Question

An ECG monitor must have an RC time constant less than $1.00\times 10^{2}\textrm{ }\mu\textrm{s}$ to be able to measure variations in
voltage over small time intervals. (a) If the resistance of the
circuit (due mostly to that of the patient’s chest) is $1.00
\textrm{ k}\Omega$ ,
what is the maximum capacitance of the circuit? (b) Would it be difficult in practice to limit the capacitance to less than the value found in (a)?

Final Answer

- $1.00\times 10^{-7}\textrm{ F}$
- It would not be difficult to create a circuit such as this. Typical capacitors have capacitances measured in nanofarads up to milifarads.

### Solution video

# OpenStax College Physics for AP® Courses, Chapter 21, Problem 70 (Problems & Exercises)

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Video Transcript

This is College Physics Answers with Shaun Dychko. The electrocardiogram monitor must have an

*RC*time constant less than 1.00 times 10 to the 2 microseconds so in other words, the maximum possible time constant is 1.00 times 10 to the 2 times 10 to the minus 6 seconds, which is 1.00 times 10 to the minus 4 seconds; substituting 10 to the minus 6 in place of the prefix 'micro' and then 10 to the minus 6 times 10 to the 2 is minus 10 to the 4. Okay! We need to figure out what the capacitance will be given that the resistance of the patient's chest is 1 kiloohms. So the maximum time constant then is that resistance multiplied by the maximum possible capacitance and we will divide both sides by*R*to solve for*C max*. So maximum capacitance is this maximum time constant divided by resistance 1.00 times 10 to the minus 4 seconds divided by 1.00 times 10 to the 3 ohms and that is 1.00 times 10 to the minus 7 farads. Now part (b) asks is it practical to have a capacitance so small and in fact it is! Capacitance is typically measured in nanofarads upto millifarads so this is 100 nanofarads and well within you know off the shelf range of possible capacitors.