Trigonometric identities -- a parallel RLC circuit connected to an AC-supply

Question

An RLC-circuit is connected to an AC-supply as in the figure below.

$I_{tot}(t)=I_0sin(\omega t+\phi)$ (denoted as $I_{ges} ( t)$ in the picture), $\phi$ is the phase angle between $V_{tot}(t)$ and $I_{tot}(t)$, respectivelly, the voltage across input terminals and the total current admitted through the circuit.

a) Determine $\phi$.

b) What current $I_\mathrm{L}(t)$(Amplitude and phase) runs through the coil $\mathrm{L}$?

Use the following information: $\mathrm{R}=10 \Omega, ~\mathrm{C}=30\mu F,~\mathrm{L}=10^{-3}H,~I_0=2A,~\omega =300\frac{1}{s}$

I was never good with LC-circuits, which is why I picked out this one out of my textbook.

How do I approach this type of exercise?

I was thinking that since it's an LC-circuit then because of Lenz's law the phase is $\phi =90°$? Is that also the case here? And the resistor $\mathrm{R}$ kind of bugs me in the circuit. Does it have any influence on the current or the phase?

How do I get the amplitude and phase in b)? Although I still think that the phase should be $90°$. But what about the amplitude?

I guess part of the current would flow through $\mathrm{R}$, right? Meaning the 'amplitude' of the current in $\mathrm{L}$ is a little less. But how would I get the value of $I_\mathrm{R}$? I don't have a value for the voltage $V$.

Sorry for my lack of work here. My knowledge on curcuits in general is really slim.

Answer 1

$$ c\dfrac{di}{dt}+\frac{v}{r}+\frac{1}{l}\int_0^tvdt=I_{t}\\$$or

by derivation we have $$c\dfrac{d^2i}{dt^2}+\frac{1}{r}\frac{dv}{dt}+\frac{1}{l}v=\frac{dI_{t} }{dt}$$ then you can put numbers , and solve diff equation

Answer 2

Write a single node equation in terms of the impedances of L and C and the resistance R.

$$\frac{V}{R} + \frac{V}{j \omega L} + \frac{V}{\frac{-j}{\omega C}} = I_0 \angle \phi$$

where j is i for math people. V is a phasor (complex number) like $I_0 \angle \phi$ which represents the amplitude and phase of the voltage across all 3 components which will depend on R,L,C, and $\omega$. $j \omega L$ is the inductive impedance, and $\frac{-j}{\omega C}$ is the capacitive impedance. Recall that the angle of a rational expression is the difference in angle between the numerator and the angle of the denominator, and the angle of $R+jX$ is $tan^{-1}(X/R)$. The magnitude is the magnitude of the numerator divided by the magnitude of the denominator, and the magnitude of $R+jX$ is $\sqrt{R^2 + X^2}$.

The magnitude and phase of the current through L is just

$$I_L = \frac{V}{j \omega L}$$

so the magnitude scales by $\omega L$, the phase shifts by 90 degrees relative to V, and of course it is at the same frequency $\omega$.