[![enter image description here][1]][1]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.

1. 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?

2. 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.

3. 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? [1]: https://i.sstatic.net/JkYrL.png

[![enter image description here][1]][1]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.

1. 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?

2. 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 lead 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.

3. 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? [1]: https://i.sstatic.net/JkYrL.png