A parallel LC-circuit is connected to an AC-supply as in the figure below.

enter image description here

\$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.


$$\begin{align} Z_R&=10 \;\Omega\\ Z_L&=j\omega L=j \cdot300\times10^{-3}=0.03j\\ Z_C&=\frac{1}{j\omega c}=-0.009j\\ \hline\\ Z_{tot}&=Z_R||Z_L||Z_C=\left({1\over10}+{1\over0.03j}-{1\over0.009j}\right)^{-1}\approx 8.7378 \angle 89.95^{\circ}\;\Omega \longrightarrow \phi=89.95^{\circ}\\ \hline\\ &\text{Using current divider theorem*, } I_L=\frac{Z_{{R||}{C}}}{Z_L||Z_{R||C}}\times I_{tot}=\\ &=\frac{{\left({1\over10}-{1\over0.009j}\right)}^{-1}}{{\left({1\over10}-{1\over0.009j}\right)}^{-1}+0.03j}\times 2\angle\phi=(2\times0.428571..)\angle(\phi-179.926..)\approx 0.857\angle {-89.98}^{\circ} \text{A} \equiv 0.857\angle{-1.57}\;\text{A}\\ &\\ &\text{Hence, } I_L(t)=0.857\sin(\omega t-1.57) \;\text{A} \Longrightarrow \text{Amplitude}=0.857 \;; \text{Phase}=-1.57\; \text{radians} \end{align}$$ *:current divider theorem

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  • \$\begingroup\$ I'm not yet familiar with impedances and how to get to the angles from that. I'm not familiar with the notation yet, is 8.7378 another expression for 89.95°, or is 8.7378\$\angle\$89.95° the expression for the resistance? \$\endgroup\$ – Rixton Jun 21 '15 at 21:43
  • \$\begingroup\$ Oooh.. I didn't know that. Go through this and see if it makes the answer clear. \$\endgroup\$ – K. Rmth Jun 22 '15 at 7:43
  • \$\begingroup\$ Okay, I just read through it and to be honest I'm still not that much more knowledgeable with the notation. Could you make an example with \$Z_{tot}\$? I think an example would make me understand things easier. I also don't know how to compute \$(\frac{1}{10}+\frac{1}{0.03j}-\frac{1}{0.009j})^{-1}\$. It's hard to understand how to do it with the j. I mean I know it's the mathematical equivalent to i, but still. \$\endgroup\$ – Rixton Jun 22 '15 at 8:17
  • \$\begingroup\$ It will be kind of hard to explain that through comments. The following gives a good explanation of basics: complex number and phasors, ac resistance, ac inductance and ac capacitance. An example is given here. This should get you right on track ;) \$\endgroup\$ – K. Rmth Jun 22 '15 at 10:13

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