# The phase between current and voltage in LC circuit

I came across this question in my text book and it says:

When does Voltage lag current by 90 degrees in an AC circuit contains only a capacitor and an inductor (its ohmic resistance is negligible).

And I thought of this:

When Xl = 2Xc.

Is my answer correct? As in the solution the book mentioned two methods and mine wasn't one of them. They are:

1. When the capcitive reactance is larger than the inductive reactance.
2. When we take the inducor out of the circuit.
• Why do you think Xc has to equal 2Xl? Show the math that leads you to that conclusion. – John D Jun 1 '17 at 20:10
• Ah, I made a mistake, excuse me, I meant that Xl = 2Xc, then the equivalent impedance for them would be (Z = Xl - Xc = Xc) then we will get a total impedance equals the capcitive reactance of the capacitor, and this will make voltage lags current by 90 degrees. – Asmaa Jun 1 '17 at 20:20
• @Asmaa Isn't worth making you struggle over. Without resistive dissipation or radiative losses, you will either have a $90^\circ$ lag or else lead and nothing in between. It's not a gradual process between the two, but a mathematically abrupt one (if you imagine a perfect lack of heat or radiative loss.) Voltage lags current by $90^\circ$ in a purely capacitive situation, so removing the inductor satisfies. But when capacitive reactance exceeds the inductive reactance, then the remaining reactance is also only capacitive. So same result. Just $X_L < X_C$, not $X_L = k\cdot X_C$. – jonk Jun 1 '17 at 20:33
• 1. When $X_C > X_L$, so $X_{Net}\ (X_C - X_L)$ is capacitive. 2. If you take inductor out of circuit, circuit is capacitive. – StainlessSteelRat Jun 1 '17 at 20:40

So $X_C$ needs to be greater than $X_L$.