Find the current using jω method

[Skarki]

Summary: Current using jω method with 3 components in the following series : Resistor, capacitor and Inductor.

I have the following information:

R=1 * 10^3 ohm
L = 0.5 H
C = 0.1 * 10^-6 F

v(t) = 230 * sqrt(2) * sin(2pi * 400 * t)

All the components are in parallel as shown in the image:

And the task is to find the expression for I(t).

What I've done so far:
I've put up expressions for Ztot = Zr + Zc + Zl

With this I tried the formula I(t) = V(t) / Ztot.

I multiplied both the numerator and denumerator with the conjugate of the denumerator. And I get an equation of the form I(t) = A+iB

And now I am stuck, any tips?

 

[berkeman]

Welcome to PF.

with 3 components in the following series

All the components are in parallel as shown in the image:

All are in series, right?

 

[Skarki]

Welcome to PF.
All are in series, right?

Yeah, I wrote wrong. All components are ofcourse in series!

 

[berkeman]

I multiplied both the numerator and denumerator with the conjugate of the denumerator. And I get an equation of the form I(t) = A+iB

Can you show us this work and your result so far?

Also, going forward it's best to learn to post in LaTeX (see the "LaTeX Guide" link below the Edit window).

 

[Skarki]

$$R=10^{3} [\Omega]$$
$$L=0.5 [H]$$
$$C=0.1 \cdot 10^{-6} [F]$$
$$U(t) = 230 \cdot \sqrt{2} \sin(2\pi \cdot 400t) [V]$$

$$Z_{R} = R [\Omega]$$
$$Z_{L} = j{\omega}L [\Omega]$$
$$Z_{C} = \frac{-j}{\omega C} [\Omega]$$

$$Z_{tot}=Z_{R} + Z_{C} + Z_{L} = R + j(\omega L - \frac{1}{\omega C})$$

$$I(t)=\frac{U(t)}{Z_{tot}} = \frac{230 \cdot \sqrt{2} \sin(2\pi \cdot 400t)}{R + j(\omega L - \frac{1}{\omega C})} = \frac{230 \cdot \sqrt{2} \sin(2\pi \cdot 400t)}{R + j(\omega L - \frac{1}{\omega C})} \cdot \frac{R - j(\omega L - \frac{1}{\omega C})}{R - j(\omega L - \frac{1}{\omega C})} = \frac{230 \cdot \sqrt{2} \sin(2\pi \cdot 400t) \cdot (R - j(\omega L - \frac{1}{\omega C})}{R^{2}+(\omega L - 1 \frac{1}{\omega C})^{2}} = \frac{230R\sqrt{2}\sin(2\pi400t)}{R^{2}+{(\omega L - \frac{1}{\omega C}})^2} - j\frac{(\omega L - \frac{1}{\omega C} )\cdot 230\sqrt{2}\sin(2\pi400t)}{R^{2}+(\omega L - 1 \frac{1}{\omega C})^{2}}$$

 

[erobz]

$$R=10^{3} [\Omega]$$
$$L=0.5 [H]$$
$$C=0.1 \cdot 10^{-6} [F]$$
$$U(t) = 230 \cdot \sqrt{2} \sin(2\pi \cdot 400t) [V]$$

$$Z_{R} = R [\Omega]$$
$$Z_{L} = j{\omega}L [\Omega]$$
$$Z_{C} = \frac{-j}{\omega C} [\Omega]$$

$$Z_{tot}=Z_{R} + Z_{C} + Z_{L} = R + j(\omega L - \frac{1}{\omega C})$$

$$I(t)=\frac{U(t)}{Z_{tot}} = \frac{230 \cdot \sqrt{2} \sin(2\pi \cdot 400t)}{R + j(\omega L - \frac{1}{\omega C})} = \frac{230 \cdot \sqrt{2} \sin(2\pi \cdot 400t)}{R + j(\omega L - \frac{1}{\omega C})} \cdot \frac{R - j(\omega L - \frac{1}{\omega C})}{R - j(\omega L - \frac{1}{\omega C})} = \frac{230 \cdot \sqrt{2} \sin(2\pi \cdot 400t) \cdot (R - j(\omega L - \frac{1}{\omega C})}{R^{2}+(\omega L - 1 \frac{1}{\omega C})^{2}} = \frac{230R\sqrt{2}\sin(2\pi400t)}{R^{2}+{(\omega L - \frac{1}{\omega C}})^2} - j\frac{(\omega L - \frac{1}{\omega C} )\cdot 230\sqrt{2}\sin(2\pi400t)}{R^{2}+(\omega L - 1 \frac{1}{\omega C})^{2}}$$

Aren't the inductor and capacitor voltages a function of time?

The Voltage Source is changing in time, and presumably so is the current. I believe the voltages across the components are all varying in time because the current is varying in time. Perhaps I'm mistaken though as that seems to be a difficult problem to solve.

 

[erobz]

As far as the algebra goes, all seems well. If this is the current (I'm not sure of what I said above is anything to worry about), then what is the issue you are concerned about? I think the next step is to find the phase angle if the objective is to get rid of the $j$.

 

[berkeman]

@Skarki -- Are you learning about how to use Phasors to solve circuit behavior at a particular frequency? If you needed to calculate the transfer function versus frequency of that circuit for the voltages between specific nodes, you would need to do a full equation with $j\omega$ carried along. But in this case, the excitation voltage is at one particular frequency, so the current flowing will also be at that same frequency, and this simplifies to a Phasor problem.

So all you need to do is calculate the complex impedance of each element at that excitation frequency and add them and then use the complex version of V=IR at that frequency to calculate the resulting current.

Does that make sense? And big props for picking up LaTeX so quickly!

And extra bonus points for carrying units along in your equations. That is a great habit to get into in your studies now and your work in the future.

https://en.wikipedia.org/wiki/Phasor

 

[Babadag]

In Wikipedia is written:

In physics and engineering, a phasor is a complex number representing a sinusoidal function whose amplitude (A), angular frequency (ω), and initial phase (θ) are time-invariant.

So, in order to work with phasors you have to use only complex numbers which are not in time domain. Then to transform √2*V*sin(2*π*400*t) into a phasor.

You may use functions in time domain but in this case you have to remain in time domain. For instance:

i(t)*R+di(t)/dt*L+∫i(t)*dt=√2*V*sin(2*π*400*t)

 

[Guineafowl]

Here’s the answer I got:

$R=1000< 0^o ~\Omega$
$X_L=j\omega L = j1256.637 ~\Omega$
$X_C=-j\frac 1 {\omega C} = -j3978.874 ~\Omega$

$Z_{TOT} = R + X_L + X_C$
$Z_{TOT} = 1000~-~j2722.237 ~\Omega ~\rm or$
$Z_{TOT} = 2900< -69.83^o ~\Omega$

$I_{(t)} = \frac {V}{Z_{TOT}} = \frac{230 < 0^o~V}{2900 < -69.83^o~\Omega}$

$I_{(t)} = 79.3\rm x 10^{-3} < 69.83^o~A_{rms}$

And, converting rms to peak:

$I_{(t)} = (\sqrt 2)79.3 \rm x 10^{-3}~sin(2\pi 400t + 69.83^o)~A$