# Source transformation with impedance being inductor or capacitor

I found a presentation that states the following about source transformation when the impedance is capacitive/inductive:

While electrical circuits books, like Nilsson's, says:

If I transform the upper-right circuit of the first image to the upper-left, assuming that, for example: $$i_s(t)=cos(\omega t)$$ $$L_p=L_s=L$$

By the first image, I have: $$e_s(t)=L\dfrac{d(i_s(t))}{dt}=-\omega L \cdot sin (\omega t)$$

By the second: $$e_s(t)=Z_L \cdot I_s=j \omega L \cdot cos(\omega t)$$

## Is the first figure's method incorrect or am I doing something wrong?

• Hint: Your final equation mixes up phasor notation and time-domain notation. If you want to compare it to the next-to-last equation, you should put it all in time domain. Commented Oct 6, 2018 at 3:15
• You're absolutely right! I think I realized the error. I'm going to answer the question. Commented Oct 6, 2018 at 3:26

In this case the error was mine. As @The Photon noted, I'm mixing phasor notation (frequency domain) with time domain notation.

In time domain, we have (as pointed in the question):

$$i_s(t)=cos(\omega t)$$

$$e_s(t)=L\dfrac{d(i_s(t))}{dt}=-\omega L \cdot sin (\omega t)$$

Going from the time domain to the frequency domain (phasor notation): $$i_s(t)=cos(\omega t) \iff I_s=1 \space \angle \space 0$$

In frequency domain, we have: $$I_s=1 \space \angle \space 0$$ $$E_s=Z_L \cdot I_s=j \omega L \cdot 1 \space \angle \space 0$$

Remembering that: $$j\omega L = \sqrt{0^2 +(\omega L)^2} \angle \space atan2 \space (\omega L,0) = \omega L \space \angle \space \frac{\pi}{2}$$         (Note: here I am converting from the rectangular phasor form to the polar phasor form)

Thus, we can write: $$E_s=Z_L \cdot I_s=\left (\omega L \space \angle \space \frac{\pi}{2} \right) \cdot \left( 1 \space \angle \space 0 \right)= \omega L \space \angle \space \frac{\pi}{2}$$

Going from frequency domain (phasor notation) to time domain: $$E_s = \omega L \space \angle \space \frac{\pi}{2} \iff e_s(t)= \omega L \cdot cos \left( \omega t + \frac{\pi}{2} \right) =-\omega L \cdot sin(\omega t)$$

## Therefore, both methods produce the same result.

Note - Remembering that from frequency to time domain we have: $$j \omega \iff \frac{d}{dt}$$

We can see that: $$E_s=Z_L \cdot I_s = j \omega L \cdot I_s = L \cdot (j \omega I_s) \iff L\frac{d(i_s(t))}{dt}=e_s (t)$$