Problem 81 - Algebra and Trigonometry

Question

Electric Circuits An electric circuit contains a battery that produces a voltage of 60 volts \((\mathrm{V})\) , a resistor with a resistance of 13 ohms \((\Omega),\) and an inductor with an inductance of 5 henrys \((\mathrm{H}),\) as shown in the figure. Using calculus, it can be shown that the current \(I=I(t)\) (in amperes, A) \(t\) seconds after the switch is closed is \(I=\frac{60}{13}\left(1-e^{-13 / 15}\right)\) (a) Use this equation to express the time \(t\) as a function of the current \(I .\) (b) After how many seconds is the current 2 \(\mathrm{A} ?\)

Step-by-Step Solution

Answer

(a) \(t = -\frac{15}{13} \ln\left(1 - \frac{13I}{60}\right)\); (b) \(t \approx 0.814\) seconds.

Step 1: Setting up the equation

We are given the equation for current as a function of time: \(I=\frac{60}{13}\left(1-e^{-13t / 15}\right)\). We need to express time \(t\) as a function of current \(I\).

Step 2: Solving for the exponential term

Rearrange the given equation to isolate the exponential term. Start by dividing both sides by \(\frac{60}{13}\): \(\frac{I}{\frac{60}{13}} = 1 - e^{-13t/15}\). Simplify the left side to get \(\frac{13I}{60} = 1 - e^{-13t/15}\).

Step 3: Rearrange to isolate the exponential

Subtract \(\frac{13I}{60}\) from 1 to get: \(e^{-13t/15} = 1 - \frac{13I}{60}\).

Step 4: Apply the natural logarithm

Apply the natural logarithm to both sides to solve for \(-\frac{13t}{15}\): \(-\frac{13t}{15} = \ln\left(1 - \frac{13I}{60}\right)\).

Step 5: Solve for time \(t\)

Multiply both sides by \(-\frac{15}{13}\) to isolate \(t\): \(t = -\frac{15}{13} \ln\left(1 - \frac{13I}{60}\right)\). This is the expression of \(t\) as a function of current \(I\).

Step 6: Calculate the time for current \(I = 2\) A

Substitute \(I = 2\) A into the expression for \(t\): \(t = -\frac{15}{13} \ln\left(1 - \frac{13 \times 2}{60}\right)\). Simplify inside the logarithm: \(t = -\frac{15}{13} \ln\left(1 - \frac{26}{60}\right)\). Calculate \(t\): \(t = -\frac{15}{13} \ln\left(\frac{34}{60}\right)\).

Step 7: Final calculation

Solve the numerical expression to find \(t\). Calculate \(\ln\left(\frac{34}{60}\right) \approx -0.707\), then \(t = -\frac{15}{13} \times -0.707\). Perform the multiplication to find \(t \approx 0.814\) seconds.

Key Concepts

Electric Circuits

In electrical engineering, electric circuits form the backbone of a wide range of applications. An electric circuit can be thought of as a pathway through which electricity flows. It usually contains components like batteries, resistors, and inductors. In this exercise, we have a circuit with three components:

• Battery: Provides a voltage of 60 volts (V).
• Resistor: Has a resistance of 13 ohms (Ω).
• Inductor: Has an inductance of 5 henrys (H).

These components work together to affect the flow of electric current, which is measured in amperes (A). Understanding how these elements interact can help you predict how the current changes over time. Using calculus, we can derive equations that describe the relationship between current and time, like the given equation for current as a function of time: \(I=\frac{60}{13}(1-e^{-13t/15})\). Such equations are crucial for designing circuits in real-world applications.

Exponential Functions

Exponential functions play a significant role in mathematical modeling, especially in electrical circuits involving inductors. The general form of an exponential function is \(f(x) = a \cdot e^{bx}\), where:

• \(a\) is a constant coefficient.
• \(e\) is the base of the natural logarithms, approximately equal to 2.718.
• \(b\) is the exponent, which is often a function of time in circuit problems.

In circuits with inductors, the current growth or decay is often exponential. In the given equation, \(I = \frac{60}{13}(1-e^{-13t/15})\), the term \(e^{-13t/15}\) represents the exponential decay of the current over time. The exponential decay implies that the rate of change of current decreases exponentially as time goes on, which is characteristic in systems where electrons face resistance or inductance. This behavior is essential for the functioning of various electronic devices.

Natural Logarithm

The natural logarithm, denoted as \(\ln(x)\), is the inverse function of the exponential function \(e^x\). It is widely used in calculus to solve equations involving exponential terms. The key property of the natural logarithm is that it allows us to "undo" an exponential function:

• If \(y = e^x\), then \(x = \ln(y)\).

In electric circuits, particularly in the given problem, we use the natural logarithm to isolate variables. When given an equation like \(e^{-13t/15} = 1 - \frac{13I}{60}\), applying \(\ln\) to both sides helps in solving for the unknown variable, \(t\), as shown:\[-\frac{13t}{15} = \ln\left(1 - \frac{13I}{60}\right)\]By rearranging and using this property, it becomes straightforward to manipulate and solve equations for different circuit parameters, proving extremely helpful in engineering disciplines.

Solving Equations

Solving equations in the context of electric circuits often involves a series of transformations and operations to isolate the variable of interest. Here's a step-by-step approach:

• Start with the equation: Here, the relationship is \(I=\frac{60}{13}(1-e^{-13t/15})\), where you need to express \(t\) as a function of \(I\).
• Isolate the exponential term: Rearrange to find \(e^{-13t/15}\) and subtract other terms to separate it.
• Apply the natural logarithm: Use the inverse property \(\ln(e^x) = x\) to eliminate the exponential, solving for the term \(-\frac{13t}{15}\).
• Solve for the variable: Multiply through by the reciprocal of coefficients to get \(t\) by itself.

This systematic approach lets you break down complex relationships in circuits into simpler, manageable parts, making it easier to solve problems. For instance, after isolating \(t\), use the expression \(t = -\frac{15}{13} \ln\left(1 - \frac{13I}{60}\right)\) for specific current values, like \(I = 2\) A, to find how long it takes the current to reach that level.