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

Source Transformation - Presentation

While electrical circuits books, like Nilsson's, says: Source Transformation - Nilsson

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?

  • 1
    \$\begingroup\$ 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. \$\endgroup\$ – The Photon Oct 6 '18 at 3:15
  • \$\begingroup\$ You're absolutely right! I think I realized the error. I'm going to answer the question. \$\endgroup\$ – Vinicius ACP Oct 6 '18 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: $$E_s=Z_L \cdot I_s = j \omega L \cdot I_s = L \cdot (j \omega I_s) \iff L\frac{d(i_s(t))}{dt}=e_s (t)$$

| improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.