LCR circuits and phase angle

[Sleepy_time]

Homework Statement

A resistor (R), capacitor (C), and inductor (L) are connected in series. What is the complex impedance, Z,
of this LCR series combination?

An AC supply of voltage V(t)=V₀e^iωt is applied across an LCR series combination.
Derive an expression for the current I(t) in the circuit. From your expression for I(t)
write down expressions for both the magnitude of the current and the tangent of the
phase angle between current and applied voltage.

Homework Equations

Z_L=iωL

Z_C=-i/(ωC)

Z_R=R

I(t)=V(t)/Z.​

where; ω=Angular frequency of the voltage and i=√(-1).

The Attempt at a Solution

For the total impedance I got:

Z=R+i(ωL-$\frac{1}{ωC}$),​

and for the current:

I=(V₀e^iωt)/(R+i(ωL-$\frac{1}{ωC}$)).​

For the tangent of the phase angle it would just be the tangent of the argument of the above equation for the current? If so, I can't get it into a "nice" equation without a trig function of another trig function. Thank you for any help.

 

[tiny-tim]

Welcome to PF!

Hi Sleepy_time! Welcome to PF!

For the tangent of the phase angle it would just be the tangent of the argument of the above equation for the current? If so, I can't get it into a "nice" equation without a trig function of another trig function.

tan of the phase angle of a + ib is simply b/a

for (a+ib)/(c+id), use the tan(θ-ψ) formula

 

[Sleepy_time]

Hi tiny-tim, thanks for the help. So what I got is shown in the attachment. Have I got it right?

 

[tiny-tim]

Hi Sleepy_time!

erm …

An AC supply of voltage V(t)=V₀e^iωt
…
the tangent of the phase angle between current and applied voltage.​

apart from that, fine!

(remember, impedance, complex current, etc are all constants! )

 

[asteriatic]

I would first like to clarify that the expressions you have derived for the total impedance and current in an LCR series combination are correct.

To find the magnitude of the current, we can use the Pythagorean theorem to find the magnitude of the complex number representing the current:

|I| = √(Re^2 + Im^2)

where Re and Im represent the real and imaginary parts of the current, respectively. Substituting in the expression for the current, we get:

|I| = √(V0^2 / [R^2 + (ωL - 1/ωC)^2])

To find the tangent of the phase angle, we can use the inverse tangent function, arctan, to find the angle whose tangent is equal to the ratio of the imaginary and real parts of the current:

tan(θ) = Im / Re

θ = arctan(Im / Re)

Substituting in the expressions for the current, we get:

θ = arctan((ωL - 1/ωC) / R)

Thus, the tangent of the phase angle between the current and applied voltage can be expressed as:

tan(θ) = (ωL - 1/ωC) / R

I hope this helps. Keep up the good work!