# Why is the Inductive reactance formula (XL=2πfL) not working for me?

I have an inductor labeled as "222" which I believe its means that its inductance value is equal to 2200 Microhenries. I double checked by measuring the inductor with and LCR meter and the value matched the label.

According to the inductor reactance formula "XL=2πfL", a frequency of around 3620 Hz impressed into a 2200 Microhenries inductor should yield around 50 ohms of inductance reactance.

To test this, I fired up my signal generator configured to output 4 volts peak to peak with a frequency of 3620 Hz. Given that my signal generator has a built in 50 ohms resistance I was expecting the voltage drop across the inductor to be 2 volts (2 volts dropped across the signal generator built in 50 ohms resistor and 2 volts dropped across the inductor 50 ohms reactance). However, the voltage drop across the inductor turned out to be 3 volts.

This means (according to me) that the impedance / reactance generated by the inductor is more like 150 ohms. Basically, 1 volts dropped across the signal generator built in resistor and and 3 volts dropped across the inductor.

Why is this happening? What am I doing wrong? Why is the formula not working for me?

Thanks.

• remember the phase angle between resistance and inductive reactance... – Brian Drummond Sep 15 '15 at 23:32
• Also, the generator's voltage setting usually refers to voltage on a 50ohm load, not the voltage before the internal resistor. – venny Sep 15 '15 at 23:59

So 2200uH is 50j at 3620 Hz.

Using AC voltage divider:

$$Vout = Vin *\frac{50}{50j + 50}$$

$$Vout = Vin *(0.5 - 0.5j)$$

$$Vout = Vin * \sqrt{2}/2 \textrm{ }[45^{\circ}]$$

$$Vout = Vin * 0.707 \textrm{ }[45^{\circ}]$$

If Vin is 4V (rms), Vout is 2.8V (rms) Your inductance has an impedance of j50 ohms at 3620 Hz. Your generator has a 50 ohm resistance at that frequency. Using the standard voltage divider equation, and taking into account the imaginary value of inductor imepedance yields a voltage ratio of 0.707. That value multiplied by your 4 volts yields an output of 2.83 volts which is close to your measured value of 3 volts. The important point is that you must take into account the phase angle of any impedances that you use as a voltage divider. In this case, your generator impedance has a 0 degree phase angle while your inductance has a 90 degree phase angle.

The internal impedance of the signal generator must be resistive rather than inductive. The applied voltage squared would be the sum of the voltage across the resistance squared and the voltage across the inductance squared.