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Ch 27: Current and Resistance
Knight Calc - Physics for Scientists and Engineers 5th Edition
Knight Calc5th EditionPhysics for Scientists and EngineersISBN: 9780137344796Not the one you use?Change textbook
All textbooksKnight Calc 5th EditionCh 27: Current and ResistanceProblem 58b
Chapter 27, Problem 58b

The total amount of charge in coulombs that has entered a wire at time t is given by the expression Q=4t−t2, where t is in seconds and t≥0. Graph I versus t for the interval 0≤t≤4 s.

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Understand the problem: The charge entering the wire is given by the equation Q = 4t - t^2, where Q is the charge in coulombs and t is the time in seconds. To graph the current (I) versus time (t), recall that current is the rate of change of charge with respect to time, i.e., I = dQ/dt.
Differentiate the given equation Q = 4t - t^2 with respect to time t to find the expression for current I. Using the power rule of differentiation, dQ/dt = d(4t)/dt - d(t^2)/dt. This simplifies to I = 4 - 2t.
Determine the interval of interest for the graph. The problem specifies the interval 0 ≤ t ≤ 4 seconds. Within this range, calculate the values of I at key points (e.g., t = 0, t = 2, t = 4) to understand the behavior of the current.
Analyze the behavior of the current. From the expression I = 4 - 2t, observe that the current decreases linearly with time. At t = 0, I = 4 A; at t = 2, I = 0 A; and at t = 4, I = -4 A. This indicates that the current changes direction after t = 2 seconds.
Plot the graph of I versus t. On the x-axis, represent time (t) in seconds, and on the y-axis, represent current (I) in amperes. Plot the points (0, 4), (2, 0), and (4, -4), and draw a straight line connecting them to represent the linear relationship I = 4 - 2t.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Charge (Q)

Charge is a fundamental property of matter that causes it to experience a force when placed in an electromagnetic field. In this context, the total charge Q is expressed as a function of time t, indicating how charge accumulates in the wire over time. The equation Q=4t−t² shows that charge can increase and then decrease, reflecting the dynamics of charge flow.
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Quadratic Functions

A quadratic function is a polynomial function of degree two, typically represented in the form f(t) = at² + bt + c. The expression Q=4t−t² is a quadratic function where the coefficient of t² is negative, indicating that the graph will be a downward-opening parabola. Understanding the properties of quadratic functions, such as their vertex and intercepts, is essential for graphing the charge over time.
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Graphing Functions

Graphing functions involves plotting points on a coordinate system to visualize the relationship between variables. For the given function Q=4t−t², one must calculate values of Q for various t within the specified interval (0≤t≤4 s) and then plot these points to create a graph. This visual representation helps in understanding how charge changes over time and identifying key features like maximum charge and when it occurs.
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Related Practice
Textbook Question

The resistivity of a metal increases slightly with increased temperature. This can be expressed as ρ=ρ₀[1+α(T−T₀)] , where T₀ is a reference temperature, usually 20°C, and α is the temperature coefficient of resistivity. For copper, α=3.9×10−3 °C−1. Suppose a 2.5-m-long, 0.40-mm-diameter copper wire is connected across the terminals of a 1.5 V ideal battery. What is the current in the wire at 20°C?

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Textbook Question

A hollow metal cylinder has inner radius a, outer radius b, length L, and conductivity σ. The current I is radially outward from the inner surface to the outer surface. Evaluate the electric field strength at the inner and outer surfaces of an iron cylinder if a=1.0 cm, b=2.5 cm, L=10 cm, and I=25 A.

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Textbook Question

The total amount of charge in coulombs that has entered a wire at time t is given by the expression Q=4t−t2, where t is in seconds and t≥0. Find an expression for the current in the wire at time .

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Textbook Question

The current supplied by a battery slowly decreases as the battery runs down. Suppose that the current as a function of time is I=(0.75A)e−t6hI=\(\left\)(0.75A\(\right\))e^{-\(\frac{t}{6h}\)}. What is the total number of electrons transported from the positive electrode to the negative electrode by the charge escalator from the time the battery is first used until it is completely dead?

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Textbook Question

The total amount of charge that has entered a wire at time t is given by the expression Q=(20C)(1−e−t2.0s)Q=\(\left\)(20C\(\right\))\(\left\)(1-e^{-\(\frac{t}{2.0s}\)}\(\right\)), where t is in seconds and t≥0. What is the maximum value of the current?

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Textbook Question

What diameter should the nichrome wire in FIGURE P27.62 be in order for the electric field strength to be the same in both wires?

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