Is the Back EMF of a Transformer a Dynamic Equilibrium?

AI Thread Summary
The back emf of a transformer's primary coil is typically equal to the supply voltage under ideal conditions, but this is a simplification as there is usually a negligible voltage drop due to resistance. Kirchhoff's voltage law supports this equality, indicating that without resistance, no current would flow. However, when considering different inductors, the rate of change of current (dI/dt) will vary even if the supply voltage remains constant. In scenarios with a constant voltage supply, there is a risk of infinite current increase if resistance is not present, while a sinewave supply prevents this blow-up due to its finite integral over time. Understanding these dynamics is crucial for transformer operation and circuit design.
haleycomet2
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Why the back emf of the primary coil always equal to the supply voltage but no greater than supply voltage(because of the large number of coil turns) when no loaded?Or it is a kind of dynamic equilibrium?If yes,how it reach equilibrium?

Thank you.
 
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haleycomet2 said:
Why the back emf of the primary coil always equal to the supply voltage but no greater than supply voltage(because of the large number of coil turns) when no loaded?Or it is a kind of dynamic equilibrium?If yes,how it reach equilibrium?

Thank you.

It is a simplifying assumption to assume that the back emf and supply voltage are the same.They are not exactly equal because there is a voltage drop across the resistance of the primary winding but usually this can be considered as negligible.
 
If you apply Kirchhoff's voltage law to primary of the transformer, you get the back emf equal to supply voltage. (neglecting resistance in primary circuit)
 
Under ideal condition(no resistance),since the back emf always equal to the supply voltage,then there would be no current flow through??
If different inductor(different L) is used in the circuit,does the dI/dt would be different even the supply voltage remains the same?
 
haleycomet2 said:
Under ideal condition(no resistance),since the back emf always equal to the supply voltage,then there would be no current flow through??
If different inductor(different L) is used in the circuit,does the dI/dt would be different even the supply voltage remains the same?

There would initially be no current. But dI/dt is nonzero, so current will begin to flow according to the appropriate differential equation. The current will increase without bound if there's no limiting resistance; Better make sure there's some resistance in the circuit, or something's going to go FSSST-BANG!

If you change L, then you change the magnitude of dI/dt.
 
ya,i just found it is clear when shown by differential equation.By the way, i am curious that is it possible to map out the trend of the unlimited increase of current?:smile:(let V is sin t and L is 1)
Thank you.
 
haleycomet2 said:
ya,i just found it is clear when shown by differential equation.By the way, i am curious that is it possible to map out the trend of the unlimited increase of current?:smile:(let V is sin t and L is 1)
Thank you.

The worry about infinite increase in current applies to a constant voltage supply (or at least one with a constant, nonzero DC component).

If the supply is a sinewave, v(t) sin(t), then its integral over time is finite. So no blow-up for sinewaves!
 
o...i see,thanks a lot..:approve:
 
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