Solving Circuits: Charge for Defibrilator w/ 6000V Capacitor

In summary, the defibrilator discharges a current through the body of a patient. The current is determined by the initial charge on the capacitor and the time interval during which the current is being discharged.
  • #1
Benny
584
0
I'm having trouble with the following question, can someone please help me with it?

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.

[tex]
L\frac{{d^2 q}}{{dt^2 }} + R\frac{{dq}}{{dt}} + \frac{q}{C} = 0
[/tex]

Using the given values I obtain:

[tex]
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
[/tex]

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

[tex]
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}
[/tex]

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 [tex]c_1 + c_2 = \frac{{24}}{{125}}...(1)[/tex].

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:

[tex]
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)
[/tex]

I get: [tex]c_1 = \frac{{36 - 63\sqrt {\frac{3}{2}} }}{{375}}[/tex]

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

From this I get:

[tex]
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}
[/tex]

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.
 
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  • #2
Benny said:
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 [tex]c_1 + c_2 = \frac{{24}}{{125}}...(1)[/tex].

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.

I don't think I even applied that 'initial condition' correctly.

You did apply it 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.

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).

The other problem I'm having is that I don't understand whether this is a boundary value problem or IVP.

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.

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?

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.

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.

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

What is a defibrillator and how does it work?

A defibrillator is a medical device commonly used in emergency situations to deliver an electric shock to the heart in order to restore normal heart rhythm. It works by sending an electric current through the body, which depolarizes the heart muscles and allows the pacemaker cells to resume their normal function.

What is a capacitor and why is it important in a defibrillator?

A capacitor is an electrical component that stores and releases a large amount of electrical energy. In a defibrillator, the capacitor is used to store the 6000V charge needed to deliver the electric shock to the heart. Without the capacitor, the defibrillator would not be able to provide the necessary energy to restart the heart.

How is the charge for a defibrillator determined?

The charge for a defibrillator is determined by several factors, including the type and severity of the heart rhythm abnormality, the age and weight of the patient, and the type of defibrillator being used. The specific charge for each patient is determined by a trained medical professional using established protocols.

What are the potential risks of using a defibrillator?

While defibrillators are generally safe and effective in restoring normal heart rhythm, there are some potential risks associated with their use. These may include burns or damage to the skin where the electrodes are placed, and potential harm to bystanders if they come into contact with the patient during the shock. However, the benefits of using a defibrillator in emergency situations far outweigh the potential risks.

How can the charge for a defibrillator be calculated in a circuit?

The charge for a defibrillator in a circuit can be calculated using the formula Q=CV, where Q is the charge in coulombs, C is the capacitance of the capacitor in farads, and V is the voltage of the capacitor in volts. In the case of a defibrillator with a 6000V capacitor, the charge would be 6000 times the capacitance of the capacitor in farads.

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