Option 3 : supply voltage is maximum

**Magnetizing inrush current:**

**Magnetizing inrush current**in transformer is the current which is drawn by a transformer at the**time of energizing**it.- To have
**minimum inrush current**in the transformer the**switch-on instant should be at maximum input voltage** - This inrush current is transient in nature and exists for
**few milliseconds**. - The inrush current may be up to
**10 times higher than the normal rated current**of the transformer. - Although the magnitude of inrush current is high it generally
**does not create any permanent fault**in the transformer as it exists for a very small time. - But still, inrush current in a power transformer is a problem, because
**it interferes with the operation of circuits**as they have been designed to function. - Some effects of high inrush current include
**nuisance fuse or breaker interruptions**, as well as**arcing and failure of primary circuit components**, such as switches. Another side effect of high inrush is the**injection of noise**.

Option 3 : Parallel resistance with a high value

**No load losses of the electrical machine:**

When the load is not connected to the electric machine, the machine draws a low value of current to keep the machine active or running

- The current which is taken by the machine is called no-load current
- Current is drawn by the core of the machine
**The equivalent circuit is represented by a high value of resistance in parallel.**- The power factor of the machine will be very low.

Option 3 : 19.6 A

**Concept:**

**Transformer on No-load:**

When the transformer is on no-load, the primary input current is not wholly reactive. The primary input current under no-load conditions has to supply due to the following

- Iron losses in the core i.e. hysteresis loss and eddy current loss
- A very small amount of copper loss in primary

The no-load primary input current I_{0} is not at 90° behind supply voltage but lags it by an angle ϕ_{0} < 90°.

No-load input power (W_{0}) is given as,

W_{0} = V_{1 }I_{0 }cos ϕ_{0}

Where,

V_{1} is the primary supply voltage

I_{0} is no-load current

cos ϕ_{0} is no-load power factor

The primary no-load current I_{0} has two-component as shown in phasor,

Where,

I_{M} is a magnetizing component of no-load current.

I_{u} is an active or working or iron loss component of no-load current.

I_{M} = I_{0} sin ϕ_{0}

I_{u} = I_{0} cos ϕ_{0}

\(I_0= \sqrt{{I_M}^2+{I_u}^2}\)

**Calculation:**

Given,

I_{0} = 20 A

cos ϕ_{0} = 0.20 lagging

∴ ϕ_{0} = cos^{-1 }(0.20) = 78.46

From the above concept,

IM = I0 sin ϕ0 = 20 × sin (78.46) = 19.6 A

I0 = 20 A

cos ϕ0 = 0.20 lagging

Iu = I0 cos ϕ0 = 20 × 0.20 = 4

\(I_0= \sqrt{{I_M}^2+{I_u}^2}\)

\(20= \sqrt{{I_M}^2+{4}^2} \)

I_{M}^{2} = 400 - 16 = 384

I_{M} = 19.6 A

When a transformer is operating on no load the primary applied voltage is approximately balanced by

a. Primary Induced emf

b. Secondary Induced emf

c. Terminal voltage across the secondary

d. Voltage drop across the resistance and reactance

Which statement/s is/are correct?Option 1 : Only a

**Operation of transformer on no load:**

When the transformer is operating at no load, the secondary winding is open-circuited, which means there is no load on the secondary side of the transformer and, therefore, current in the secondary will be zero.

While primary winding carries a small current I_{0} called no-load current which is 2 to 10% of the rated current.

The no-load current consists of two components:

**Reactive or magnetizing component (I _{m}):**

It is in quadrature with the applied voltage V_{1}. It produces flux in the core and does not consume any power.

**Active or power component (I _{w}):** It is also known as a working component. It is in phase with the applied voltage V

**Phasor diagram:**

- The function of the magnetizing component is to produce the magnetizing flux, and thus, it will be in phase with the flux.
- Induced emf in the primary and the secondary winding lags the flux ϕ by 90 degrees.
- The primary copper loss is neglected, and secondary current losses are zero as I
_{2}is zero. - Therefore, the current I
_{0}lags behind the voltage vector V_{1}by an angle ϕ_{0}called the no-load power factor angle and is shown in the phasor diagram. **The applied voltage V**because the difference between them is negligible at no load._{1}is drawn equal and opposite to the induced emf E_{1}- Active component I
_{w}is drawn in phase with the applied voltage V_{1}. - The phasor sum of magnetizing current I
_{m}and the working current I_{w}gives the no-load current I_{0}.

Therefore, when a transformer is operating on no load the primary applied voltage is approximately balanced by primary induced emf.

