## Chapter 5

## Q. 5.7

**Defibrillator**

After being shocked by the power line, the careless worker from Example 5.6 lies motionless on the ground. “Ma, his heart ain’t beating. Get my defibrillator,” yells the farmer, while attempting cardiopulmonary resuscitation (CPR) on him.

A defibrillator contains a large capacitor that is charged to a high voltage. This voltage is discharged when the pads are pressed on the patient’s chest. The scenario shown in Figure 5.22 can be modeled as that shown in Figure 5.23. Note that before t = 0, S1 and S2 have been closed and opened, respectively, for a long time period. Usually, defibrillators have an inductor in series with the patient, but it has been omitted to simplify this problem.

a. The farmer shocks the worker across his chest with the defibrillator. What is the current through the worker’s chest at t = 0?

b. How long does it take until the voltage across his chest drops to v_o(t) =\frac{1}{2} V_{ in}?

## Step-by-Step

## Verified Solution

a. At t = 0, the capacitor has been fully charged to v_o(0)=5 kV. Thus, I=v_o(0) / R=5000 / 50,000=0.1 A

b. Applying KCL to the parallel capacitor-resistor (i.e., Node A in Figure 5.23) for t > 0:

0=\frac{v_0(t)}{50 k \Omega}+25 \mu F \frac{dv_0(t)}{dt} \rightarrow 0=\frac{v_0(t)}{1.25}+\frac{dv_0(t)}{dt}or:

\frac{dv_0(t)}{v_0(t)}=\frac{-dt}{1.25}Integrating both sides of this equation yields:

\ln \frac{v_0(t)}{5000}=\frac{-t}{1.25}This leads to:

v_0(t)=5000 e^{-\frac{-t}{1.25}}Now, substituting v_{o}(t)=0.5 \times V_{in}=0.5 \times 5000=2500 V :

2500=5000 e^{\frac{-t}{1.25}}Next, find the corresponding time t required to drop to 2500 V, which is:

t = -1.25ln0.5 = 0.8664 s