Hi I have the following question:

V = 120V, I = 10 A, I lags V by 60degrees


simulate this circuit – Schematic created using CircuitLab

The load consists of a resistor and an inductor in series. What is the reactance of the inductor?

The way I have tried to solve it:

Since the current lags the voltage 60 degrees: \$ I = 10 \angle 60 \$

\$V = 120\angle0 V\$

\$V = IZ\$

\$Z = \frac{V}{I}\$ (1)

\$= \frac{120\angle0}{10 \angle 60}\$

\$ = 6-10.4j\$

Hence the reactance \$X_L = -10.4 \$

I know this is wrong, because by convention inductors have positive reactance! So where am i going wrong?

I could use power formulas for S, P and Q to solve this, but i would like to know why doesn't the above work? Should I be using the conjugate in (1), it's the only thing I can think of.


  • \$\begingroup\$ Feels ill posed, with too many variables. \$\endgroup\$ – Scott Seidman Apr 27 '16 at 13:28
  • \$\begingroup\$ I have edited it to make it a bit clearer now if that helps. thanks! \$\endgroup\$ – kbro Apr 27 '16 at 13:45
  • \$\begingroup\$ If you expect to figure out the inductance, you need to specify the frequency, and probably the resistance. \$\endgroup\$ – Scott Seidman Apr 27 '16 at 13:46
  • \$\begingroup\$ I am only trying to calculate the reactance X_L. You are correct if I was calculating the Inductance. \$\endgroup\$ – kbro Apr 27 '16 at 13:50
  • \$\begingroup\$ If you make voltage the reference at 0°, then to make current -60° because I lags V by 60°. \$ 10 \measuredangle 60 \Omega \$ means leading. \$\endgroup\$ – StainlessSteelRat Apr 27 '16 at 18:45

i would like to know why doesn't the above work?

It's not \$\angle{60}\$, it's \$\angle{-60}\$ as the denominator: -

enter image description here

The supply voltage can be regarded as at 0 degrees and the current is lagging 60 degrees behind at 300 degrees. Hence the angle of the current is 300 degrees or -60. See this website which also shows this picture: -

enter image description here

  • \$\begingroup\$ Are you able to explain why we make it -60 degrees when we put it in the denominator? \$\endgroup\$ – kbro Apr 27 '16 at 13:42
  • \$\begingroup\$ No I think I got that wrong so I'm deleting until my brain wakes up. \$\endgroup\$ – Andy aka Apr 27 '16 at 13:43
  • \$\begingroup\$ OK I think I sorted it. \$\endgroup\$ – Andy aka Apr 27 '16 at 13:52
  • \$\begingroup\$ Ahh yes, that does make sense when you put it like that. Thanks! \$\endgroup\$ – kbro Apr 27 '16 at 14:00

In phasor form you get the impedance Z = 20<-60. When you convert it to complex form it will become 20cos(-60)+20sin(-60) j = 10-17.32 j. On complex plane it means 10 units in horizontal and 17.32 units in negative vertical direction.


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