Differentials/ second order/ electrical vibration

[ttttrigg3r]

Homework Statement

A simple series circuit has an inductor of 1 henry, a capacitor of 10^-6 farads, and a resistor of 1000 ohms. The initial charge on the capacitor is zero. If a 12V battery is connected is connected to the circuit, and the circuit is closed at t=0, find the charge on the capacitor 1 second later and the steady state charge.

From this I know initial current =0

The Attempt at a Solution

Q''+1000Q'+10^6Q=0
Q(t)=C1(e^(-500t))cos(500sqrt(3)t)+C2(e^(-500t))sin(500sqrt(3)t)
I got that far. now if I used Q(0)=0 and Q'(0)=0 I get C1=C2=0 which I think is wrong. Can someone help? Thank you in advance.

 

[mighty2000]

Hello, thank you for your question. Based on the given information, I can help you find the charge on the capacitor after 1 second and the steady state charge.

First, we can use the formula Q=CV to find the initial charge on the capacitor. Since the initial voltage is 0, the initial charge on the capacitor will also be 0.

Next, we can use the formula Q(t)=Q0(e^(-t/RC)) to find the charge on the capacitor after 1 second. Plugging in the values, we get Q(1)=0(e^(-1/1000*10^-6))=0. This means that after 1 second, the charge on the capacitor will still be 0.

Finally, to find the steady state charge, we can use the formula Q=CV. Since the circuit is closed, the capacitor will eventually become fully charged and the current will stop flowing. This means that the steady state charge will be equal to the maximum charge that the capacitor can hold, which is given by Q=CV=12V*10^-6F=0.012C.

In summary, after 1 second, the charge on the capacitor will be 0 and the steady state charge will be 0.012C. I hope this helps! Let me know if you have any further questions.