# How to find general solution of an L-R-C series circuit with inductance L = 1/5 henry, resistance R=4 ohms, and capacitance c=1/520 farad?

Consider an L-R-C series circuit with inductance L = 1/5 henry, resistance R=4 ohms, and capacitance c=1/520 farad. Find the general form of the charge on the capacitor if this is a free series circuit.

I got q(t) = e^(-10t)(c1(cos(50t) + c2(sin(50t)), but the answer that my friend's teacher provided is q(t)=e^(-3t)(c1(cos(50t)+c2(sin(50t)). Who is correct?

For the R-L-C circuit, using KVL, the voltage across the capacitor is given by:

$$LC \dfrac{d^2V_c}{dt^2} + RC \dfrac{dV_c}{dt} + V_c(t) = 0$$

This gives us:

$$\dfrac{1}{2600} \dfrac{d^2V_c}{dt^2} + \dfrac{4}{520} \dfrac{dV_c}{dt} + V_c(t) = 0$$

Solving this yields:

$$V_c(t) = e^{-10t} \left(c_1 \cos(50 t) + c_2 \sin(50 t) \right)$$

• Got it. Thank you so much!! – Joshua Ree Nov 20 '13 at 7:13
• You are very welcome. I wondered if somehow your friend messed up one of the parameters or signs. Regards – Amzoti Nov 20 '13 at 7:15
• @amWhy: I have seen that too. I never downvote as a general rule of thumb. My downvotes (small number) were due mostly to spam, but there were some offensive posts I DV too). I wish people would at least give a reason for the DV. How is anyone supposed to understand an objection otherwise. Regardless, you will soon reach an awesome milestone! Thanks for the support! – Amzoti Nov 20 '13 at 15:44
• Thanks my dear friend for your support. My writing is not good to be involved in a discussion like you have here. Many thanks my dear Amzoti. :-) – mrs Nov 21 '13 at 5:48
• @B.S.: You are very welcome my friend! When people pick on my friends here and say their solution is wrong, it gets me fired up! :-) – Amzoti Nov 21 '13 at 7:44