Transformer concepts (magnetic induction)

1. Aug 30, 2010

Terocamo

1. The problem statement, all variables and given/known data
The transfromer equation is given by$$\frac{Vp}{Vs}$$=$$\frac{Np}{Ns}$$ for an ideal transformer.
This equation is correct only if there are perfect flux linkage, the resistance of both primary and secondary coil is small and carries negligible current.
The notes mention here if the secondary coil is in open-circuit, the latter requirement about resistance is well met.

However I dont understand how opening the circuit of secondary coil will affect the current in the primary coil. Also if there is no current in the secondary coil, can it produce magnetic field?

2. Aug 30, 2010

n.karthick

if the secondary coil is in open-circuit, the resistance of primary winding still affects the flow of current and Ip will not be 90 deg with Vp.

if secondary circuit is opened, current still flows in primary and it will produce magnetic field which is responsible for Vs

3. Aug 30, 2010

Terocamo

Can you explain in more detail about how the secondary coil affect the current and power of the primary coil?

4. Aug 30, 2010

n.karthick

If secondary current flows (or increases) in a transformer, it creates (more) flux which is opposite to flux created by primary current. Hence the resultant flux is reduced momentarily. But the primary coil immediately draws more current from source to produce same amount of flux so that Vp (voltage across primary) is unaltered. Since current from source is increased, now source is supplying power to the transformer which is in turn transformed to secondary.

5. Aug 30, 2010

Terocamo

Thz, I think I finally understand now, it is very helpful.

6. Aug 30, 2010

Terocamo

I have another problem. Why is the equation P=V^2/R not used to calculate the power loss in high voltage transmission cables? And why is I^2R okay to use?

7. Aug 31, 2010

n.karthick

You can use both but in P=V^2/R, V is the voltage drop in transmission cable and not the transmission voltage. I prefer I^2R since if I measure the current I can find the loss.