Question

A 500 ohm resistor and a 1.2-mH inductor are connected in parallel to a 12-V,40-kHz source?
Would the impedence be 258 ohms?

Answer 1

Impedance can be calculated as:\[Z=\frac{ R·i \omega L }{ R+i \omega L }\rightarrow \left| Z \right|=\frac{ R \omega L }{ \sqrt{R^2+ \omega^2 L^2} }=\frac{ 500·(2· \pi ·40·10^{3})·1.2·10^{-3} }{ \sqrt{500^2+(2· \pi ·40·10^{3})1.2·10^{-3})^2}}\] If you do the calculations you will get 258.25 ohms

Answer 2

thanks for confirming that

Answer 3

u r welcome