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Chapter 21: Q32CQ (page 773)

When making an ECG measurement, it is important to measure voltage variations over small time intervals. The time is limited by the \(RC\) constant of the circuit-it is not possible to measure time variations shorter than \(RC\). How would you manipulate \(R\) and \(C\) in the circuit to allow necessary measurements?

Short Answer

Expert verified

Putting resistances in parallel and capacitors in series.

Step by step solution

01

Discharging of a capacitor

The capacitor discharges if the plates of a charged capacitor are connected by a conducting wire. Once more, there is a current because there is a movement of charge across the wires. \({\bf{q}} = {\bf{QeCR}} - {\bf{t}}\).

02

Explanation

For measuring smaller time periods, \(R\)and \(C\) should be made smaller. This can be done by putting resistances in parallel and capacitors in series. This will reduce the effective time period of the circuit.

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Most popular questions from this chapter

An ECG monitor must have an \(RC\) time constant less than \(1.00 \times {10^2}{\rm{ }}\mu s\) 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鈥檚 chest) is\(1.00{\rm{ }}k\Omega \), 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)?

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