# Power calculation and Ohm's Law for Reactance

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.

Cheers!

• Feels ill posed, with too many variables. – Scott Seidman Apr 27 '16 at 13:28
• I have edited it to make it a bit clearer now if that helps. thanks! – kbro Apr 27 '16 at 13:45
• If you expect to figure out the inductance, you need to specify the frequency, and probably the resistance. – Scott Seidman Apr 27 '16 at 13:46
• I am only trying to calculate the reactance X_L. You are correct if I was calculating the Inductance. – kbro Apr 27 '16 at 13:50
• 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. – 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: - 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: - • Are you able to explain why we make it -60 degrees when we put it in the denominator? – kbro Apr 27 '16 at 13:42
• No I think I got that wrong so I'm deleting until my brain wakes up. – Andy aka Apr 27 '16 at 13:43
• OK I think I sorted it. – Andy aka Apr 27 '16 at 13:52
• Ahh yes, that does make sense when you put it like that. Thanks! – 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.