A heart pacemaker consists of a switch, a battery of constant voltage E0 , a capacitor with constant capacitance C

A heart pacemaker consists of a switch, a battery of constant voltage E₀ , a capacitor with constant capacitance C, and the heart as a resistor with constant resistance R. When the switch is closed, the capacitor charges; when the switch is open, the capacitor discharges, sending an electrical stimulus to the heart.During the time the heart is being stimulated, the voltage E across the heart satisfies the linear differential equation

$\f[\frac{dE}{dt}=-\frac{1}{RC}E\f]$

Solve the DE, subject to E(4) =E ₀ .

I’ll start attacking this problem by observing that the given differential equation is separable. So i’ll do the separation as follows:

$\f[\frac{dE}{E}=-\frac{1}{RC}dt\f]$ .

I’ll then integrate both sides of the equation as follows:

$\f[\int \frac{dE}{E}=-\int \frac{1}{RC}dt\f]$

On integration we get:

$\f[ln E=-\frac{t}{RC}+k\f]$

Here , k is a constant of integration. I’ll then make E(t) the subject of the formula by introducing an exponential on both sides, so that:

$\f[E(t)=e^{-\frac{t}{RC}+k}\f]$

The exponential on the right side of the equation can be broken down into two terms using the laws of indices as follows:

$\f[e^{-\frac{t}{RC}+k}=e^{-\frac{t}{RC}}*e^{k}\f]$

But e^k is just a constant. Therefore ,we can rewrite the above equation as follows:

$\f[e^{-\frac{t}{RC}}*e^{k}=Ae^{-\frac{t}{RC}}\f]$

Here,

$\f[e^{k}=A\f]$

The expression for E(t) then becomes;

$\f[E(t)=Ae^{-\frac{t}{RC}}\f]$

We are given the following boundary condition: E(4) =E ₀. Plugging this into E(t),

$\f[E_{0}=Ae^{-\frac{4}{RC}}\f]$

Solving for A we get:

$\f[A=E_{0}e^{\frac{4}{RC}}\f]$

Therefore E(t) becomes:

$\f[E(t)=E_{0}e^{\frac{4}{RC}}*e^{-\frac{t}{RC}}\f]$

Merging the two exponential terms ,we get the following:

$\f[E(t)=E_{0}e^{-\frac{1}{RC}(t-4)}\f]$

That is the solution for E(t) in terms of the given constants and time t. The negative sign on the exponential term indicates a ‘decay’ in the voltage when the capacitor is discharging.

A heart pacemaker consists of a switch, a battery of constant voltage E0 , a capacitor with constant capacitance C

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