Question

A defibrillator delivers a jolt of electrical energy to restart a heart that is either stopped or undergoing ventricular defibrillation -- a rapid irregular beating. This energy is supplied by a charged capacitor and applied via paddles placed on a patient's body(one on each side of the heart). Consider a 20.7 μF capacitor charged to 2210 V in a defibrillator and used on a patient. After 9.85×10^-2 seconds the voltage across the defibrillator's paddles drops to 49.6 V.

Find the effective resistance of the body between the defibrillators paddles. Hint: assume the voltage across the paddles is the same as the voltage stored on the capacitor.

Answer 1

First we will discuss about the discharging of capacitor:

We will assume that, initially the capacitor is charged and having a total of Q charge on it (i.e. at t = 0 seconds). But when we close the switch 'S' the capacitor will start discharging due to the load 'R' connected in series. Now, after some time 't' the capacitor will only have 'q' charge left on it.

Now applying Kirchhoff's voltage law in the circuit, we get

................. .....(1)

(Where C is the capacitance of the capacitor)

And (Charge is decreasing with respect to time so the current is shown with negative sign.)

Putting value of 'I' in equation (1), we get

  

or,

or,   

or,

or,   

or,

or, .....................................(2)

From equation (2) we can calculate the potential difference across the two plates of capacitor after some time 't'.

Dividing both sides of equation (2) by 'C' , we get

or,    ........................(3)

(where V₀ is the potential difference across the two plates of capacitor at t = 0 seconds i.e. initially)

After rearranging we can write equation (3) as

or,

or, .............................(4)

According to the question,   , , and

Putting these values in equation (4), we get

  

or,

or,

or,

or,   

Hence, the effective resistance of the body between the defibrillators is nearly .