A parallel LC-circuit is connected to an AC-supply as in the figure below.
\$I_{tot}(t)=I_0sin(\omega t+\phi)\$, \$\phi\$ is the phase angle between \$V_{tot}(t)\$ and \$I_{tot}(t)\$
a) Determine \$\phi\$.
b) What current \$I_L(t)\$(Amplitude and phase) runs through the coil L?
Use the following information: \$R=10 \Omega, ~C=30\mu F,~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 \$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 R, right? Meaning the 'amplitude' of the current in L is a little less. But how would I get the value of \$I_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.
Best Answer
$$\begin{align} Z_R&=10 \;\Omega\\ Z_L&=j\omega L=j \cdot300\times10^{-3}=0.03j\\ Z_C&=\frac{1}{j\omega c}=-0.009j\\ \hline\\ Z_{tot}&=Z_R||Z_L||Z_C=\left({1\over10}+{1\over0.03j}-{1\over0.009j}\right)^{-1}\approx 8.7378 \angle 89.95^{\circ}\;\Omega \longrightarrow \phi=89.95^{\circ}\\ \hline\\ &\text{Using current divider theorem*, } I_L=\frac{Z_{{R||}{C}}}{Z_L||Z_{R||C}}\times I_{tot}=\\ &=\frac{{\left({1\over10}-{1\over0.009j}\right)}^{-1}}{{\left({1\over10}-{1\over0.009j}\right)}^{-1}+0.03j}\times 2\angle\phi=(2\times0.428571..)\angle(\phi-179.926..)\approx 0.857\angle {-89.98}^{\circ} \text{A} \equiv 0.857\angle{-1.57}\;\text{A}\\ &\\ &\text{Hence, } I_L(t)=0.857\sin(\omega t-1.57) \;\text{A} \Longrightarrow \text{Amplitude}=0.857 \;; \text{Phase}=-1.57\; \text{radians} \end{align}$$ *:current divider theorem