How does the turns ratio affect the current and voltage in a transformer?

[hylian_loach]

I had trouble with transformers and ohm's law, and discussing it with friends, I got this out of them:

In the formula V=IR, the V represents the potential difference (aka. Voltage Drop).

However, in the formula regarding transformers P=VI, the V represents the induced emf, while in the formula regarding power dissipation in wires P=IV, the V represents the voltage drop between two points

When increasing the voltage to reduce energy loss, the current decreases and therefore the potential difference between the first transformer and the next is much smaller. This leads to a larger voltage at the second transformer, which means less energy has been lost.

I feel like I need to ask because I have (numerous times) accidentally used induced EMF in the formula V=IR, and therefore, end up with a different resistance in the primary and secondary coils, which is, obviously strange.
My teacher claimed that the resistance in the wire does change between the primary and secondary, and I did not feel the need to argue that such a magnitude in change in resistance was impossible.

So, is the above quote true? Because although this may not be examinable, it is very confusing, and correcting any errors would help clear this up! Thanks in advance.

 

[rbj]

if you have an "ideal transformer" that doesn't, itself, consume power (converting it to heat) or store energy (like a capacitor), then the power going in must equal the power coming out of the transformer. both the power going in and power coming out are the product of their voltage times current. that, plus the voltage ratio that is equal to the "turns ratio" of the windings is what explains how a transformer "transforms" an effective resistance from what it really is on the secondary winding side to what it appears to be on the primary side.

 

[hylian_loach]

if you have an "ideal transformer" that doesn't, itself, consume power (converting it to heat) or store energy (like a capacitor), then the power going in must equal the power coming out of the transformer. both the power going in and power coming out are the product of their voltage times current. that, plus the voltage ratio that is equal to the "turns ratio" of the windings is what explains how a transformer "transforms" an effective resistance from what it really is on the secondary winding side to what it appears to be on the primary side.

But how can the resistance of a wire change? If the same material conductor is used, shouldn't the resistance appear to be equal on either side of the transformer?

 

[Dale]

But how can the resistance of a wire change? If the same material conductor is used, shouldn't the resistance appear to be equal on either side of the transformer?

The resistance of a wire is not simply dependent on the material, it is also dependent on the length and diameter of the wire. On the high-voltage side there are typically many more turns, so the length of the wire is longer and therefore the resistance is higher. However, this is actually not the relevant effect in an ideal transformer because in ideal transformers assume 0 resistance in the coils themselves (as mentioned by rbj).

Let's say that you have an appliance that uses 10 A of current at 100 V. Such an appliance would be using 1 kW of power. From the perspective of the power company this looks like a 100V/10A = 10 Ohm load. Now, let's say that we put an ideal transformer between with a voltage step-up factor of 10. This is an ideal transformer so the wires of the transformer itself have zero resistance. No power is lost in the transformer, only in the appliance. However, now, due to the transformer the power company delivers that same 1 kW of power as 1000V and 1A and this looks like a 1000V/1A = 1000 Ohm load.

The point is that it is not the resistance of the transformer itself that is different from one side to the other, it is the effective resistance of the load that is changed. It has nothing to do with changing the resistance of the wires in the transformer, it also has nothing to do with changing the resistance of the load itself. It has to do only with the induction of a different voltage on the other side of the transformer.

 

[BriK]

The resistance of a wire won't change unless you change it physically in some manner...such as in heating it, or lengthening it, etc...

Maybe impedence versus resistance...

 

[rbj]

But how can the resistance of a wire change? If the same material conductor is used, shouldn't the resistance appear to be equal on either side of the transformer?

The resistance of a wire won't change unless you change it physically in some manner...such as in heating it, or lengthening it, etc...

Maybe impedence versus resistance...

you guys are talking past each other. the resistance of the wire in the transformer does not change appreciably and changes in small amounts only if conditions (such as the wire's temperature) changes. hot metal has more resistivity than cold metal. i s'pose the resistance of transformer winding might change significantly in a "runaway" situation where it is being driven by a current source and the heating of the wires causes bigger resistance which causes more heating, etc. when the transformer is about to transform into a bomb. but I'm not assuming that case either.

hylian_loach, do you know about the concepts "Voltage-controlled Voltage Sources" and "Current-controlled Current Sources" and the like? (you can also have "Current-controlled Voltage Sources" and "Voltage-controlled Current Sources" if you like.)

you can model a resistor as a Current-controlled Voltage Source or Voltage-controlled Current Source where the controlling equation is just Ohm's Law:

$$v(t) = R \ i(t)$$
or
$$i(t) = \frac{1}{R} \ v(t)$$

but the ideal transformer is essentially two controlled sources:

$$v_2(t) = \frac{N_2}{N_1} \ v_1(t)$$

and

$$i_1(t) = \frac{N_2}{N_1} \ i_2(t)$$

where N₁/N₂ is the "turns ratio" of the transformer's secondary winding to primary winding. now, if you hang a resistor, R₂, off of the secondary winding and drive the primary with a suitable AC voltage source, you get:

$$v_2(t) = R_2 \ i_2(t)$$
or
$$i_2(t) = \frac{1}{R} \ v_2(t)$$

but, because
$$v_2(t) = \frac{N_2}{N_1} \ v_1(t)$$
and
$$i_1(t) = \frac{N_2}{N_1} \ i_2(t)$$

you can solve for i₁(t) in terms of v₁(t). what is that solution and what does that solution look like?