An electric device delivers a current of [tex]$15.0 A$[/tex] for 30 seconds. How many electrons flow through it?
Answer (2)
Define the piecewise function f ( t ) as 1 for 0 < t < 1 and 0 for 1"> t > 1 .
Apply the Laplace transform definition: L { f ( t )} = ∫ 0 ∞ e − s t f ( t ) d t .
Evaluate the integral: ∫ 0 1 e − s t d t = s 1 − e − s .
State the Laplace transform and its convergence condition: 0"> L { f ( t )} = s 1 − e − s for s > 0 .
Explanation
Problem Analysis We are asked to find the Laplace transform of the function f ( t ) defined as:
Function Definition 1 \end{cases}"> f ( t ) = { 1 , 0 , 0 < t < 1 t > 1
Laplace Transform Definition The Laplace transform is defined as:
Laplace Transform Formula F ( s ) = L { f ( t )} = ∫ 0 ∞ e − s t f ( t ) d t
Substitution Substituting the given function f ( t ) into the Laplace transform integral, we get:
Integral Setup L { f ( t )} = ∫ 0 ∞ e − s t f ( t ) d t = ∫ 0 1 e − s t ( 1 ) d t + ∫ 1 ∞ e − s t ( 0 ) d t
Simplification Simplifying the integral, we have:
Simplified Integral L { f ( t )} = ∫ 0 1 e − s t d t
Evaluation Now, we evaluate the integral:
Integral Result ∫ 0 1 e − s t d t = [ − s 1 e − s t ] 0 1 = − s 1 e − s ( 1 ) − ( − s 1 e − s ( 0 ) ) = − s 1 e − s + s 1 = s 1 − e − s
Laplace Transform Result Thus, the Laplace transform of f ( t ) is:
Final Laplace Transform L { f ( t )} = s 1 − e − s
Convergence Condition The domain of F ( s ) is the set of values of s for which the integral converges. Since the integral ∫ 0 1 e − s t d t is a definite integral with finite limits, it converges for all real values of s except possibly s = 0 . However, lim s → 0 s 1 − e − s = 1 , so the Laplace transform is defined for s = 0 as well. The question specifies 0"> s > 0 .
Final Answer Therefore, the Laplace transform of the given function is:
Final Answer with Condition 0"> L { f ( t )} = s 1 − e − s , s > 0
Examples
Laplace transforms are incredibly useful in electrical engineering. Imagine you have a circuit with a switch that turns on a voltage source at a specific time. The voltage can be modeled as a piecewise function, much like the one in our problem. By using Laplace transforms, engineers can easily analyze the circuit's behavior, such as determining the current flow or voltage response, without having to solve complex differential equations directly in the time domain. This makes circuit design and analysis much more efficient!
The total charge that flows through the device is 450 coulombs over 30 seconds. This corresponds to approximately 2.81 x 10^21 electrons, calculated using the current and the charge of a single electron. The steps involved calculating total charge from current and time, and then converting that charge into the number of electrons using the elementary charge value.
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