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Given the following circuit diagram
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I know that in order to find either values, I can apply...

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or ...

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But I am having some trouble understanding how I can get rid of the imaginary term j in both cases, even with conjugate root theorem I can't seem to rid of j totally.

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  • \$\begingroup\$ It should end up in form \$x+jy\$ electrical4u.com/rl-parallel-circuit \$\endgroup\$
    – Tony Stewart EE75
    Mar 4, 2017 at 15:45
  • \$\begingroup\$ You'll still have \$j\$ in the equation because the voltage across the inductor is till a phasor. It will have a phase shift. \$\endgroup\$
    – Envidia
    Mar 7, 2017 at 21:19

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You can't "just get rid of" the imaginary term. You're finding the phasor voltage across the elements, and that will be a complex number.

If you just want the amplitude of the sinusoidal voltage across one of the elements, take the magnitude of its phasor voltage, using the usual way of getting the magnitude of a complex number.

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  • \$\begingroup\$ Does it mean that I treat j as i and enter it in my calculator to obtain a complex number in the polar/rectangular form? \$\endgroup\$
    – stephchia
    Mar 4, 2017 at 17:16
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    \$\begingroup\$ j is just an EE's way of writing what the mathematicians and physicists write as i, without getting confused with the variable used for current. \$\endgroup\$
    – The Photon
    Mar 5, 2017 at 2:11
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The voltages sum to $$V_S = V_R + V_L$$ and because they carry the same current, are perpendicular to one another $$ V_R = R \cdot I$$ and $$V_L = j\omega L \cdot I= j X_L \cdot I.$$

Draw the vectors and you will see!

Note that the actual voltages measured will have no \$j\$ in them!

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