Compute the capacitance of the capacitor in the defibrillator and calculate the magnitude of the average
current flowing during the pulse.
Hint: The relation between the stored charge (Q), and the voltage across the capacitor is given by
In a charged capacitor, positive charges are on one side of the plate, and negative charges are on the other.
The amount of energy (E) stored in such a configuration is given by
Currents in the range of a few amperes flowing in the region of the heart can cause death within a few minutes. In this
connection, a large current of about 10 A is often less dangerous than a 1-A current. When the smaller current passes through the heart,
it may tetanize only part of the heart, thereby causing a desynchronization of the heart action; this condition is called fibrillation. The
movements of the heart become erratic and ineffective in pumping blood. Usually fibrillation does not stop when the current source is
removed. A large current tetanizes the whole heart, and when the current is discontinued the heart
may resume its normal rhythmic activity.
Fibrillations often occur during a heart attack and during cardiac surgery. The tetanizing effect of large currents can be used to
synchronize the heart. A clinical device designed for this purpose is called a defibrillator. A capacitor in this device is charged to about
6000 V and stores about 200 J of energy. Two
electrodes connected to the capacitor through a switch are placed on the chest. When the switch is closed, the capacitor rapidly
discharges through the body. The current pulse lasts about 5 msec, during which the heart is tetanized. After the pulse, the heart may
resume its normal beat. Often the heart must be shocked a few times before it resynchronizes
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