# Voltage and current during the first cycle of RL circuit

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.

• If we connect and ideal current source with an ideal capacitor then the voltage that was built by the positive current will be cancelled out by the voltage built by the negative current. - no it won't when the sinewave begins at t = 0. There is no cancellation; it is integration and not cancellation @Alex Apr 7, 2021 at 10:40
• No it's not the same current @Alex - clearly it's not else the graph wouldn't show what you are asking about. Apr 7, 2021 at 10:43
• If you shorted the resistor in the RL circuit, the current would never go negative so, it's not the resistor that causes the asymmetry but the process of integration. It's not cancellation; it's integration in every sense of the word both mathematically and in engineering terms. Apr 7, 2021 at 10:51
• The natural state of affairs after a long length of time is for the waveform to lose the transient and become symmetrical about 0 volts. The resistor ensures this. The resistor is responsible for allowing the transient to settle. Without the resistor there is a permanent transient and the current never settles. Apr 7, 2021 at 10:59
• The resistor will burn more energy in a half cycle that is bigger than a half cycle that is smaller hence, it will equalize half cycles to the same amplitude in the fullness of time. Apr 7, 2021 at 11:07

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:

$$\hat{\overline{\text{I}}_\text{in}}=\frac{\hat{\text{u}}}{\sqrt{\text{R}^2+\left(\omega\text{L}\right)^2}}\tag5$$