Question

A $150\mu F$defibrillator capacitor is charged to $1500V$. When fired through a patient’s chest, it loses$95\%$ of its charge in $40ms$What is the resistance of the patient’s chest?

Short answer

$R=89\Omega$

Step-by-step solution

Given information.

We need to find the resistance of the patient's chest.

Simplification.

When the switch is off, the capacitor discharge and the current flows through the resistor. To discharge the capacitor, the charges flow through the circuit in time . the initial charge is related to final charge Q by equation
in the form

$Q=Q_0{e}^{-t}/RC$ Where $R$is the resistance and t.$C$is the capacitance. Let us solve equation(1) for R to get it by

$R=\frac{-t}{CIn(Q/Q_0)}$

The capacitor losses $95\%$of its charge, so the final charge is $5\%$of the initial one $Q=0.05Q_{0,}$so we plug the values for $t$.localid="1649245614284" $C$and$Q$into equation to get $R$

$$\begin{gathered} R=\frac{-t}{CIn(Q/Q_0)} \\ =\frac{-40\times {10}^{-3_s}}{(150\times {10}^{-6}F)\left(In\right(0.05\left)\right)} \\ =89\Omega \end{gathered}$$