# Circuits - charge

1. Oct 5, 2005

### Benny

A defibrilator discharges a current through the body of a patient. It consists of an open circuit containing a capacitor of 32 microfarads, an inductor of 0.05H with resistance of 50 ohms, and the patient has a resistance of 50 ohms when the device is discharged through them. Initially the capacitor is charged to 6000V, Find the initial charge on the capacitor, the current during discharge.

$$L\frac{{d^2 q}}{{dt^2 }} + R\frac{{dq}}{{dt}} + \frac{q}{C} = 0$$

Using the given values I obtain:

$$0.05\frac{{d^2 q}}{{dt^2 }} + 100\frac{{dq}}{{dt}} + \frac{q}{{32 \times 10^{ - 6} }} = 0 \to \frac{{d^2 q}}{{dt^2 }} + 2000\frac{{dq}}{{dt}} + 625000q = 0$$

Using the quadratic formula on the characteristic equation gives me two real roots and I obtain the general solution as:

$$q\left( t \right) = c_1 e^{\left( { - 1000 - 500\sqrt {\frac{3}{2}} } \right)t} + c_2 e^{\left( { - 1000 + 500\sqrt {\frac{3}{2}} } \right)t}$$

The numbers already look difficult to deal with so I suspect that there might be an error in my working but I haven't been able to pick one out yet. I have two undetermined constants but I can only extract one initial condition from the stem of the question, q = CV so q(0) = 32 microfarads * 6000 = (24/125). So one equation is $$c_1 + c_2 = \frac{{24}}{{125}}...(1)$$.

I don't think I even applied that 'initial condition' correctly. I often have trouble extracting relevant parts of wordy problems. :uhh: I suspect that I might have used an incorrect 'initial condition' because from what I've just done, I could have obtained the initial charge without even solving the DE. The other problem I'm having is that I don't understand whether this is a boundary value problem or IVP. I don't see anything I can apply to the derivative of q(t), q'(t). Perhaps q'(0) = I(0) = V/R = 6000/100 = 60?

In that case I would have:

$$60 = c_1 \left( { - 1000 - 500\sqrt {\frac{3}{2}} } \right) + \left( {\frac{{24}}{{125}} - c_1 } \right)\left( { - 1000 + 500\sqrt {\frac{3}{2}} } \right)$$

I get: $$c_1 = \frac{{36 - 63\sqrt {\frac{3}{2}} }}{{375}}$$

So $$c_2 = \frac{{24}}{{125}} - c_1 \to c_2 = \frac{{36 + 63\sqrt {\frac{3}{2}} }}{{375}}$$.

From this I get:

$$q\left( t \right) = \left( {\frac{{36 - 63\sqrt {\frac{3}{2}} }}{{375}}} \right)e^{\left( { - 1000 - 500\sqrt {\frac{3}{2}} } \right)t} + \left( {\frac{{36 + 63\sqrt {\frac{3}{2}} }}{{375}}} \right)e^{\left( { - 1000 + 500\sqrt {\frac{3}{2}} } \right)t}$$

This is just my working to show that I've done something. The numbers in the questions we get are normally fudged so that we get 'reasonable' numbers to work with so I'm pretty sure that I went wrong with the initial conditions or something earlier on.

Also the question also asks for a sketch of the current curve "especially for the first 8ms." Is there some kind of significance associated with that time interval? I just can't see it. This whole question is confusing me. Any help with this question would be great thanks.

Last edited: Oct 5, 2005
2. Oct 5, 2005

### Tom Mattson

Staff Emeritus
For a second initial condition, ask yourself what the current is before the paddles of the defibrilator touch the patient. Remember that the patient completes the circuit, so that prior to contact the circuit is open.

You did apply it correctly.

Of course you could have determined the initial charge without solving the DE. In fact you cannot solve the IVP without knowing the initial charge, and one other piece of data (such as the initial current).

It's an IVP simply because you're working in the time domain, and so the values your conditions at t=0 are "initial" values.

No, as I said earlier the problem statement tells you that the circuit is initially open. You're just throwing numbers together here, and it's not correct.

I can't tell, because I haven't verified your numbers (don't have a calculator on me). But your procedure looks correct, except for your determination of the initial current.

3. Oct 6, 2005

### Benny

Ok I think I can see what is needed now. Thanks for the help.