RL circuit with AC source transient I have a simulation of RL circuit connected with an ac source. Any one can tell me please that why the peak of the current during the first positive half cycle is higher than the peak of current during the first negative half cycle although both halves of the cycle come from the same voltage source and current goes through the same components in both positive and negative half cycles?

 A: Yes, this is dc offset to satisfy the physics. The current through an inductor cannot change instantaneously ($v=L\frac{di}{dt}$).  The dc transient takes you from one steady-state condition to the next.
If you solve the differential equation of the circuit you will find that the current is, $i(t)=i_p + i_c$, where $i_p$ is the particular solution (also called the forced response or steady-state solution) and where $i_c$ is the complementary solution (also called the natural response).
The complementary solution will be of the form,
$$i_c(t)=Ae^{-\frac{t}{\tau}}$$
where $\tau$ is the circuit time constant (L/R).
The particular solution will be of the form,
$$i_p(t)=\frac{V_M}{\sqrt{R^2+(\omega L)^2}}sin[\omega t + \theta - tan^{-1}(\frac{\omega L}{R})]$$
where $V_M$ is the amplitude and $\theta$ is the phase angle of an applied sinusoidal voltage, $V_M sin(\omega t + \theta)$, and $\omega$ is the voltage source's frequency in radians/second.
You can solve for $A$ by noticing the condition that right before the switch is closed at time $t=0^{-}$ the current is zero.
Play around with $\theta$ and you can see all possible results from a full positive offset to a full negative offset.
I prefer to solve circuits like this with Laplace transform methods but it doesn't matter.
