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So, here's the circuit: enter image description here

How can I make phasor diagrams for the circuit in case when $$I_C<I_L$$ and $$I_C>I_L$$

Edit: What made me confused is the current conditions, it's actually voltage $$U_C<U_L$$ and $$U_C>U_L$$

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  • \$\begingroup\$ One will cause a lag of current (C) or lead (L), of which the exact phasor is dependent on the combination of both. I am not sure if you can give a phasor diagram without knowing things like the values of each and the winding resistance of the inductor. Though I could be wrong. \$\endgroup\$ May 18 '15 at 20:16
  • \$\begingroup\$ Oh sry, I forgot to mention the values: $L=100mH$, $C=10nF$, $R=100Ω$ I'm not sure what I need to calculate for phasor diagrams and how do I know if current leads or lags? \$\endgroup\$
    – A6SE
    May 18 '15 at 20:22
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What @EMFields said is correct, i think you are getting confused between the current through each element and the values of each element.

Because they are in series the current through each element will be equal. But the Value of each element will affect that very current.First convert the inductance and capacitance into reactances, and find the overall impedance. $$X_l=\omega L$$

$$X_c = \frac{1}{\omega C}$$

$$Z = R+j(X_l-Xc)$$

The sign of the imaginary part will tell you if the circuit is predominantly inductive or predominantly capacitive . And hence if the current will lag the voltage or lead the voltage respectively.

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  • \$\begingroup\$ I understand EM fields answer, but a phasor diagram usually represents the phase shift from the reactive components, thus I believe your answer is more complete. \$\endgroup\$ May 19 '15 at 0:05
  • \$\begingroup\$ That phase shift is not the current through the circuit but voltage at each point in the circuit ie. After the resistor, after the inductor and after the capacitor at each of the nodes. \$\endgroup\$
    – Sada93
    May 19 '15 at 0:09
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Since all of the circuit elements are in series, It = Ir = Ic = Il.

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  • \$\begingroup\$ Precisely my thoughts dude +1 \$\endgroup\$
    – Andy aka
    May 18 '15 at 21:10

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