4.6 Dc circuits containing resistors and capacitors  (Page 4/9)

Calculating time: RC Circuit in a heart defibrillator

A heart defibrillator is used to resuscitate an accident victim by discharging a capacitor through the trunk of her body. A simplified version of the circuit is seen in [link] . (a) What is the time constant if an $\mathrm{8.00-\mu F}$ capacitor is used and the path resistance through her body is $\text{1.00}\times {10}^3\Omega$ ? (b) If the initial voltage is 10.0 kV, how long does it take to decline to $5.00\times {10}^2\text{V}$ ?

Strategy

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.

Solution for (a)

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

Solution for (b)

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

Discussion

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 [link] , to compensate for magnetic and AC effects that will be covered in Magnetism .