Electronic – Why have resistors in a circuit down-stepping 230VAC

is there any power loss?(compare to step up transformer,actually I see there is some thin and thick copper wire binding in the transformer. Will this affect the power loss?

Primary and secondary share the same core so, the main thing to consider is whether the core saturates more when driven backwards. Here's a sort of worked example: -

If the primary winding is 10 henries inductance, on 240V AC supply the current will be: -

\$\dfrac{240V}{2\pi f L}\$ = 76 mA.

Let's say the transformer had 1000 turns on the primary so ampere turns are 76.

The 12V secondary has one twentieth of the primary turns i.e. 50 and its inductance will be \$20^2\$ times smaller at 25 mH. Now, if you applied 12V AC to the secondary you'd get a magnetization current of: -

\$\dfrac{12V}{2\pi f L}\$ = 1528 mA.

Ampere turns are 50 x 1.528 = 76 (i.e. just the same)

Could someone explain the physics behind v1/v2=N1/N2

In a transformer electric energy is converted into magnetic energy and then back to electric energy.

The formula V1 / V2 = N1 / N2 applies to this process without any losses.

Losses can occur due to:

• Resistance of the transformer wires

• Saturation of the magnetic core of the transformer

• Inductance of the transformer (a transformer is also a coupled inductor)

• frequency of the AC signal fed to the transformer

So the formula V1 / V2 = N1 / N2 does apply but it only applies to an ideal transformer.

Also an ideal transformer does not lose any power nor will it add any. So that means:

P1 = P2 and also V1 * I1 = V2 * I2

If you would connect a current source to one side of the transformer then the relation will still hold however you must make sure that the current on the other side of the transformer can flow so you must apply a load or short it.

This is the opposite of connecting a voltage source at the primary side, you are expected to have a load at the secondary side or leave it open, but you're not allowed to short the V2 side !