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Chapter 1: Problem 9

The current through a given circuit element is given by \(i(t)=2 e^{-t}\) A. As usual, time \(t\) is in seconds. Find the net charge that passes through the element in the interval from \(t=\) 0 to \(t=\infty\). (Hint: Current is the rate of flow of charge. Thus, to find charge, we must integrate current with respect to time.)

Short Answer

Expert verified
The net charge that passes through the element from \(t = 0\) to \(t = \infty\) is 2 C.

Step by step solution

01

Understand the Relationship Between Charge and Current

The current function \(i(t)\) describes how much charge flows per unit time. To find the total charge that flows over a period of time, we need to integrate the current over that time interval. In mathematical terms, the charge \(q(t)\) in a time interval from \(t_0\) to \(t_1\) is given by: \[ q(t) = \int_{t_0}^{t_1} i(t) \, dt \] In this case, we are integrating from \(t = 0\) to \(t = \infty\).
02

Set up the Integral for the Total Charge

To find the total charge that passes through the element, we need to solve the integral: \[ Q = \int_{0}^{\infty} i(t) \, dt = \int_{0}^{\infty} 2e^{-t} \, dt \]
03

Solve the Integral

The integral is \( \int_{0}^{\infty} 2e^{-t} \, dt \). To solve this, recognize that \( \int e^{-t} \, dt = -e^{-t} + C \), where \( C \) is the constant of integration. So, \[ \int_{0}^{\infty} 2e^{-t} \, dt = 2 \left[ -e^{-t} \right]_{0}^{\infty} = 2 \left( 0 - (-1) \right) = 2 \times 1 = 2 \]
04

Calculate the Definite Integral from Limits 0 to Infinity

Evaluating the integral at the limits, \[ \lim_{t \to \infty} (-2e^{-t}) = 0 \] and at \(t = 0\), \(-2e^{-0} = -2\). So, the definite integral results in \(0 - (-2) = 2\). Therefore, the total charge \(Q\) is 2 C.

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Key Concepts

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

Exponential Decay in Circuit Analysis
In circuit analysis, exponential decay describes how certain quantities, like current, decrease over time. The function given, \( i(t) = 2e^{-t} \), is an example of exponential decay. Here, the current (*t*) diminishes as time progresses. This behavior is common in circuits involving capacitors and resistors, where the initial charge or current fades away exponentially.

Understanding exponential decay is crucial because it informs us how quickly or slowly the energy in a circuit decreases. The "\( e^{-t} \)" factor denotes the pace of decay. As time increases, \( e^{-t} \) rapidly approaches zero, illustrating a swift decline in current.

In practical applications, this concept helps in predicting the lifespan of electronic components and understanding how long it takes for a circuit to reach a certain state after power is applied or removed.
Using Integration to Calculate Total Charge
Integration is a fundamental technique in calculus used to sum up small quantities to find a total. When calculating the total charge passing through a circuit element, we integrate the current function over a specific time interval.

The process involves setting up an integral of the form \( Q = \int i(t) \, dt \). For our situation, the integral becomes \( Q = \int_{0}^{\infty} 2e^{-t} \, dt \). This expression indicates we are summing up all infinitesimally small contributions of current over an infinite time span.

To solve the integral, we use the fact that the anti-derivative of \( e^{-t} \) is \(-e^{-t} \), and then apply the limits from 0 to infinity. This mathematical tool allows us to compute the net charge that has flowed through the circuit during the given time frame, capturing the entire history of current flow.
Understanding Charge Calculation in Circuits
Charge calculation is a critical aspect of circuit analysis. It quantifies the total electrons that have moved through a circuit element over time, measured in coulombs (C). The relationship between current \( i(t) \) and charge \( q(t) \) is pivotal here, where current is the rate at which charge moves.

The exercise provided \( i(t) = 2e^{-t} \) as a current function, which we integrated over time to determine the charge. The definite integral \( \int_{0}^{\infty} 2e^{-t} \, dt \) told us how much charge had passed through from time zero to infinity, yielding a total of 2 coulombs.

This calculation gives us insight into how circuits behave over time, providing both a theoretical and practical understanding essential for designing and analyzing electronic systems. It reveals how even though the current may decay exponentially, the total charge accumulated over time can still be significant.

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Most popular questions from this chapter

Consider a circuit containing four nodes labeled \(a, b, c\), and \(d\). Also, we know that \(v_{a b}=12 \mathrm{~V}, v_{c b}=-4 \mathrm{~V}\), and \(v_{d a}=8 \mathrm{~V}\). Determine the values of \(v_{a c}\) and \(v_{c d}\). (Hint: Draw a picture showing the nodes and the known voltages.)

Which of the following are self-contradictory combinations of circuit elements? a. A 12-V voltage source in parallel with a 2 -A current source. b. A 2-A current source in series with a 3-A current source. c. A 2-A current source in parallel with a short circuit. d. A 2-A current source in series with an open circuit. e. A 5-V voltage source in parallel with a short circuit.

Name eight subdivisions of electrical engineering.

Figure P1.28 shows an ammeter (AM) and voltmeter (VM) connected to measure the current and voltage, respectively, for circuit element \(A\). When current actually enters the + terminal of the ammeter the reading is positive, and when current leaves the \(+\) terminal the reading is negative. If the actual voltage polarity is positive at the \(+\) terminal of the VM, the reading is positive; otherwise, it is negative. (Actually, for the connection shown, the ammeter reads the sum of the current in element \(A\) and the very small current taken by the voltmeter. For purposes of this problem, assume that the current taken by the voltmeter is negligible.) Find the power for element \(A\) and state whether energy is being delivered to element \(A\) or taken from it if: a. the ammeter reading is \(+2 \mathrm{~A}\) and the voltmeter reading is \(-25 \mathrm{~V}\); b. the ammeter reading is \(-2\) A and the voltmeter reading is \(+25 \mathrm{~V}\); c. the ammeter reading is \(-2 \mathrm{~A}\) and the voltmeter reading is \(-25 \mathrm{~V}\).

In the fluid-flow and electric circuit analogy, suggest the analogous quantities for: a. a conductor; b. a constriction; c. a pump; and d. current flow.
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