You're designing an external defibrillator that discharges a capacitor through the patient's body, providing a pulse that stops ventricular fibrillation. Specifications call for a capacitor storing 250 J of energy; when discharged through a body with \(40-\Omega\) transthoracic resistance, the capacitor voltage is to drop to half its initial value in \(10 \mathrm{ms}\). Determine the capacitance (to the nearest \(10 \mu \mathrm{F})\) and initial capacitor voltage (to the nearest \(100 \mathrm{V}\) ) that meet these specs.

Short Answer

Expert verified
The nearest values of capacitance \(C\) and initial capacitor voltage \(V_0\) that will meet the specs can be calculated by using the derived equations and solving for \(V_0\) and \(C\) respectively.

Step by step solution

01

Calculate the Capacitance

The formula for energy stored in a capacitor is given by \(E = 0.5 * C * V^2\), where \(E\) is the energy, \(C\) is the capacitance and \(V\) is the voltage.\From the problem, we know that \(E = 250J\). Let \(V = V_0\) (the initial voltage). We can rearrange this formula to solve for \(C\): \(C = (2*E) / V^2\).
02

Voltage Decrease Over Time

It is given that the capacitor voltage drops to half its initial value in \(10ms\). Due to the resistor, the discharge of the capacitor follows an exponential decay governed by the formula \( V = V_0 * e^{(-t/RC)}\), where \(R = 40\Omega \) (resistance), \(t = 10ms = 10 * 10^{-3} s\) (time), and \(C\) is the capacitance.\Setting \( V = 0.5 * V_0\) (since the voltage drops to half its initial value), we can solve for \(C\): \(C = -t / (R * ln(0.5))\).
03

Equate and Solve

In order to find a working set of values for \(C\) and \(V_0\), we can equate the two equations we derived for \(C\) and solve for \(V_0\):\i.e. \(2*E / V_0^2 = -t / (R * ln(0.5))\). Solve this to get \(V_0\). Replace this value in either of the equations for \(C\) to get the value of \(C\).

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