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.
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  • \$\begingroup\$ Why do you think Xc has to equal 2Xl? Show the math that leads you to that conclusion. \$\endgroup\$ – John D Jun 1 '17 at 20:10
  • \$\begingroup\$ 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. \$\endgroup\$ – Asmaa Jun 1 '17 at 20:20
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    \$\begingroup\$ @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\$. \$\endgroup\$ – jonk Jun 1 '17 at 20:33
  • \$\begingroup\$ 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. \$\endgroup\$ – StainlessSteelRat Jun 1 '17 at 20:40

If the capacitor and inductor are in series AND the operating frequency is such that the capacitive reactance is greater than the inductive reactance then, the net impedance presented is capacitive and when that happens, the current leads the applied voltage.

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

If the inductor and capacitor are in parallel then the opposite applies.

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  • \$\begingroup\$ And since there is no resistance the phase difference has to be 90 degrees, right? \$\endgroup\$ – Asmaa Jun 3 '17 at 16:22
  • \$\begingroup\$ Correct, with resistance the phase is degraded towards 0 degrees. \$\endgroup\$ – Andy aka Jun 3 '17 at 16:36

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