- #1

- 7

- 0

By what factor is the voltage multiplied?

V[tex]_{s}[/tex]

------=?

V[tex]_{P}[/tex]

Honestly I am not sure exactly where to start in solving this problem.

- Thread starter orionj
- Start date

- #1

- 7

- 0

By what factor is the voltage multiplied?

V[tex]_{s}[/tex]

------=?

V[tex]_{P}[/tex]

Honestly I am not sure exactly where to start in solving this problem.

- #2

- 7

- 0

Any ideas?

- #3

- 10

- 0

For instance if you plug your mobile phone supply to the power socket and not plug the mobile phone to it the power that goes through transformer is nearly 0!

And then if you connect your mobile phone to it the charging power goes through transformer to your phone. So the transformer power is not 0!

Generally every transformer has a parameter - nominal power which should not be exceeded unless you want to burn the device. If you say '74 W transformer' for electrical engineer it means that the transformer has NOMINAL POWER 74W.

But in this case the given power is not nominal power of the transformer (it is a little bit confusing). If so we would have too few data to solve the problem.

74W (i think) is the power that actually goes through transformer to some (unknown) loading.

The equations you should use are:

[tex] \frac{V_{out}}{V_{in}} = n [/tex] (ideal transformer equation)

[tex] P_{in} = V_{in}I_{in} [/tex] (input power)

[tex] P_{in} = V_{out}I_{out} [/tex] (output power)

And of course input power is equal to output power (there is no loss in transformer)

- #4

- 129

- 3

In a perfect universe, maybe. In real life, losses occur....And of course input power is equal to output power (there is no loss in transformer)

See: http://www.tpub.com/content/doe/h1011v4/css/h1011v4_56.htm

- #5

- 84

- 0

- #6

- 10

- 0

Yeah right :PIn a perfect universe, maybe. In real life, losses occur.

See: http://www.tpub.com/content/doe/h1011v4/css/h1011v4_56.htm

And you will have capacitance between windings, magnetisation current, and nonlinearities due to core permability change.

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