Problem 87 - Algebra and Trigonometry

Question

An electric circuit contains a battery that produces a voltage of 60 volts (V), a resistor with a resistance of 13 ohms \((\Omega),\) and an inductor with an inductance of 5 henrys (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 \(I=\frac{60}{13}\left(1-e^{-13 / 5}\right) .\) (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

The current is 2 A after approximately 0.858 seconds.

Step 1: Express the equation for current I

We start with the given equation for current \(I\): \[ I = \frac{60}{13}\left(1 - e^{-13t / 5}\right) \]

Step 2: Rearrange for e-term

Subtract \(\frac{60}{13}\) from both sides to isolate the exponential term:\[ \frac{13I}{60} = 1 - e^{-13t / 5} \] Then, rearrange to solve for the exponential term:\[ e^{-13t / 5} = 1 - \frac{13I}{60} \]

Step 3: Solve for time t

Take the natural logarithm of both sides to solve for \(t\):\[ -\frac{13t}{5} = \ln\left(1 - \frac{13I}{60}\right) \]Multiply both sides by \(-\frac{5}{13}\) to solve for \(t\):\[ t = -\frac{5}{13} \ln\left(1 - \frac{13I}{60}\right) \]

Step 4: Substitute I = 2 A to find time

Insert \(I = 2\) A into the expression from Step 3:\[ t = -\frac{5}{13} \ln\left(1 - \frac{13 \times 2}{60}\right) \]Simplify the expression inside the logarithm:\[ t = -\frac{5}{13} \ln\left(1 - \frac{26}{60}\right) \]Further simplify:\[ t = -\frac{5}{13} \ln\left(\frac{34}{60}\right) \]Evaluate the expression for \(t\).

Key Concepts

Understanding Exponential Functions

Exponential functions are a key tool in mathematics, especially when it comes to modeling growth and decay processes, such as in electrical circuits. In our exercise, the exponential function appears in the expression for current \(I\) over time \(t\): \[I = \frac{60}{13}\left(1 - e^{-\frac{13t}{5}}\right)\] This formula models how the current increases as a circuit is powered up.

• The term \(1 - e^{-\frac{13t}{5}}\) describes how the current approaches its maximum value over time.
• As \(t\) increases, \(e^{-\frac{13t}{5}}\) approaches zero, leading \(I\) to approach its maximum value of \(\frac{60}{13}\).

It's important to understand that \(e\) is the base of natural logarithms, roughly equal to 2.718. Exponential decay, as expressed by the \(e^{-kt}\) term, means that the rate of current increase is initially fast but slows down as it reaches a steady state. This mathematical property makes exponential functions ideal for modeling time-dependent phenomena in physics and engineering.

Solving Logarithmic Equations

Logarithmic equations are the inverse of exponential equations, and we use them to solve for specific variables within exponential contexts. In the given exercise, we need to solve for time \(t\) when current \(I\) reaches a specific value: \[t = -\frac{5}{13} \ln\left(1 - \frac{13I}{60}\right)\]

• The natural logarithm \(\ln\) is the inverse function of the exponential function \(e^x\).
• Taking the logarithm helps to isolate variables in equations where they are exponentiated.

Here's how we approach solving these:1. Ensure the equation is clearly rearranged to isolate the \(e\)-term. 2. Take the natural logarithm of both sides to eliminate the \(e\).3. Solve for the desired variable.For example, when \(I = 2\) A, substituting gives:\[t = -\frac{5}{13} \ln\left(1 - \frac{26}{60}\right)\] This requires us to compute the logarithm of the remaining expression, enabling us to solve for the time \(t\) accurately.

Current in Electrical Circuits

In electrical circuits, the current \(I\) is an essential parameter indicating the flow of electric charge through a conductor. The current's behavior over time is crucial for understanding how circuits function under different conditions. In our case, the equation \[I = \frac{60}{13}\left(1 - e^{-\frac{13t}{5}}\right)\] describes how the current increases as the circuit components like resistors and inductors interact.

• Voltage \(V\) provided by the battery drives the current through the circuit.
• Resistance \(R\) and inductance \(L\) affect how quickly the current reaches its peak.
• The exponential term reflects how the inductor initially resists changes in current flow.

These components work together to determine the rate and stability at which current reaches its steady state. Ohm's Law, \(V = IR\), is often applied but when inductors are involved, the current's dynamic behavior over time also factors in, necessitating calculus to accurately model. Recognizing how current changes with time in a circuit allows for designing and analyzing more effective electrical systems, which is critical for both practical and theoretical applications.