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Here's the circuit:

schematic

simulate this circuit – Schematic created using CircuitLab

$$I=2.25,0° mA$$ What's the $$n=N1/N2$$ of the transformer so that on R3 most power is dissipated. Power? U1?

What I don't know is what to do with the transformer. When it's in parallel I know. I just need help with that, else I think I would manage.

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  • \$\begingroup\$ Like Russel McMahon said, it's an autotransformer, but your drawing obscures the fact. A more conventional way to draw it would be as one coil with a tap in the middle, and a double bar that signifies the iron core. commons.wikimedia.org/wiki/File:Autotransformer.png \$\endgroup\$ Commented Jun 19, 2015 at 16:59

1 Answer 1

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What I don't know is what to do with the transformer. When it's in parallel I know. I just need help with that, else I think I would manage.

It's called an "auto transformer.
It obeys the usual rules for an ideal transformer

Call the voltage at the right hand end of R2 "Vin"

Short:

  • Vin is divided between N1 & N2 in the ratio of their turns
    (as in any transformer) so -

  • \$ Vout = \dfrac{N2}{N1+N2} \$

Longer:

$$ Vn2 = Vn1 \times N2/N1 \tag1 $$

So $$ Vn1 = Vn2 \times N1/N2 \tag2 $$

By inspection

$$ Vin = Vn1 + Vn2 \tag3 $$

So

$$ Vn1 = Vin - Vn2 ... \tag4 $$

From 2 & 4

$$ Vn2 \times N1/N2 = Vin - Vn2 $$ $$ Vn2 \times (N1/N2 +1 ) = Vin $$ $$ Vn2 = \dfrac{Vin}{N1/N2 + 1} $$ $$ Vn2 = \dfrac{Vin}{(N1 + N2) / N2} $$
$$ Vn2 = \dfrac{Vin \times N2}{N1 + N2} $$

i.e., MUCH more simply,
Vin is divided across N1 and N2 in the ratio of their turns.

Example

If N1:N2 = 1:9 Vout is 90% of Vin.
If N1:N2 = 9:1 Vout is 10% of Vin.


Wikipedia - autotransformer

Autotransformer basics

Autotransformers

Many examples

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