Transformers and basic inductor physics

Question

1) Why does resistor reduce the lag in a RL circuit? I understand why current lags by 90° with the voltage across the inductor. (After watching this) But I don't understand why that lag, would be diminished by the presence of a resistor, the resistor would just decrease the amplitude of the current. I thought that, there would be lesser voltage across the inductor in a RL circuit, as some would fall across the R, and hence the change would also be less which produces smaller opposition. But then if I apply that reduced voltage across L alone in a separate circuit, there's a solid 90° lag, it's not reduced. If there were a capacitor, it would push the current more, and therefore it would reduce the lag, by why does the resistor reduce the lag in a RL circuit?

2) In equivalent circuit of a transformer, there are two inductances, $L_p$ and $L_s$. These are called Leakage inductances, as the name suggests due to the leakage flux from both the coils. Are these self inductance? Cause, when we calculate the voltage across an inductor we include two components, one is due to self induction, $-L \frac{di}{dt}$ and the other due to mutual $M \frac{di}{dt}$. Are $L_p$ and $L_s$ only due to leakage flux or they generally represent self inductances of both the coils? If yes, what about the rest of the flux, won't that induce emf?

3) In the book I'm following,

In an ideal transformer, assuming there's no leakage of flux, the induced emf in the primary coil will be given by $$e_1 = \dfrac{d\lambda_1}{dt} = N_1\dfrac{d\phi}{dt}$$ and for an ideal transformer $$v_1=e_1$$and thus $e_1$ and therefore $\phi_1$ must be sinusoidal of frequency $f$ Hz, the same as that of the voltage source. So, $$\phi = \phi_{max} \sin\omega t \implies e_1=N_1\dfrac{d\phi}{dt}=N_1\omega\phi_{max} \cos\omega t$$ Therefore, induced emf leads the flux by $90^\circ$

How can the induced emf leads the flux by $90^\circ$? An induced emf is created only when there's a change in the flux, which is produced only when there's a current flow, which means the flux has to be produced first to create an induced emf.

Answer 1

I will explain to you why the induced emf leads the flux by 90 degree.

Imagine we have ac source connected to RC circuit.the voltage across the capacitor lags the currenrt by 90 degree.this means that at time t=0 the voltage is zero and the current is a maximum.does this mean that before i close the switch there is a current in the circuit(i expect you to think that way according to your question) ?

No that does not happen.the solutions you are using in soving your circuits are called the steady state solutions and there is a transient phenomenon that occur when you close the switch of your circuit and this phenomenon dies out and you get the steady state solution.

I think your problem is that you have not studied yet the transient phase of ac circuits.

Answer 2

The current always lags the voltage in an (pure) inductor. When you put an R in series, if the R is large, then the actual current that flows is dominated by the resistance (the inductance has little effect). In that case, the current and voltage at the terminals of the series R-L circuit are nearly in phase.

It may be easier to consider the V as a function of the current: The vector sum of the voltage across the R and the L add to the applied voltage. If the R >> the inductance, then you can see that the voltage across the inductor will be small, so it won't significantly affect the combined voltage vector. However, the voltage across the inductor (itself) if 90 º leading from the total system current.