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Chapter 26: Problem 1

A steady current of \(0.50\) A flows through a wire. How much charge passes through the wire in one minute? Because \(I=q / t\), it follows that \(q=I t=(0.50 \mathrm{~A})(60 \mathrm{~s})=30 \mathrm{C}\). (Recall that \(1 \mathrm{~A}=1 \mathrm{C} / \mathrm{s}\).)

Short Answer

Expert verified
30 Coulombs.

Step by step solution

01

Understanding the Formula

The formula used here is based on the definition of electric current, which is the flow of electric charge. The formula is \( I = \frac{q}{t} \), where \( I \) is the current, \( q \) is the charge, and \( t \) is the time.
02

Rearranging the Formula

To find the total charge (\( q \)), we rearrange the formula to solve for \( q \). We get \( q = I \times t \). This tells us that charge, when current and time are known, is calculated by multiplying the two.
03

Substituting Values

Substitute the given values into the rearranged formula. The current \( I \) is \( 0.50 \) A, and the time \( t \) is \( 60 \) seconds (since 1 minute equals 60 seconds). So, the formula becomes \( q = 0.50 \times 60 \).
04

Performing the Calculation

Calculate the product to find the total charge, \( q = 0.50 \times 60 = 30 \) C. This means that 30 Coulombs of charge pass through the wire in one minute.

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Key Concepts

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

Charge Calculation
When dealing with electric circuits, it is often important to calculate the amount of electric charge that passes through a wire over a period of time. This is known as charge calculation.
In the context of the formula for current, which is given by \( I = \frac{q}{t} \), we can rearrange this formula to solve for the charge \( q \). The rearranged formula is:
  • \( q = I \times t \)
Here,
  • \( q \) represents the charge, measured in Coulombs (C).
  • \( I \) is the current, which is the flow of charge per unit time, measured in Amperes (A).
  • \( t \) is the time in seconds (s).
This calculation is essential in many electrical applications, such as determining how much charge a battery can deliver over a certain time period. By multiplying the current by the time, one can determine the total amount of charge that has moved through the wire.
Electric Charge
Electric charge is a fundamental property of matter that causes it to experience a force when placed in an electromagnetic field. It is the basic unit that makes up electric current when charges are in motion. The unit for electric charge is the Coulomb (C).
Charges are typically either positive or negative. In terms of particles:
  • Protons carry a positive charge.
  • Electrons carry a negative charge.
In conductors like metals, charges are carried mainly by electrons moving through the material. The concept of electric charge is crucial because it explains why electrical phenomena occur and how electrical equipment operates. Understanding charge behavior helps us design circuits and electronic devices effectively.
Current Formula
The formula for electric current provides the relationship between current, charge, and time. Electric current is defined as the rate at which charge flows through a surface. This can be mathematically expressed as:
  • \( I = \frac{q}{t} \)
Where:
  • \( I \) is the current, expressed in Amperes (A).
  • \( q \) is the electric charge, in Coulombs (C).
  • \( t \) is the time, in seconds (s).
This formula means that the current flowing through a circuit depends on how much charge flows past a point in the circuit per second. A higher current indicates more charge passing through the wire in a given time. Understanding this concept is foundational for exploring more complex electrical theories and real-world applications like electric motors or generators.

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Most popular questions from this chapter

A dry cell has an emf of \(1.52 \mathrm{~V}\). Its terminal potential drops to zero when a current of 25 A passes through it. What is its internal resistance? As is shown in Fig. \(26-3\), the battery acts like a pure emf \(\varepsilon\) in series with a resistor \(r\). We are told that, under the conditions shown, the potential difference from \(A\) to \(B\) is zero. Therefore, $$ 0=+\varepsilon-I r \text { or } 0=1.52 \mathrm{~V}-(25 \mathrm{~A}) r $$ from which the internal resistance is \(r=0.061 \Omega\).

A battery has an emf of \(13.2 \mathrm{~V}\) and an internal resistance of \(24.0\) \(\mathrm{m} \Omega\). If the load current is \(20.0 \mathrm{~A}\), find the terminal voltage.

The resistivity of aluminum is \(2.8 \times 10^{-8} \Omega \cdot \mathrm{m}\). How long a piece of aluminum wire \(1.0 \mathrm{~mm}\) in diameter is needed to give a resistance of \(4.0 \Omega\) ?

Number 10 wire has a diameter of \(2.59 \mathrm{~mm}\). How many meters of number 10 aluminum wire are needed to give a resistance of \(1.0\) \(\Omega ? \rho\) for aluminum is \(2.8 \times 10^{-8} \Omega \cdot \mathrm{m}\). From \(R=\rho L / A\) $$ L=\frac{R A}{\rho}=\frac{(1.0 \Omega)(\pi)\left(2.59 \times 10^{-3} \mathrm{~m}\right)^{2} / 4}{2.8 \times 10^{-8} \Omega \cdot \mathrm{m}}=0.19 \mathrm{~km} $$

A battery charger supplies a current of \(10 \mathrm{~A}\) to charge a storage battery that has an open-circuit voltage of \(5.6 \mathrm{~V}\). If the voltmeter connected across the charger reads \(6.8 \mathrm{~V}\), what is the internal resistance of the battery at this time?
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