Option 4 : has small magnitude and low power factor

__Transformer on no-load:__

Circuit diagram for a transformer on load:

Where,

V1 is the applied primary voltage

IW is the working component of current through R0 (Magnetizing resistance)

Iμ is the magnetizing component of current through X0 (magnetizing reactance)

N1 and N2 are primary and secondary turns ratio

- In the case of no-load, the secondary terminal of the transformer is open.
- There is no path available for the current to flow on the secondary side.
- Hence, the transformer does not draw current from the source.
**A small magnitude of current flows through the primary transformer (no-load current I0) called excitation current (used for excitation of the core).**- No-load current (I0) is further divided into Iμ and IW

Phasor Diagram when transformer on no load:

Where,

E1 is the primary induced EMF

V1 is the primary terminal voltage

E2 is the secondary induced EMF

The magnetizing component of current is very much higher compared to the working component of current (Iμ > IW)

**No Load Power Factor, cos ϕ0 = (Iμ / I0) ≈ 0.2 to 0.25**

Hence, the Power factor of a transformer on no load is poor due to the magnetizing reactance of the transformer.

**So no-load current of a transformer has a small magnitude and low power factor.**

A single-phase transformer is switched on to an AC supply. In order to have minimum inrush current switch should be closed at?

Option 1 :

Maximum supply voltage

- Magnetizing inrush current in transformer is the current which is drawn by a transformer at the time of energizing it.
- This inrush current is transient in nature and exists for few milliseconds.
- The inrush current may be up to 10 times higher than normal rated current of transformer.
- Although the magnitude of inrush current is high but it generally does not create any permanent fault in transformer as it exists for very small time.
- But still inrush current in power transformer is a problem, because it interferes with the operation of circuits as they have been designed to function.
- Some effects of high inrush current include nuisance fuse or breaker interruptions, as well as arcing and failure of primary circuit components, such as switches. Another side effect of high inrush is the injection of noise.
- To have minimum in rush current in the transformer the switch-on instant should be at maximum input voltage.

Option 3 : 200 V

For the transformer when excited from different frequencies, the ratio of voltage to frequency always remains constant.

\(\frac{{{V_1}}}{{{f_1}}} = \frac{{{V_2}}}{{{f_2}}}\)

So, applying this concept for the HV side:

\(\frac{{400}}{{50}} = \frac{{{V_2}}}{{40}}\)

Therefore, the voltage on the HV side is,

V_{2} = 320 V

And Voltage on LV side using transformer by using turns ratio,

Turn ratio, \(a = \frac{{{V_{HV}}}}{{{V_{LV}}}}\)

Therefore, V_{LV} = 320 / 2 = 160 V

**But we are asked the voltage at the LV side at no-load.**

A no-load test is performed at the Rated Voltage and Rated Frequency condition.

Therefore, **V _{LV} = 200 V, remains the same.**

Option 3 : supply voltage is maximum

**Magnetizing inrush current:**

**Magnetizing inrush current**in transformer is the current which is drawn by a transformer at the**time of energizing**it.- To have
**minimum inrush current**in the transformer the**switch-on instant should be at maximum input voltage** - This inrush current is transient in nature and exists for
**few milliseconds**. - The inrush current may be up to
**10 times higher than the normal rated current**of the transformer. - Although the magnitude of inrush current is high it generally
**does not create any permanent fault**in the transformer as it exists for a very small time. - But still, inrush current in a power transformer is a problem, because
**it interferes with the operation of circuits**as they have been designed to function. - Some effects of high inrush current include
**nuisance fuse or breaker interruptions**, as well as**arcing and failure of primary circuit components**, such as switches. Another side effect of high inrush is the**injection of noise**.

Option 3 : Parallel resistance with a high value

**No load losses of the electrical machine:**

When the load is not connected to the electric machine, the machine draws a low value of current to keep the machine active or running

- The current which is taken by the machine is called no-load current
- Current is drawn by the core of the machine
**The equivalent circuit is represented by a high value of resistance in parallel.**- The power factor of the machine will be very low.

Option 1 : Magnetizing reactance of the transformer

__Transformer on no-load:__

Circuit diagram for a transformer on load:

Where,

V_{1} is the applied primary voltage

I_{W} is the working component of current through R_{0} (Magnetizing resistance)

I_{μ } is the magnetizing component of current through X_{0} (magnetizing reactance)

N_{1} and N_{2} are primary and secondary turns ratio

- In the case of no-load, the secondary terminal of the transformer is open.
- There is no path available for the current to flow on the secondary side.
- Hence, the transformer does not draw current from the source.
- A small ampere of current flows through the primary transformer (no-load current I
_{0}) called excitation current (used for excitation of the core). - No-load current (I
_{0}) is further divided into I_{μ }and I_{W}

**Phasor Diagram when transformer on no load:**

Where,

E_{1} is the primary induced EMF

V_{1} is the primary terminal voltage

E_{2} is the secondary induced EMF

The magnetizing component of current is very much higher compared to the working component of current (I_{μ} > I_{W})

Power factor cos ϕ_{0} = (I_{μ }/ I_{0}) ≈ 0.2 to 0.25

**So, the Power factor of a transformer on no load is poor due to the magnetizing reactance of the transformer.**