Tag: electro-magnetism

This question is already present in ID: 177624.

Transformers are found everywhere alternating current is used. This includes both large power stations to the power cord for our laptop computer. A transformer is an electrical device that trades voltage for current in a circuit, while not affecting the total electrical power. This means it takes high-voltage electricity with a small current and changes it into low-voltage electricity with a large current, or vice-versa. That is, step up or step down the voltage in a circuit. 
A step-up transformer turns low-voltage electricity into high-voltage electricity while dropping the current. A step-down transformer changes high-voltage electricity into low-voltage electricity. 
For a transformer to work, the current in one coil has to somehow make current flow in the other coil and the circuit it's connected to. A DC current in one coil will make a magnetic field on the other coil, but a magnetic field by itself won't drive any electrons around. A changing magnetic field, however, does create an electric force which will accelerate those electrons in the other coil into carrying a current. This process is described by Faraday's law of induction. We get a changing field from an AC current, since the current which makes the field is changing. Hence, DC current cannot be used as a source for a generator.

A transformer takes high-voltage electricity with a small current and changes it into low-voltage electricity with a large current, or vice versa. That is, step up or step down the voltage in a circuit. A step-up transformer turns low-voltage electricity into high-voltage electricity while dropping the current. A step-down transformer changes high-voltage electricity into low-voltage electricity.
If we change the number of turns in the coils we change the induced emf. This allows you to change (transform) the voltage from the primary to the secondary coil.

$\text{Power source}=EI=240\times 0.7=166$
$\Rightarrow \eta =\dfrac{\text{Output}}{\text{Input}}=\cfrac{140}{166}$ $\Rightarrow \eta=83.3\%$

A transformer can't  work on dc current because in dc current, neither magnitude nor direction change, so, there is no change in flux, so, no emf is induced in secondary coil hence no voltage or current is there in secondary coil.

Step up transformer increases the voltage in secondary coil and decreases current in secondary coil. Transformer works only on ac current because in DC current, there is no change of flux associated with the coil

The efficiency of the transformer is defined as ratio of useful power output to the input power. The two being measured is the same unit. Its unit is either watts or kw. Transformer efficiency is denoted by $\eta$

$\eta = \dfrac{\text{output power}}{\text{input power}} = =\dfrac{\text{output power}}{\text{input power + losses}}$

$\eta  = \dfrac{\text{output power}}{\text{input power + iron losses of copper + losses}}$

$\eta = \dfrac{V _2I _2 \cos \phi _2}{V _2 I _2 \cos \phi _2 + P _i + P _c}$

Where

$V _2  = $ secondary terminal voltage

$I _2 = $ Full load secondary current.

$\cos \phi _2 = $ Power factor of the load

$P _i = $ Iron losses = constant loss = exactly current loss + hysteresis loss

$P _c = $ full load copper loss

The efficiency is a function of load i.e. lead current $I _2$ assuming $\cos \phi$ constant. The secondary terminal voltage $V _2$ is also assumed constant.

$\dfrac{d \eta}{d I _2} = 0$

Now $\eta = \dfrac{V _2I _2\cos \phi _2}{ V _2I _2\cos \phi _2 + P _i + I _2^2 R _{2e} }$

$\dfrac{d \eta}{d I _2} = \dfrac{d}{d I _2}(\dfrac{V _2I _2\cos \phi _2}{ V _2I _2\cos \phi _2 + P _i + I _2^2 R _{2e} }) = 0$

$( V _2I _2\cos \phi _2 + P _i + I _2^2 R _{2e}) \dfrac{d}{d I _2}( V _2I _2\cos \phi _2) – (V _2I _2\cos \phi _2) \dfrac{d}{d I _2}( V _2I _2\cos \phi _2 + P _i + I _2^2 R _{2e}) = 0$

$( V _2I _2\cos \phi _2 + P _i + I _2^2 R _{2e})(V _2 \cos \phi _2) – (V _2 I _2 \cos \phi _2)(V _2 \cos \phi _2 + 2 I _2 R _{2e}) = 0$

Cancelling $V _2 \cos \phi _ 2$ from both the terms we get

$ V _2I _2\cos \phi _2 + P _i + I _2^2 R _{2e} – V _2I _2\cos \phi _2 – 2I^2 R _{2e} = 0 $

$P _i – I^2 R _{2e} = 0$

$P _i = I^2 R _{2e} = P _{Cu}$

$P _i = P _{Cu}$

 So, condition to achieve maximum efficiency is that

Copper Loss = Iron Loss

$\mu =\frac { { V } _{ s }{ I } _{ s } }{ { V } _{ p }{ I } _{ p } } =0.8\Rightarrow { I } _{ p }=\frac { { V } _{ s }{ I } _{ s } }{ { V } _{ p }\mu  } =\frac { 440\times 2 }{ 220\times .8 } =5A$

Power = Voltage x Current
As voltages are high so currents are lower. This is done to reduce the resistive losses.
Resistance Loss = ${ { I } _{ rms }^{ 2 } }R$
So, lower current implies less power loss. 
As resistance is inversely proportional to their diameter; hence decreasing their diameter by x will increase resistance by $x^2$ and hence increasing power loss hence not done.
As the voltage required for domestic and commercial uses are pretty less then they can be easily brought down by step-down transformers. 

Answer is C.

Slip-ring commutator merely maintains a connection between the moving rotor and the stationary stator.
Slip rings are useful for DC motors where the current has to change direction every half revolution whereas split rings are used to make constant current connection to a commutator, which is what we want for an AC motor.
Hence, if the slip rings are not present, then the armature stops its rotation.

A transformer can not step up a d.c. input so output potential here will be zero. No potential will be induced in the secondary coil.