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:

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