# Why in a voltage driven parallel RLC circuit (at resonance), the current of the inductor has a DC component?

I'm refreshing my knowledge on LC/RLC circuits, and an interesting thing came up that puzzled me.

I'm doing simulation on parallel a RLC circuit (in LTSpice), which is driven with a sine voltage source, at the resonating frequency of the LC tank. The puzzling part is that the current flowing out of the voltage source, as well as the current flowing through the inductor (90 degrees out of phase of corse) have a DC value of Vout,max/R. For the component values that I've chosen - L=100u ; C=10n ; R=100R, the output current of the voltage source has a DC value of 100mA, as well as the current through the inductor. The current through the capacitor, and through the output load resistor swing ±100mA.

Here is a picture of the simulation output :

The same thing happens if I try to simulate just the LC tank. The output current of the voltage source has a DC value of 100mA (with an extremely small AC waveform), and the current of the inductor has a 100mA DC value :

What's the reason behind this ? Regards!

• You know you are using 1p as a resistor value, which I think in LTspice works out to 1E-12 ohms? I just want to make sure you want that. – jonk Aug 15 '16 at 20:41
• I'm using the 1p resistors in order to see the current (in the direction I want) which is flowing. The 1p resistors are there only for that reason – Aleks Aug 15 '16 at 20:50
• Yes, I thought so. I just wanted to make sure we were both on the same page about it. – jonk Aug 15 '16 at 20:50
• Yeah. LTSpice drives me nuts while showing the output current of a voltage source with a 180degrees phase shift, and I can rotate the resistor any way I like :-) – Aleks Aug 15 '16 at 20:52
• It's a transient, initiated when the sine wave input is switched on. If you wait for steady state it will disappear. – Chu Aug 16 '16 at 6:41

Just so we are on the same page... You have your resonant tank (which might be closer to 159155Hz) in parallel with a resistor of $100\Omega$. That pretty much means $\infty$ for the tank, leaving the $100\Omega$ resistor. So the DC current from the source isn't surprising. Removing the resistor just leaves the tank. But since it is $\infty\Omega$ at resonance, again it is no surprise that there is then zero DC current. I'm guessing that you already know this stuff. The only reason I'm bringing it up is to make it clear that we both see the same thing here.