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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?
It's not limited to inductors. It happens with capacitors when driven with a sinusoidal current that begins at a phase angle of 0°: -
It quite literally is the process of mathematical integration at work.
If you integrate a sinewave that begins at t=0 then you will get the same effect.
It's math and, inductors and capacitors do math; they integrate.
There is no circuit trick; if you mathematically integrated a sinewave that starts from 0° you will get a waveform that is a bit top-heavy and gradually settles in time due to the presence of the resistor.
Well, using Laplace transform we can see that:
$$\text{I}_\text{in}\left(t\right)=\mathscr{L}_\text{s}^{-1}\left[\frac{\hat{\text{u}}\omega}{\text{s}^2+\omega^2}\cdot\frac{1}{\text{R}+\text{sL}}\right]_{\left(t\right)}\tag1$$
Using the convolution property of the Laplace transform, we can write:
$$\text{I}_\text{in}\left(t\right)=\int_0^t\mathscr{L}_\text{s}^{-1}\left[\frac{\hat{\text{u}}\omega}{\text{s}^2+\omega^2}\right]_{\left(\tau\right)}\cdot\mathscr{L}_\text{s}^{-1}\left[\frac{1}{\text{R}+\text{sL}}\right]_{\left(t-\tau\right)}\space\text{d}\tau=$$ $$\int_0^t\hat{\text{u}}\sin\left(\omega\tau\right)\cdot\frac{\exp\left(-\frac{\text{R}}{\text{L}}\cdot\left(t-\tau\right)\right)}{\text{L}}\space\text{d}\tau=\frac{\hat{\text{u}}}{\text{L}}\int_0^t\sin\left(\omega\tau\right)\exp\left(-\frac{\text{R}}{\text{L}}\cdot\left(t-\tau\right)\right)\space\text{d}\tau=$$ $$\frac{\hat{\text{u}}}{\text{R}^2+\left(\omega\text{L}\right)^2}\cdot\left(\omega\text{L}\left(\exp\left(-\frac{\text{R}t}{\text{L}}\right)-\cos\left(\omega t\right)\right)+\text{R}\sin\left(\omega t\right)\right)\tag2$$
And, when we have:
$$\text{I}_\text{in}\left(t\right)=\frac{\hat{\text{u}}}{\text{R}^2+\left(\omega\text{L}\right)^2}\cdot\left(\omega\text{L}\left(\underbrace{\exp\left(-\frac{\text{R}t}{\text{L}}\right)}_{=\space\text{K}\left(t\right)}-\cos\left(\omega t\right)\right)+\text{R}\sin\left(\omega t\right)\right)\tag3$$
We, can see that when \$t\to\infty\$ we have \$\text{K}\left(t\right)=0\$. So, without the transient part we get:
$$\overline{\text{I}}_\text{in}\left(t\right)=\frac{\hat{\text{u}}}{\text{R}^2+\left(\omega\text{L}\right)^2}\cdot\left(\omega\text{L}\left(0-\cos\left(\omega t\right)\right)+\text{R}\sin\left(\omega t\right)\right)=$$ $$\frac{\hat{\text{u}}\left(\text{R}\sin\left(\omega t\right)-\omega\text{L}\cos\left(\omega t\right)\right)}{\text{R}^2+\left(\omega\text{L}\right)^2}\tag4$$
The amplitude of the resulting current is given by:
$$\hat{\overline{\text{I}}_\text{in}}=\frac{\hat{\text{u}}}{\sqrt{\text{R}^2+\left(\omega\text{L}\right)^2}}\tag5$$
Read this question and my answer:
Transients in series RL circuit with AC source
It's the transient term of the current that adds up to the steady state term of the current.
