# Inductance

#### solozz_z

Joined Oct 24, 2019
3
For the below circuit,

can someone please explain how 1/j became e^(-j*pi/2) in the above equation?

thanks!!

#### ericgibbs

Joined Jan 29, 2010
9,370
Hi solo,
As this is Homework show us your attempt, then we can help.
E

#### solozz_z

Joined Oct 24, 2019
3
Hi Eric,
Thanks for the reply. I was able to figure out the first step as below:

I am stuck after this.

#### Papabravo

Joined Feb 24, 2006
12,774
$$\frac{1}{j}\;=\;\frac{j}{j^{2}}\;=\;-j\;=\;cos(-\frac{\pi}{2})\;+\;jsin(-\frac{\pi}{2})\;=\;\large{ e^{\frac{-j\pi}{2}}}$$

Last edited:
$$\frac{1}{j}\;=\;\frac{j}{j^{2}}\;=\;-j\;=\;cos(-\frac{\pi}{2})\;+\;jsin(-\frac{\pi}{2})\;=\;\large{ e^{\frac{-j\pi}{2}}}$$