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# Motor Starter, Transformer Assignment Help

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Electrostatics - Motor Starter, Transformer

A starter is a device which is used for starting a d.c motor safely its function is t introduce a suitable resistance in the circuit at the time of starting of the motor. The resistance decreases gradually and reduced to zero when the motor runs at full speed.

Infect resistance f armature of d.c motor is kept low (to reduce the copper losses) anthem armature is stationary there is no back therefore when full operating voltage is applied the current through armature coil may become so large (I = V/R) that the motor may burn. A starter is needed to avoid this.

The essential parts of a starter are shown in fig to start the motor, the handle H is brought in contact with the point A, so that its entire resistance R is in series with the motor M therefore, the initial current becomes small. The electromagnet E gets magnetized due to the passage of cure. It attracts the iron piece I attached to the handle H. as the handle moves towards E the series resistance of the starter decreases gradually. The current through the motor increases accordingly when H just touches E resistance R of the starter is out of circuit. The current becomes maximum and the armature rotates at full speed.

When the main supply is switched off electromagnet loses its magnetism and it can no longer hold H. by the action of the spring S, the handle comes to the off position.

**Transformer **

A transformer is an electrical device which is used of changing the voltages.

A transformer which increases the voltages is called a step up transformer. A transformer which decreases the voltages is called step done transformer

**Principle: **A transformer is based on the principle of mutual induction wherever the amount of magnetic flux linked with a coil changes and is induced in the neighboring coil.

**Construction: **A transformer consist of a rectangular soft iron core made of laminated sheets well insulated from one another the coils **P**_{1} P_{2} and **S**_{1} S_{2 }are wound on the same core but the well insulated from each other note that both the coils. Are insulated form the core the source of alternating (to be transformed) is connected to **P**_{1} P_{2} the primary coil and a load resistance R is connected to **S**_{1} S_{2} the secondary coil through an open switch’s. Thus there can be no current through the secondary coil so long as the switch is open.

For an ideal transformer we assume that the resistances of the primary and secondary windings are negligible. Further the energy loss due to magnetic hysteresis in the iron core is also negligible. Well designed high capacity transformers may have energy losses as low as 1%.

**Theory and working**

Let the alternating supplied by the source connected to primary be

**E = E**_{0} sin wt

As we have assumed the primary to be a pure inductance the sinusoidal primary current Ip lags the primary voltage E by **90**. The primary’s power factor **cos ∅ = cos 90° = 0** therefore no power is dissipated in primary.

The alternating primary current induces an alternating magnetic flux ∅n in the iron core. Because the core extends through the secondary winding the induced flux also extends through the turns of secondary.

According to faraday law of induction the induced per tam (e) is same for both the primary and secondary. Also the voltage E across the primary is equal to the induced in the primary and the voltage E across the secondary is equal to the induced in the secondary. Thus

**E **_{turn} = d∅B /dt = Ep / np = E / n

Here n; n represent total number of turns in primary and secondary coils respectively

Es = Ep n/np

If ns > np ; E > E the transformer is a step up transformer similarly when n < n ;E < E the device is called a step down transformer.

M /n = K represents transformation ration.

Note that this relation is based on three assumptions

The primary resistance and current are small

There is no leakage of flux the same flux links both the primary and secondary coils,

The secondary current is small.

Now the rate at which the generator/source transfers energy to the primary =** Ip Ep** the rate at which the primary then transfers energy to the secondary (via the alternating magnetic field linking the two coils) is** I E**

As we assume that no energy in lost along the way conservation of energy requires that

I**p Ep = I E**

∴ I = Ip Ep / E

Is = Ip np / ns = Ip / K

For a step up transformer **E > E**

K > 1 ∴ I < Ip

Secondary current is weaker when secondary voltage is higher whatever we gain in voltage we lose in current in the same ratio.

The reverse is true for a step down transformer.

**Ip = Is (n /np) = E / R (n / np)**

Using eqn we get

**Ip = I/R Ep (ns / np) (ns / np)**

Ip = 1 / R (n / np) Ep

This eqn has the form

**Ip = Ep / R**

Where the equivalent resistance **R** is

**R = (np / ns) 2 R **

Thus R is the value of load resistance as seen by the source/ generator the source/ generator produce current Ip and voltage Ep as if it were connected to a resistance R

Not suggests that for maximum transfer of energy form an device to a resistive load the resistance fo device and resistance of load must be equal.

We can match the impedances of the two devices by coupling them through a transformer with a suitable turn ratio (n/n)

Thus impedance matching is yet another important function of the transformer.

Efficiency of a transformer is defamed as the ratio of output power to the input power.

