0
\$\begingroup\$

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.

\$\endgroup\$
0
\$\begingroup\$

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

\$\endgroup\$
  • \$\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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.