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This is what the textbook says:

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I don't understand why does ωLIm equal Vm so if someone knows please explain it to me

Thanks

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  • \$\begingroup\$ You need to have some linear algebra background. A superficially explanation: Since V=L di/dt and d/dt=s=jw, | V / i | =| Z | = | L d/dt | = Lw. So, | Z | . Im = Vm. \$\endgroup\$ – Rohat Kılıç Jan 21 '17 at 14:27
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There is nothing to it really. Once you differentiate Im * L * Sin(wt) wrt to t you get w * L * Im * cos(wt) which is equal to Vm * cos(wt) if YOU let w * L * Im = Vm.

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They're just defining it to be that, in equation 16-14. Later they'll go back and show that this motivates the phasor analysis of linear circuits.

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I think you don't have a problem with the \$v_L = L \frac{d i_L}{d t}\$. Here \$v_L\$ and \$i_L\$ are total instantaneous signals. Then the it assumes that the input signal is a sinusoidal \$ i_L = I_M \sin(\omega t) \$ with magnitude \$I_M\$. Then it just takes the derivate of it and use the trigonometrical identity to obtain a simpler formula. As a consequence, current and voltage have orthogonal phase.

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