**N = output power / input power = E, I / Ep Ip**

In an ideal transformer where there is no power loss **n = 1 (100%)** however practically there are many energy losses. Hence efficiency of a transformer in practice is less than one (less than **100%**)

Motor Starter, Transformer Assignment Help, Motor Starter, Transformer Homework Help, Motor Starter, Transformer Tutors, Motor Starter, Transformer Solutions, Motor Starter, Transformer Tutors, Electrostatics Help, Physics Tutors, Motor Starter, Transformer Questions Answers

Infect resistance f armature of d.c motor is kept low (to reduce the copper losses) anthem armature is stationary there is no back therefore when full operating voltage is applied the current through armature coil may become so large (I = V/R) that the motor may burn. A starter is needed to avoid this.

The essential parts of a starter are shown in fig to start the motor, the handle H is brought in contact with the point A, so that its entire resistance R is in series with the motor M therefore, the initial current becomes small. The electromagnet E gets magnetized due to the passage of cure. It attracts the iron piece I attached to the handle H. as the handle moves towards E the series resistance of the starter decreases gradually. The current through the motor increases accordingly when H just touches E resistance R of the starter is out of circuit. The current becomes maximum and the armature rotates at full speed.

When the main supply is switched off electromagnet loses its magnetism and it can no longer hold H. by the action of the spring S, the handle comes to the off position.

**Transformer**

A transformer is an electrical device which is used of changing the voltages.

A transformer which increases the voltages is called a step up transformer. A transformer which decreases the voltages is called step done transformer

**Principle:**A transformer is based on the principle of mutual induction wherever the amount of magnetic flux linked with a coil changes and is induced in the neighboring coil.

**Construction:**A transformer consist of a rectangular soft iron core made of laminated sheets well insulated from one another the coils

**P**and

_{1}P_{2}**S**

_{1}S_{2 }are wound on the same core but the well insulated from each other note that both the coils. Are insulated form the core the source of alternating (to be transformed) is connected to

**P**

_{1}P_{2}the primary coil and a load resistance R is connected to

**S**the secondary coil through an open switch’s. Thus there can be no current through the secondary coil so long as the switch is open.

_{1}S_{2}For an ideal transformer we assume that the resistances of the primary and secondary windings are negligible. Further the energy loss due to magnetic hysteresis in the iron core is also negligible. Well designed high capacity transformers may have energy losses as low as 1%.

**Theory and working**

Let the alternating supplied by the source connected to primary be

**E = E**

_{0}sin wtAs we have assumed the primary to be a pure inductance the sinusoidal primary current Ip lags the primary voltage E by

**90**. The primary’s power factor

**cos ∅ = cos 90° = 0**therefore no power is dissipated in primary.

The alternating primary current induces an alternating magnetic flux ∅n in the iron core. Because the core extends through the secondary winding the induced flux also extends through the turns of secondary.

According to faraday law of induction the induced per tam (e) is same for both the primary and secondary. Also the voltage E across the primary is equal to the induced in the primary and the voltage E across the secondary is equal to the induced in the secondary. Thus

**E**

_{turn}= d∅B /dt = Ep / np = E / nHere n; n represent total number of turns in primary and secondary coils respectively

Es = Ep n/np

If ns > np ; E > E the transformer is a step up transformer similarly when n < n ;E < E the device is called a step down transformer.

M /n = K represents transformation ration.

Note that this relation is based on three assumptions

The primary resistance and current are small

There is no leakage of flux the same flux links both the primary and secondary coils,

The secondary current is small.

Now the rate at which the generator/source transfers energy to the primary =

**Ip Ep**the rate at which the primary then transfers energy to the secondary (via the alternating magnetic field linking the two coils) is

**I E**

As we assume that no energy in lost along the way conservation of energy requires that

I

**p Ep = I E**

∴ I = Ip Ep / E

Is = Ip np / ns = Ip / K

∴ I = Ip Ep / E

Is = Ip np / ns = Ip / K

For a step up transformer

**E > E**

K > 1 ∴ I < Ip

K > 1 ∴ I < Ip

Secondary current is weaker when secondary voltage is higher whatever we gain in voltage we lose in current in the same ratio.

The reverse is true for a step down transformer.

**Ip = Is (n /np) = E / R (n / np)**

Using eqn we get

**Ip = I/R Ep (ns / np) (ns / np)**

Ip = 1 / R (n / np) Ep

Ip = 1 / R (n / np) Ep

This eqn has the form

**Ip = Ep / R**

Where the equivalent resistance

**R**is

**R = (np / ns) 2 R**

Thus R is the value of load resistance as seen by the source/ generator the source/ generator produce current Ip and voltage Ep as if it were connected to a resistance R

Not suggests that for maximum transfer of energy form an device to a resistive load the resistance fo device and resistance of load must be equal.

We can match the impedances of the two devices by coupling them through a transformer with a suitable turn ratio (n/n)

Thus impedance matching is yet another important function of the transformer.

Efficiency of a transformer is defamed as the ratio of output power to the input power.

**N = output power / input power = E, I / Ep Ip**

In an ideal transformer where there is no power loss

**n = 1 (100%)**however practically there are many energy losses. Hence efficiency of a transformer in practice is less than one (less than

**100%**)