Question

problem 19.53
electrocardiographs are often connected as shown in (figure 1). the lead wires to the legs are said to be capacitively coupled. a time - constant of 3.0 s is typical and allows rapid changes in potential to be recorded accurately.
part a
if (c = 3.5 mu f), what value must (r) have? hint: consider each leg as a separate circuit.
express your answer to two significant figures and include the appropriate units.

Explanation:

Step1: Recall the formula for time - constant

The time - constant $\tau$ of an RC circuit is given by $\tau = RC$.

Step2: Rearrange the formula to solve for R

We can rewrite the formula as $R=\frac{\tau}{C}$.

Step3: Substitute the given values

We know that $\tau = 3.0\ s$ and $C = 3.5\times10^{- 6}\ F$. So $R=\frac{3.0}{3.5\times10^{-6}}$.

Step4: Calculate the value of R

$R=\frac{3.0}{3.5\times10^{-6}}\approx8.6\times10^{5}\ \Omega$.

Answer:

$8.6\times10^{5}\ \Omega$