Question

A 100 μF defibrillator capacitor is charged to 1500 V. When fired through a patient's chest, it loses 95% of its charge in 40 ms.
What is the resistance of the patient's chest?   

Answer 1

Concepts and reason The concept used to solve this problem is the charge in a discharging capacitor.

First, use the expression for charge in a capacitor at a time for discharging to obtain an expression for resistance.

Finally, use the obtained expression for resistance to find the resistance of the patient's chest.

Fundamentals For discharging, the charge is given by the following expression:

\(Q_{\mathrm{r}}=Q_{i} e^{-t / R C}\)

Here, the final charge is \(Q_{\mathrm{r}}\), the initial charge is \(Q_{i}\), the time taken is \(t\), the resistance is \(R\), and the capacitance is \(C\).

For discharging, the charge is given by the following expression:

\(Q_{\mathrm{r}}=Q_{i} e^{-t / R C}\)

\(\frac{Q_{\mathrm{r}}}{Q_{\mathrm{i}}}=e^{-t / R C}\)

Take \(\ln\) on both sides

\(\ln \left(\frac{Q_{\mathrm{r}}}{Q_{\mathrm{i}}}\right)=\ln \left(e^{-t / R C}\right)\)

\(=\frac{-t}{R C}\)

Rearrange the above expression to obtain resistance.

\(R=\frac{-t}{\ln \left(\frac{Q_{\mathrm{f}}}{Q_{\mathrm{i}}}\right) C}\)

\(95 \%\) of the charge in the capacitor is discharged. Hence, the following expression is obtained:

\(Q_{\mathrm{r}}=5 \% Q_{\mathrm{i}}\)

\(=\frac{5}{100} Q_{i}\)

\(=0.05 Q\)

The resistance is given by the following expression:

\(R=\frac{-t}{\ln \left(\frac{Q_{\mathrm{f}}}{Q_{\mathrm{i}}}\right) C}\)

Substitute \(40 \mathrm{~ms}\) for \(t, 0.05 Q\), for \(Q_{r}\), and \(100 \mu \mathrm{F}\) for \(C\)

\(R=\frac{-\left(40 \mathrm{~ms}\left(\frac{10^{-3} \mathrm{~s}}{1 \mathrm{~ms}}\right)\right)}{\ln \left(\frac{0.05 Q_{i}}{Q_{i}}\right)\left(100 \mu \mathrm{F}\left(\frac{10^{-6} \mathrm{~F}}{1 \mu \mathrm{F}}\right)\right)}\)

\(=133.5 \Omega\)

Ans:

The resistance of the patient's chest is 13