A heart defibrillator being used on a patient has an RC time constant of $\mathbf{10}\mathbf{.}\mathbf{0}\mathbf{ms}$due to the resistance of the patient and the capacitance of the defibrillator. (a) If the defibrillator has an$\mathbf{8}\mathbf{.}\mathbf{00}\mathbf{-}\mathbf{\mu F}$capacitance, what is the resistance of the path through the patient? (You may neglect the capacitance of the patient and the resistance of the defibrillator.) (b) If the initial voltage is$\mathbf{12}\mathbf{.}\mathbf{0}\mathbf{KV}$, how long does it take to decline to$\mathbf{6}\mathbf{.}\mathbf{00}\mathbf{\times }\mathbf{102}\mathit{V}$?

Defibrillator: A device used in hospitals to control the movements of the heart muscles by delivering a controlled electric shock to the heart.

Capacitance: The proportion of a system's change in electric charge to its corresponding change in electric potential.

Voltage: A unit of measurement for electrical force (volts).

The formula for discharging capacitor,

$\mathrm{V}\left(\mathrm{t}\right)={\mathrm{V}}_{\mathrm{i}}{\mathrm{e}}^{-\mathrm{t}/\mathrm{\tau }}$

The formula for resistance,

$\tau =RC$

(a)

The defibrillator has a capacitance as, $\mathrm{C}=8\mathrm{\mu F}\left(\frac{{10}^{-6}\mathrm{F}}{1\mathrm{\mu F}}\right)=8\times {10}^{-6}\mathrm{F}$and time constant of $\tau =10.0ms\left(\frac{{10}^{-3}s}{1ms}\right)=1.00\times {10}^{-2}$the resistance and capacitance of the patient are ignored.

The patient's resistance can be calculated as,

$\tau =RC$

$$\begin{gathered} \mathrm{R}=\frac{\mathrm{\tau }}{\mathrm{C}} \\ =\frac{1.00\times {10}^{-2}\mathrm{s}}{8\times {10}^{-8}\mathrm{F}} \\ =1250\mathrm{\Omega }\left(\frac{1\mathrm{k\Omega }}{1000\mathrm{\Omega }}\right) \\ =1.25\mathrm{k\Omega } \end{gathered}$$

Therefore, the resistance of the path through the patient is$1.25k\Omega$.

(b)

Calculate the timerequired for the voltage to drop to$\mathrm{V}\left(\mathrm{t}=\mathrm{T}\right)=\mathrm{V},=600\mathrm{V}$.

If the initial voltage is:$V=12kV\left(\frac{1000V}{1kV}\right)=1.2\times {10}^{4}V$.

The formula for a discharging capacitor is used,

$$\begin{gathered} \mathrm{V}\left(\mathrm{t}\right)={\mathrm{V}}_{\mathrm{i}}{\mathrm{e}}^{-\mathrm{t}/\mathrm{\tau }} \\ {\mathrm{e}}^{-\mathrm{t}/\mathrm{\tau }}=\frac{{\mathrm{V}}_{\mathrm{f}}}{{\mathrm{V}}_{\mathrm{i}}} \\ -\frac{\mathrm{T}}{\mathrm{\tau }}=\ln \left(\frac{{\mathrm{V}}_{\mathrm{f}}}{{\mathrm{V}}_{\mathrm{i}}}\right) \\ \mathrm{T}=\mathrm{\tau ln}\left(\frac{{\mathrm{V}}_{\mathrm{i}}}{\mathrm{Vf}}\right) \end{gathered}$$

Substituting the given data in above expression, will give,

$$\begin{gathered} \mathrm{T}=\left(1.0\times {10}^{-2}\mathrm{s}\right)\times \ln \left(\frac{1.0\times {10}^{-4}\mathrm{s}}{600\mathrm{V}}\right) \\ =0.3\mathrm{s}\left(\frac{1000\mathrm{ms}}{1\mathrm{s}}\right) \\ =30\mathrm{ms} \end{gathered}$$

Hence, it takes $30\mathrm{ms}$to decline to$600V$.