# 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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### 1 Answer

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`