Chapter 19, Problem 69GP

To determine

The resistance of the pacemaker.

Answer to Problem 69GP

The resistance of the pacemaker is $1.11\times {10}^5\Omega$ .

Explanation of Solution

Given info:

The capacitance of the capacitor in pacemaker is, $C=7.5\text{μF}=7.5\times {10}^{-6}\text{F}$ .

The beat rate of the pacemaker is, $t=72\text{beats}/\min =\frac{72\text{beats}}{\left(1\min \times \frac{60\text{s}}{1\min }\right)}=1.2{\text{s}}^{-1}$ .

The final voltage of the pacemaker is, $V=0.63V_0$ .

Here, $V_0$ is the initial voltage of the pacemaker.

Formula Used:

The expression to calculate the time constant of the pacemaker is,

  $\tau =\frac{1}{t}$

The expression to calculate the resistance of the pacemaker is,

  $R=\frac{\tau }{C}$

Calculation:

Substitute the values in the above expression.

  $\begin{array}{c}\tau =\frac{1}{1.2{\text{s}}^{-1}}\\ =0.83\text{s}\end{array}$

Substitute the values in the above expression.

  $\begin{array}{c}R=\frac{0.83\text{s}}{7.5\times {10}^{-6}\text{F}}\\ =1.11\times {10}^5\Omega \end{array}$

Conclusion:

Thus, the resistance of the pacemaker is $1.11\times {10}^5\Omega$ .