100 Ω / 137.4 Ω

Cos θ = 0.73

P = V x I x Cos θ

P = 11kV x 80A x 0.73

 P = 642.4kW 

I.e. The rated Power is appropriate when transformer is operated on the rated frequency of 50Hz.

• Related Post: Can We Operate a 60Hz Transformer on 50Hz Supply Source and Vice Versa?

50 Hz Transformer Operated on 5Hz 

If the frequency is too low, primary will have insufficient reactance and too much primary current will flow, producing considerable copper losses (P = I²R). The transformer may start to smoke and burn with blast leading to dangerous fire.

Transformer with same rating is connected to the 5Hz supply source. We will do the same calculation to find the current in case of lower frequency than rated frequency of 50Hz.

Inductive Reactance = X_L = 2πfL = 2 x 3.1415 x 5 x 0.3

X_L = 9.42 Ω

and

Impedance Z = √ (R²+X_L²)

Z = √ (100²+9.42²)

Z = 100.44 Ω

The current in the transformer primary

I = 11kV / 100.44 Ω

 I = 109.52 A 

Power factor = Cos θ = R/Z

Cos θ = 100 Ω / 100.44 Ω

Cos θ = 0.9

P = V x I x Cos θ

P = 11kV x 109.52 x 0.9

 P = 1084kW 

The power is much more than the rated power of the transformer due to high current, high magnetizing current and more power flux. This will cause insulation losses and transformer may stat to smoke due to low inductive reactance to oppose the flow of large current.

Related Post: Why Transformer Does not Work on DC Supply instead of AC?

50 Hz Transformer Operated on 500Hz

If the frequency is too much high as compared to the rated frequency, the inductive reactance of the primary will prevent the primary from drawing sufficient power. The hysteresis losses and eddy current losses will be excessive.

The same transformer is connected to the 500Hz frequency supply. Let’s do the same calculation as above to find the current in case of higher frequency.

Inductive Reactance = X_L = 2πfL = 2 x 3.1415 x 500 x 0.3

X_L = 942.4 Ω

and

Impedance Z = √ (R²+X_L²)

Z = √ (100²+942.4²)

Z = 947.7 Ω

The current in the transformer primary

I = 11kV / 947.7 Ω

 I = 11.6 A 

Power factor = Cos θ = R/Z

Cos θ = 100 Ω / 947.7 Ω

Cos θ = 0.1

P = V x I x Cos θ

P = 11kV x 11.6A x 0.1

 P = 12.76kW 

• Related Post: Why Flux in Primary and Secondary Winding is Always Equal?

The amount of transferred power is too much low as compared to the rated power in case of higher frequency of 500Hz.

As mentioned above, when current is reduced through high inductive reactance (due to high frequency where X_L = 2πfL), power will be reduced because current is directly proportional to the power. In addition, eddy current and hysteresis losses will be excessive.

The reason behind this story is that:

I = V/Z

Where

Z = √ (R²+X_L²)

But

X_L = 2πfL i.e. XL ∝ f

i.e.

X_L ∝ 1/I

And P ∝ I

and

Φ_Max ∝ V and I.  … (Φ_Max = – V_M / ωN_P)

Related Posts:

• Why Current Increases When Capacitance Increases or Capacitive Reactance Decreases?
• Why Current Decreases When Inductance or Inductive Reactance Increases?
• In a Capacitive Circuit, Why the Current Increases When Frequency Increases?
• In an Inductive Circuit, Why the Current Increases When Frequency Decreases?
• Why Power Factor Decreases When Inductance or Inductive Reactance Increases?
• Why Power Factor Decreases When Capacitive Reactance Increases or Capacitance Decreases?