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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) Current is \(1.5 \times 10^{-11} \, \text{A}\). (b) About \(4.69 \times 10^7\) ions flow in 0.50 s.

Step by step solution

01

Use Ohm's Law to Find Current

Ohm's Law states that the current \(I\) is the ratio of the potential difference \(V\) to the resistance \(R\). The formula is \(I = \frac{V}{R}\). Here, \(V = 75 \, \text{mV} = 75 \times 10^{-3} \, \text{V}\) and \(R = 5.0 \times 10^9 \, \Omega\). Substitute these values into the formula: \[ I = \frac{75 \times 10^{-3} \, \text{V}}{5.0 \times 10^9 \, \Omega} \approx 1.5 \times 10^{-11} \, \text{A}. \]
02

Determine Number of Ions in 0.50 s

To calculate the number of ions, use \(I = n \cdot q / t\), where \(n\) is the number of ions, \(q = e = 1.6 \times 10^{-19} \, \text{C}\) is the charge of one ion, and \(t = 0.50 \, \text{s}\). Rearrange to find \(n\): \[ n = \frac{I \cdot t}{q} = \frac{(1.5 \times 10^{-11} \, \text{A}) \cdot (0.50 \, \text{s})}{1.6 \times 10^{-19} \, \text{C}}. \] Calculate to find \(n \approx 4.69 \times 10^7\) ions.

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Key Concepts

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

Current Calculation
When dealing with electrical circuits, we often need to calculate the current, which is essentially the flow of electrical charge. Ohm's Law provides a straightforward relationship to find this current. The law states that the current \( I \) through a conductor between two points is directly proportional to the voltage \( V \) across the two points and inversely proportional to the resistance \( R \) of the conductor. This can be written as:
\[I = \frac{V}{R}\]- **Voltage (V):** The electric potential difference between two points, unit is volts (V).- **Resistance (R):** The measure of how much an object opposes the passage of electrons, unit is ohms (Ω).- **Current (I):** The flow of electric charge, unit is amperes (A).To solve a typical problem, you simply rearrange the equation and input the known values. For example, if the voltage across a biological cell is 75 mV, and resistance is \(5.0 \times 10^{9} \Omega\), substitute these values to find the current.
By calculating \(I = \frac{75 \times 10^{-3} \, \text{V}}{5.0 \times 10^{9} \, \Omega} \), you find a very small current of approximately \(1.5 \times 10^{-11} \, \text{A}\). This tells us how much electric charge flows through the cell's resistance barrier.
Electric Potential Difference
Electric potential difference, commonly known as voltage, is a fundamental concept in understanding electrical systems. It represents the capability to move charge from one point to another. Here's what you need to know:
- **Definition:** Voltage is the work done to move a electric charge between two points in an electric field.- **Units:** Voltage is measured in volts (V).- **Measurement:** You measure the potential difference using a voltmeter connected across the component.The voltage between two points is often likened to the pressure in a water system. Just like how pressure pushes water through pipes, voltage pushes electrical charges through a conductor. In the context of a biological cell, where we have resistive elements such as cell walls, the potential difference is what causes ions to move across the membrane. A potential difference of 75 mV across a cell membrane indicates the energy available to drive the movement of ions like \(\text{Na}^{+}\) within the living system.
Resistance in Biological Systems
Resistance isn't just a topic for physics and engineering. It plays a vital role in biological systems as well. Biological cells have membranes that act like resistive barriers, controlling the flow of ions and maintaining essential functions. Here's how it works:
- **Biological Relevance:** Cell membranes resist the flow of ions, just as resistors control current in electric circuits.- **Importance:** This resistance is crucial for nerve impulse conduction, muscle contraction, and maintaining cellular homeostasis.- **Factors Affecting Resistance:** Membrane composition, ion channel availability, and temperature all contribute to the overall resistance.In the given problem, the resistance between the membranes of a cell is exceedingly high, \(5.0 \times 10^{9} \Omega \). Despite these high resistance values, minute levels of current still pass through, which are significant for cellular processes. Understanding this helps explain how cells control their internal environments and respond to external stimuli efficiently, maintaining the delicate balance required for life's processes.

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Most popular questions from this chapter

Three resistors, \(25,45,\) and \(75 \Omega,\) are connected in series, and a \(0.51-\) A current passes through them. What are (a) the equivalent resistance and (b) the potential difference across the three resistors?

Two resistances, \(R_{1}\) and \(R_{2},\) are connected in series across a \(12-\mathrm{V}\) battery. The current increases by \(0.20 \mathrm{A}\) when \(R_{2}\) is removed, leaving \(R_{1}\) connected across the battery. However, the current increases by just \(0.10 \mathrm{A}\) when \(R_{1}\) is removed, leaving \(R_{2}\) connected across the battery. Find (a) \(R_{1}\) and \((b) R_{2}\).

A large spool in an electrician's workshop has \(75 \mathrm{m}\) of insulationcoated wire coiled around it. When the electrician connects a battery to the ends of the spooled wire, the resulting current is \(2.4 \mathrm{A} .\) Some weeks later, after cutting off various lengths of wire for use in repairs, the electrician finds that the spooled wire carries a \(3.1-\) A current when the same battery is connected to it. What is the length of wire remaining on the spool?

A circuit contains a 48-V battery and a single light bulb whose resistance is \(240 \Omega .\) A second, identical, light bulb can be wired either in series or in parallel with the first one (see the figure). Concepts: (i) How is the power \(P\) that is delivered to a light bulb related to the bulb's resistance \(R\) and the voltage \(V\) across it? (ii) When there is only one bulb in the circuit, what is the voltage across it? (iii) The more power delivered to a bulb, the brighter it is. When two bulbs are wired in series, does the brightness of each bulb increase, decrease, or remain the same relative to the brightness of the bulb in the single-bulb circuit? (iv) When two bulbs are wired in parallel, does the brightness of each bulb increase, decrease, or remain the same relative to the brightness of the bulb in the single-bulb circuit? Calculations: Determine the power delivered to a single bulb when the circuit contains (a) only one bulb, (b) two bulbs in series and (c) two bulbs in parallel. Assume that the battery has no internal resistance.

A tungsten wire has a radius of \(0.075 \mathrm{mm}\) and is heated from 20.0 to \(1320^{\circ} \mathrm{C} .\) The temperature coefficient of resistivity is \(\alpha=4.5 \times\) \(10^{-3}\left(\mathrm{C}^{\circ}\right)^{-1} .\) When \(120 \mathrm{V}\) is applied across the ends of the hot wire, a current of \(1.5 \mathrm{A}\) is produced. How long is the wire? Neglect any effects due to thermal expansion of the wire.

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