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A 500 ohm resistor and a 1.2-mH inductor are connected in parallel to a 12-V, 40-kHz...

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t4trendesetter | (Level 1) Honors

Posted September 4, 2013 at 7:11 PM via web

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A 500 ohm resistor and a 1.2-mH inductor are connected in parallel to a 12-V, 40-kHz source. How do I find the phase angle (theta)?

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valentin68 | College Teacher | (Level 3) Associate Educator

Posted September 5, 2013 at 6:16 AM (Answer #1)

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The circuit is RL parallel, it means that both R and L components are subjected to the same voltage drop U. In any alternative circuit, the current on the inductor I(L) is lagging the voltage U with 90 degree, while the current on the resistor I(R) is in phase with the voltage U.

For a diagram of the currents through R and L and voltage U please see attached figure. The total current in the circuit I is therefore the vectorial sum of I(L) and I(R).

Vectorial I= I(L)+ I(R)

In absolute value `I = sqrt(I(L)^2+ I(R)^2)`

where `I(L) = U/(omega*L) = U/(2*pi*F*L) = 12/(2*pi*40*10^(3)*1.2*10^(-3))=0.0398 A`

`I(R) = U/R =12/500 =0.024 A`

Thus the value of the phase angle comes from

`tan(phi)= (I(L))/(I(R)) = -1.658`

`phi =-58.909 degree =+301.09 degree`

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