Problem 28
Question
The rate of change of electric charge with respect to time is called current. Suppose that \(\frac{1}{3} t^{3}+t\) coulombs of charge flow through a wire in \(t\) seconds. Find the current in amperes (coulombs per second) after 3 seconds. When will a 20 -ampere fuse in the line blow?
Step-by-Step Solution
Verified Answer
The current after 3 seconds is 10 amperes. The fuse will blow at approximately \(t = \sqrt{19}\) seconds, which is about 4.36 seconds.
1Step 1: Understand the Function
The function given is \(Q(t) = \frac{1}{3} t^{3}+t\), which represents the charge in coulombs as a function of time \(t\) in seconds. We need to find the current, which is the rate of change of charge with respect to time.
2Step 2: Calculate the Current
Current \(I(t)\) is the derivative of the charge function \(Q(t)\) with respect to time. So, we find \(I(t) = \frac{d}{dt} \left( \frac{1}{3} t^{3} + t \right)\). Using the power rule of differentiation, we get \(I(t) = t^2 + 1\).
3Step 3: Evaluate Current at t = 3 Seconds
To find the current at \(t = 3\) seconds, substitute \(t = 3\) into the expression for \(I(t)\). This gives \(I(3) = 3^2 + 1 = 9 + 1 = 10\) amperes.
4Step 4: Determine When the Fuse Will Blow
The fuse blows when the current reaches 20 amperes. Set \(I(t) = 20\) and solve for \(t\). So, \(t^2 + 1 = 20\). Solving for \(t\), we get \(t^2 = 19\), thus \(t = \sqrt{19}\).
Key Concepts
Electric ChargeDifferentiationCurrent in Amperes
Electric Charge
Electric charge is a fundamental property of matter, represented by the symbol \(Q\) in equations.
It is the amount of charge, typically measured in coulombs, that passes a point in a circuit.
Charge is often caused by the transfer of electrons, which carry a tiny amount of negative charge.
Understanding this concept sets the groundwork for differentiating the function to find the rate at which the charge flows, otherwise known as current.
It is the amount of charge, typically measured in coulombs, that passes a point in a circuit.
Charge is often caused by the transfer of electrons, which carry a tiny amount of negative charge.
- In physics and electrical engineering, understanding charge is essential for studying electricity and circuits.
- Charge can be positive or negative, depending on the balance of protons (positive) and electrons (negative)
- One coulomb correlates to a very large number of electrons, approximately \(6.242 imes 10^{18}\).
Understanding this concept sets the groundwork for differentiating the function to find the rate at which the charge flows, otherwise known as current.
Differentiation
Differentiation is a core concept in calculus that helps to find the rate of change of a function.
In this context, we use it to find how quickly the electric charge changes with respect to time.
In mathematical terms, if \(Q(t)\) is the function representing electric charge, then the current \(I(t)\) is determined by differentiating \(Q(t)\) with respect to time \(t\).
In this context, we use it to find how quickly the electric charge changes with respect to time.
In mathematical terms, if \(Q(t)\) is the function representing electric charge, then the current \(I(t)\) is determined by differentiating \(Q(t)\) with respect to time \(t\).
- By applying the power rule, differentiation of a term \(\frac{1}{3}t^3\) gives \(t^2\).
- The differentiation of a simple term like \(t\) results in \(1\).
- Combining these, the derivative, which is \(I(t)\), becomes \(t^2 + 1\).
Current in Amperes
Current is the flow of electric charge per unit time, measured in amperes (A).
One ampere is equivalent to one coulomb of charge passing through a point per second.
In the exercise, we find the current by differentiating the electric charge function \(Q(t)\), resulting in \(I(t) = t^2 + 1\).
To find the specific current value at a given moment, we substitute the time \(t\) into \(I(t)\). For instance:
One ampere is equivalent to one coulomb of charge passing through a point per second.
In the exercise, we find the current by differentiating the electric charge function \(Q(t)\), resulting in \(I(t) = t^2 + 1\).
To find the specific current value at a given moment, we substitute the time \(t\) into \(I(t)\). For instance:
- At \(t = 3\) seconds, the current is \(I(3) = 3^2 + 1 = 10\text{ A}\).
- To determine when a 20 ampere fuse will blow, set \(t^2 + 1 = 20\) and solve for \(t\).
- It yields \(t = \sqrt{19}\), which indicates the time in seconds when the current would reach 20 amperes, potentially causing the fuse to blow.
Other exercises in this chapter
Problem 28
\(\lim _{h \rightarrow 0} \frac{(3+h)^{2}+2(3+h)-15}{h}\)
View solution Problem 28
$$ \text { In Problems } 1-44, \text { find } D_{x} y \text { using the rules of this section. } $$ $$ y=\left(x^{4}-1\right)\left(x^{2}+1\right) $$
View solution Problem 28
In Problems 21-28, find the indicated derivative. $$ \frac{d y}{d t}, \text { where } y=\left[\sin t \tan \left(t^{2}+1\right)\right] $$
View solution Problem 29
Show that the curves \(y=\sqrt{2} \sin x\) and \(y=\sqrt{2} \cos x\) intersect at right angles at a certain point with \(0
View solution