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I am having trouble finding the differential equation of a mixed RLC-circuit, where C is parallel to RL. It is a steady-state sinusoidal AC circuit. I need it to determine the Power Factor explicitly as a function of the components. I know I am supposed to use the KCL or KVL, but I can't seem to derive the correct one.

I am actually a math student, so I apologize if this question is straight forward.

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  • \$\begingroup\$ Are you required to do it via differential equations or can you use frequency response (or Laplace) representations of the components? \$\endgroup\$ – Chu May 24 '16 at 0:18
  • \$\begingroup\$ Unfortunately I have to do it using differential equations. I can use $$u(t) = V_m cos(\omega t), i(t) = I_m cos(\omega t - \phi)$$, etc., or phasor representation. \$\endgroup\$ – user260710 May 24 '16 at 0:33
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The voltage, \$ v\$, across \$\small C\$ is equal to the the voltage across the \$\small R, L\$ series combination.

The current in \$\small C\$ is \$i_1=\small C\large \frac{dv}{dt}\$

The current in \$\small RL\$ is related to \$v\$ by; \$ v=\small R\large i_2+\small L\large\frac{di_2}{dt}\$, where \$i_2\$ is the current through \$\small R\$ and \$\small L\$.

Solve both ODEs for \$ i_1\$ and \$ i_2\$, and the total current in the \$\small RLC\$ circuit is then \$i=i_1 + i_2\$.

Let \$\small t\rightarrow \infty\$ in the real parts of the exponentials to remove any transients, and you now have \$ i\$ and \$v\$ in the required steady-state sinusoidal forms.

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  • \$\begingroup\$ But current resistor = current inductor? So the total current is $$i(t) = i_c + i_r$$ So why is the current in the RL-circuit related by that equation? I would think it is related by $$R i_2 = L \frac{di_2}{dt}$$ \$\endgroup\$ – user260710 May 24 '16 at 2:16
  • \$\begingroup\$ I assume your question means that L and R are in series, and this series combination is in parallel with C. The voltage across a series circuit is the sum of the voltages across each individual element. Try two resistors in series and apply Ohm's Law. \$\endgroup\$ – Chu May 24 '16 at 6:50
  • \$\begingroup\$ Oops yes that was what I meant. After you edited your original post I found the correct total current and PF. Thank you very much! \$\endgroup\$ – user260710 May 26 '16 at 21:30

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