Answer to Question #164687 in Electric Circuits for Elizabeth

Question

A heart defibrillator being used on a patient has an RC time constant of 10.0 ms due to the resistance of thepatient and the capacitance of the defibrillator. (a) If the defibrillator has a capacitance of 8.00μF, what is the resistance of the path through the patient?(You may neglect the capacitance of the patient and the resistance of thedefibrillator.) (b) If the initial voltage is 12.0 kV, how long does it take to decline to 6.00 × 10^2V?

Answer 1

$a)$

$\tau =RC=10.0\ ms=10\times 10^{-3}\ s=0.01 s$

$C=8.00\ \mu =8\times 10^{-6}F$

$R=\frac{0.01}{8\times 10^{--6}}=1250\ \Omega$

$b)$ $V_o=12\ KV=12,000V;V=600V$

While discharging,

$V=V_o e^{-\frac{t}{\tau}}$

$\implies 600=12000 e^{-\frac{t}{0.01}}$

$\implies t=0.03 \ s$