Calculating Time: RC Circuit in a Heart Defibrillator

Since the resistance and capacitance are given, it is straightforward to multiply them to give the time constant asked for in part (a). To find the time for the voltage to decline to $5.00\times {10}^2\text{V}$, we repeatedly multiply the initial voltage by 0.368 until a voltage less than or equal to $5.00\times {10}^2\text{V}$ is obtained. Each multiplication corresponds to a time of $\tau$ seconds.

The time constant $\tau$ is given by the equation $\tau =\text{RC}$. Entering the given values for resistance and capacitance (and remembering that units for a farad can be expressed as $\mathrm{s}/\Omega$) gives

In the first 8.00 ms, the voltage (10.0 kV) declines to 0.368 of its initial value. That is:

(Notice that we carry an extra digit for each intermediate calculation.) After another 8.00 ms, we multiply by 0.368 again, and the voltage is

Similarly, after another 8.00 ms, the voltage is

So after only 24.0 ms, the voltage is down to 498 V, or 4.98% of its original value.Such brief times are useful in heart defibrillation, because the brief but intense current causes a brief but effective contraction of the heart. The actual circuit in a heart defibrillator is slightly more complex than the one in Figure 21.39, to compensate for magnetic and AC effects that will be covered in Magnetism.