Problem — Phasor diagram (XL greater than XC)

Problem

Draw the phasor diagram for a series RLC circuit with $X_L>X_C$. Indicate the three phasors $V_R$, $V_L$, $V_C$ taking the current phasor $I_0$ as reference, and construct graphically the resultant phasor $V_0$ with the phase shift angle $\varphi$.

Solution

In a series branch the current is the same in every element, so we choose the phasor $I_0$ as reference (along the real axis). The voltages are arranged as follows:

• $V_R=R\,I_0$ is in phase with the current (along $+\text{Re}$);
• $V_L=X_L\,I_0$ leads by $90^\circ$ ($+\text{Im}$);
• $V_C=X_C\,I_0$ lags by $90^\circ$ ($-\text{Im}$).

Phasor diagram with $X_L>X_C$: the resultant $V_0$ is rotated by $\varphi>0$ relative to $I_0$.

The phasors $V_L$ and $V_C$ are opposite: their sum equals $(X_L-X_C)I_0$ and, since $X_L>X_C$, points towards $+\text{Im}$. Adding $V_R$ vectorially to this component gives the resultant phasor $V_0$, tilted by an angle $\varphi>0$ above the real axis: the voltage leads the current (equivalently: the current lags). $\tan\varphi=\frac{X_L-X_C}{R}>0$

The modulus of the resultant is $V_0=I_0\sqrt{R^2+(X_L-X_C)^2}=Z\,I_0$

$\ev{V_R\parallel I_0,\quad V_L\ \text{at }+90^\circ,\quad V_C\ \text{at }-90^\circ,\quad V_0=Z\,I_0\ \text{with }\varphi>0}$

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Topics: Electromagnetic induction Concepts: Alternating current Skills: Phasor method