/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 50 A heart pacemaker fires 72 times... [FREE SOLUTION] | 91Ó°ÊÓ

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A heart pacemaker fires 72 times a minute, each time a 25.0 -nF capacitor is charged (by a battery in series with a resistor) to 0.632 of its full voltage. What is the value of the resistance?

Short Answer

Expert verified
The value of the resistance in the heart pacemaker circuit is approximately \(5231.9\, \Omega\).

Step by step solution

01

Identify the relevant formula

The formula we need for capacitor charging through a resistor is \[V(t) = V_{max}(1 - e^{(-\frac{t}{RC})})\], where V(t) is the voltage across the capacitor at time t, V_max is the maximum voltage of the capacitor when fully charged, R is the resistance, C is the capacitance, and t is the time taken to charge the capacitor.
02

Gather information from the exercise and find relevant quantities

From the exercise, we know that: 1. The capacitance is 25.0 nF, which in farads is \(C = 25.0\times10^{-9} F\) 2. The capacitor is charged to 63.2% of its full voltage in each firing, so \(V(t) = 0.632V_{max}\) 3. The pacemaker fires 72 times a minute, which means the charging time for one firing is \(\frac{1}{72} min = \frac{60}{72} sec\). That's our t value.
03

Rearrange the formula to solve for R

We need to rearrange the equation \[V(t) = V_{max}(1 - e^{(-\frac{t}{RC})})\] to solve for R. Divide both sides of the equation by V_max: \[\frac{V(t)}{V_{max}} = 1 - e^{(-\frac{t}{RC})}\] Now, we know that \(\frac{V(t)}{V_{max}} = 0.632\) so, \[0.632 = 1 - e^{(-\frac{t}{RC})}\] Subtract 0.632 from both sides of the equation: \[-0.368 = -e^{(-\frac{t}{RC})}\] Now, we divide both sides of the equation by -1: \[0.368 = e^{(-\frac{t}{RC})}\] Next, take the natural logarithm of both sides: \[\ln{0.368} = -\frac{t}{RC}\] Now, we want to solve for R: \[R = -\frac{t}{C\ln{0.368}}\]
04

Solve for R using the values we found, and state the result

Now, we can plug in the values we found for t and C: \[R = -\frac{(60/72)}{25.0\times10^{-9}\ln{0.368}}\] Next, calculate the value of R: \[R \approx 5231.9\, \Omega\] Thus, the value of the resistance in the heart pacemaker circuit is approximately \(5231.9\, \Omega\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Capacitance
Capacitance is a property of a capacitor which tells us how much electric charge it can store for a given voltage. It is denoted as 'C' and its unit is the farad (F). To visualize this, think of a capacitor like a small battery, where capacitance represents the size of the battery. The larger the capacitance, the more charge it can hold for a certain voltage level. Capacitors come in various sizes and can be made out of different materials, but they all serve the purpose of storing and releasing electric charge.

In the example of the pacemaker, a 25.0-nanofarad (nF) capacitor is used. This relatively small capacitance indicates that the capacitor is designed for low-current, high-frequency applications such as in medical devices, where precise timing and energy control are crucial.
Time Constant (RC Circuit)
The time constant of an RC circuit, typically signified by the Greek letter 'tau' (Ï„), is a measure of the time it takes for the voltage across the capacitor to either charge or discharge through the resistor by approximately 63.2% of the difference between the initial and final value. This value is calculated as the product of the resistance 'R' and the capacitance 'C' and is represented by the formula Ï„ = R*C.

The time constant is crucial in understanding how fast a circuit responds to changes. A higher time constant means the circuit takes longer to respond (slow charging/discharging), while a lower time constant means it responds more quickly. For the heart pacemaker, the time for each charge cycle is crucial since it must reliably time the firing rate for heartbeats.
Exponential Decay in Circuits
Exponential decay in circuits occurs when a capacitor discharges through a resistor over time. The voltage across the capacitor doesn't drop linearly; it decreases at a rate proportional to its current value, which leads to an exponential decay curve. This pattern is described by the formula V(t) = V_max(1 - e^(-t/RC)), where V(t) is the voltage at time 't', V_max is the maximum initial voltage, and RC is the time constant.

This behavior is significant because it shows that as time goes on, the rate of discharge slows down, and the voltage decreases ever more gradually. It's like rolling a ball uphill that slows down as the slope becomes steeper. In our pacemaker example, understanding the exponential decay allows for precise calculations of the required firing rates and timings essential for the proper function of the device.
Natural Logarithm in Physics
The natural logarithm, denoted as 'ln', is a mathematical function that is fundamentally connected to the concept of exponential growth and decay. It is the inverse operation of exponentiation to the base 'e', where 'e' is an irrational and transcendental number approximately equal to 2.71828. In physics, taking the natural logarithm of both sides of an equation is often used to isolate variables in exponential decay equations.

For example, in RC circuits, taking the natural logarithm allows us to solve for variables such as resistance 'R' or time 't'. In the pacemaker calculation, after applying the natural logarithm to the exponential expression, we could isolate 'R' and find its value by manipulating the equation. Understanding how to work with natural logarithms is an essential skill in physics for dealing with exponential relationships in real-world situations.

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Most popular questions from this chapter

A flashing lamp in a Christmas earring is based on an \(R C\) discharge of a capacitor through its resistance. The effective duration of the flash is 0.250 s, during which it produces an average 0.500 W from an average 3.00 V. (a) What energy does it dissipate? (b) How much charge moves through the lamp? (c) Find the capacitance. (d) What is the resistance of the lamp? (since average values are given for some quantities, the shape of the pulse profile is not needed.)

The timing device in an automobile's intermittent wiper system is based on an \(R C\) time constant and utilizes a \(0.500-\mu \mathrm{F}\) capacitor and a variable resistor. Over what range must \(R\) be made to vary to achieve time constants from 2.00 to \(15.0 \mathrm{s}\) ?

Three \(300-\Omega\) resistors are connect in series with an AAA battery with a rating of 3 AmpHours. (a) How long can the battery supply the resistors with power? (b) If the resistors are connected in parallel, how long can the battery last?

An 1800-W toaster, a 1400-W speaker, and a 75-W lamp are plugged into the same outlet in a 15 -A fuse and 120-V circuit. (The three devices are in parallel when plugged into the same socket.) (a) What current is drawn by each device? (b) Will this combination blow the \(15-\mathrm{A}\) fuse?

Show that if two resistors \(R_{1}\) and \(R_{2}\) are combined and one is much greater than the other \(\left(R_{1} \gg R_{2}\right)\) their series resistance is very nearly equal to the greater resistance \(R_{1}\) and (b) their parallel resistance is very nearly equal to smaller resistance \(R_{2}\).

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