Determining the coil inductance of a relay when the relay is on
You don't specify which relay you're using so I'm taking this one as an example.  

Rated voltage = 110V AC
Current = 21mA @ 60Hz
Coil resistance = 932\$\Omega\$    

Now the total impedance 

\$ Z = \dfrac{110V}{21mA} = 5240\Omega \$  

That's the resistive part (932\$\Omega\$) with the reactive part at 90°. Then the reactance 

\$ X_L = \sqrt{Z^2 - R^2} = \sqrt{5240^2 - 932^2} = 5154\Omega \$

Then, since 

\$ X_L = 2 \pi f L \$

\$ L = \dfrac{X_L}{2 \pi f} = \dfrac{5154\Omega}{ 2 \pi \mbox{ } 60Hz} = 13.7H \$

That's a pretty high value, but if you do the same calculation for 50Hz, where the current is 24.2mA, you get a comparable value: 14.1H.