# Second order series circuit initial voltage

1. May 15, 2012

### stef6987

1. The problem statement, all variables and given/known data
How to find the initial capacitor voltage if the capacitor is in series with a 1.5 ohm resistor, these 2 are in parallel with a reversed polarity 6v DC source, (reversed polarity - the negative terminal is pointing up) i assume the voltage across the resistor and capacitor should be -6 volts because of the polarity of the source, the switch is opened at t = 0 nd the 6volt source is disconnected leaving a second order series circuit with another 12v source, another 6ohm resistor and 1.25H inductor (these 3 are in series with the 6v source at t<0 due to the open circuiting circuiting of the capacitor no current goes through the 1.5ohm and capacitor) when t<0 i could find the initial current from simple kvl: -6 - 12 + 6i = 0 ( the inductor is short circuited) hence i(0-) = i(0+) = 3 amps. for t>0 we hve the series second order circuit. check my solutions to see if im correct.

2. Relevant equations

3. The attempt at a solution

i(0+) = 3amps, if im right the voltage accross the 1.5 ohm resistor and capacitor is -6 volts or 6volts? we can find the voltage over the resistor using V=ir = 3*1.5=4.5 volts and subtract this from the -6 or 6 volts? and that would be the initial capacitor voltage, am i correct? thankyou

Last edited: May 15, 2012
2. May 15, 2012

### Staff: Mentor

I'm not "getting" your description of the circuit layout. Would it be possible to post a diagram?

3. May 15, 2012

### stef6987

I'll have to get 10 posts first! haha but i will do that ASAP. sorry for the bad explanation

4. May 15, 2012

### stef6987

ok here we go, heres the circuit, i need to find the initial capacitor voltage.

i found i(0-) to be 3Amps (correct me if im wrong) so when the switch is opened we have a second order RLC circuit, but i'm not sure how to find the initial capacitor voltage, can we say the voltage across the 1.5ohm resistor and capacitor is -6 ohms?

5. May 16, 2012

### Staff: Mentor

No, but you can say that it's -6 Volts

When the capacitor potential is -6V then there will be no current through its series resistor because the potential at both of its ends will be the same. That is, the branch is in equilibrium when the capacitor potential matches that of the potential applied across the branch.

6. May 16, 2012

### stef6987

so you saying that Vc(0-) = -6V? Is this beacuse their is no current flowing through the 1.5ohm resistor, and at t<0 the capacitor is open circuited, so the initial voltage at the capacitor must be -6 which is equal to the branch with the switch? my tutor today said it was 0 volts :\ but to be quite honest i don't think he had a great idea haha.

7. May 16, 2012

### Staff: Mentor

Yes. If the switch is closed (for a long time) before t = 0, then the capacitor must be charged to the potential across its branch.

8. May 16, 2012

### stef6987

thanks alot! simple as it seems, i thought that the potential drop across the 1.5ohm and capacitor would total -6volts, but infact the drop across the capacitor is the total drop across the branch, i didn't think bout it like that hehe thanks alot!