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Suppose that the resistance between the walls of a biological cell is \(5.0 \times 10^{9} \Omega .\) (a) What is the current when the potential difference between the walls is 75 \(\mathrm{mV}\) ? (b) If the current is composed of \(\mathrm{Na}^{+}\) ions \((q=+e),\) how many such ions flow in 0.50 \(\mathrm{s} ?\)

Short Answer

Expert verified
(a) The current is \(1.5 \times 10^{-11} \text{ A}\). (b) Approximately \(4.69 \times 10^{7}\) ions flow.

Step by step solution

01

Understanding the Given Data

First, let's identify the given values for part (a): the resistance \( R \) is \( 5.0 \times 10^{9} \Omega \), and the potential difference \( V \) is \( 75 \text{ mV} = 75 \times 10^{-3} \text{ V} \).
02

Applying Ohm's Law

Ohm's Law states that \( V = I \times R \), where \( V \) is the potential difference, \( I \) is the current, and \( R \) is the resistance. We can rearrange this to find the current \( I = \frac{V}{R} \).
03

Calculating the Current

Substitute the given values into the equation: \[ I = \frac{75 \times 10^{-3}}{5.0 \times 10^{9}} \]. Calculating this gives \[ I = 1.5 \times 10^{-11} \text{ A} \].
04

Determine Charge per Ion

The charge of a \(\mathrm{Na}^{+}\) ion is the elementary charge, \( q = e \approx 1.6 \times 10^{-19} \text{ C} \).
05

Calculate the Number of Ions

The total charge \( Q \) that flows in \( 0.50 \text{ s} \) is given by \( Q = I \times t \). Substituting the values, \[ Q = 1.5 \times 10^{-11} \text{ A} \times 0.50 \text{ s} = 7.5 \times 10^{-12} \text{ C} \].
06

Finding the Number of Ions

The number of \(\mathrm{Na}^{+}\) ions \( n \) is given by \( n = \frac{Q}{q} \). Substitute the values to get \[ n = \frac{7.5 \times 10^{-12}}{1.6 \times 10^{-19}} \approx 4.69 \times 10^{7} \].

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Key Concepts

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

Resistance
Resistance is a fundamental concept in understanding how electricity flows through materials. It acts like a gatekeeper, determining how easily electric current can pass. Measured in ohms (\(\Omega\)), resistance depends on the material's properties, length, and cross-sectional area.
Resistors can be likened to pipes that control the flow of water. The narrower or longer the pipe, the harder it is for water to flow through. Similarly, the higher the resistance, the lower the current for a given potential difference. Ohm's Law, which we'll touch on later, ties resistance with current and potential difference: \( V = I \times R \).
Understanding resistance helps in designing circuits and managing current flow to prevent damage or inefficiency.
Potential Difference
Potential difference, often called voltage, is like the pressure pushing the current through a circuit. It's measured in volts (\(V\)) and is denoted in equations as \(V\).
Think of potential difference as the energy each charge carries as it moves. Within a cell or any electrical device, this difference in electrical charge between two points encourages electrons to flow, resulting in a current.
The potential difference is essential as it determines how much work can be done as the charge moves from one point to another. Ohm's Law connects it with current and resistance, showing its pivotal role in circuit behavior.
Electric Current
Electric current is the rate of flow of electric charge through a conductor, measured in amperes (\(A\)). It's like the flow of water in a river, transporting energy from one place to another.
Current can be either direct (DC) or alternating (AC). In direct current, charges flow steadily in one direction, much like in batteries. In alternating current, the direction alternates, as seen in most household electricity.
Using Ohm's Law, \( I = \frac{V}{R} \), we can determine how different factors, such as resistance and voltage, affect the current. Understanding current is crucial for safely and efficiently powering electrical devices.
Elementary Charge
The elementary charge is the most basic unit of electric charge, denoted by \(e\), approximately equal to \(1.6 \times 10^{-19}\) coulombs (\(C\)).
This charge corresponds to the charge of a single proton or the negative of a single electron, the smallest units of charge in nature. In many electrical calculations, understanding this base unit helps relate atomic-scale processes to larger, measurable electric currents.
For ions like \(\mathrm{Na}^+\), which are simply atoms with a net charge due to a missing electron, the elementary charge helps us calculate the number of ions moving in a given electric current, as seen in solving problems involving current flow through resistive materials.

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Most popular questions from this chapter

The rear window defogger of a car consists of thirteen thin wires (resistivity \(=88.0 \times 10^{-8} \Omega \cdot \mathrm{m}\) ) embedded in the glass. The wires are connected in parallel to the \(12.0-\mathrm{V}\) battery, and each has a length of 1.30 \(\mathrm{m} .\) The defogger can melt \(2.10 \times 10^{-2} \mathrm{kg}\) of ice at \(0^{\circ} \mathrm{C}\) into water at \(0^{\circ} \mathrm{C}\) in two minutes. Assume that all the power delivered to the wires is used immediately to melt the ice. Find the cross-sectional area of each wire.

A galvanometer with a coil resistance of 9.00\(\Omega\) is used with a shunt resistor to make a nondigital ammeter that has an equivalent resistance of 0.40\(\Omega\) . The current in the shunt resistor is 3.00 \(\mathrm{mA}\) when the galvanometer reads full scale. Find the full-scale current of the galvanometer.

The rear window of a van is coated with a layer of ice at \(0^{\circ} \mathrm{C}\) . The density of ice is 917 \(\mathrm{kg} / \mathrm{m}^{3}\) . The driver of the van turns on the rear-window defroster, which operates at 12 \(\mathrm{V}\) and 23 \(\mathrm{A}\) . The defroster directly heats an area of 0.52 \(\mathrm{m}^{2}\) of the rear window. What is the maximum thickness of ice coating this area that the defroster can melt in 3.0 minutes?

When a "dry-cell" flashlight battery with an internal resistance of 0.33\(\Omega\) is connected to a \(1.50-\Omega\) light bulb, the bulb shines dimly. However, when a lead-acid "wet-cell" battery with an internal resistance of 0.050\(\Omega\) is connected, the bulb is noticeably brighter. Both batteries have the same emf. Find the ratio \(P_{\text { wet }} / P_{\text { dry }}\) of the power delivered to the bulb by the wet-cell battery to the power delivered by the dry-cell battery.

An \(86-\Omega\) resistor and a \(67-\Omega\) resistor are connected in series across a battery. The voltage across the \(86-\Omega\) resistor is 27 V. What is the voltage across the \(67-\Omega\) resistor?

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