Question

The equation governing the amount of current \(I\) (in amperes) after time \(t\) (in seconds) in a simple RL circuit consisting of a resistance \(R\) (in ohms), an inductance \(L\) (in henrys), and an electromotive force \(E\) (in volts) is $$I=\frac{E}{R}\left[1-e^{-(R / L) t}\right]$$ If \(E=12\) volts, \(R=10\) ohms, and \(L=5\) henrys, how long does it take to obtain a current of 0.5 ampere? Of 1.0 ampere? Graph the equation.

Short answer

0.231 seconds for 0.5 A, 0.896 seconds for 1.0 A.

Step-by-step solution

Write down the given equation

The given equation for the current in the RL circuit is: \[ I = \frac{E}{R}\left[1 - e^{-(R / L)t}\right] \] We need to use this equation to find the time \( t \) for the given current values.

Substitute known values into the equation

Given values are: - \( E = 12 \) volts - \( R = 10 \) ohms - \( L = 5 \) henrys Substitute these into the equation: \[ I = \frac{12}{10}\left[1 - e^{-(10 / 5)t}\right] \] Simplify the fraction: \[ I = 1.2\left[1 - e^{-2t}\right] \]

Solve for the time \( t \) when the current is 0.5 amperes

Set \( I = 0.5 \) and solve for \( t \): \[ 0.5 = 1.2\left[1 - e^{-2t}\right] \] Divide both sides by 1.2: \[ \frac{0.5}{1.2} = 1 - e^{-2t} \] \[ \frac{5}{12} = 1 - e^{-2t} \] Rearrange to isolate the exponential term: \[ e^{-2t} = 1 - \frac{5}{12} \] \[ e^{-2t} = \frac{7}{12} \] Apply the natural logarithm to both sides: \[ -2t = \ln\left(\frac{7}{12}\right) \] Solve for \( t \): \[ t = -\frac{\ln\left(\frac{7}{12}\right)}{2} \] \[ t \approx 0.231 \text{ seconds} \]

Solve for the time \( t \) when the current is 1.0 ampere

Set \( I = 1.0 \) and solve for \( t \): \[ 1.0 = 1.2\left[1 - e^{-2t}\right] \] Divide both sides by 1.2: \[ \frac{1.0}{1.2} = 1 - e^{-2t} \] \[ \frac{5}{6} = 1 - e^{-2t} \] Rearrange to isolate the exponential term: \[ e^{-2t} = 1 - \frac{5}{6} \] \[ e^{-2t} = \frac{1}{6} \] Apply the natural logarithm to both sides: \[ -2t = \ln\left(\frac{1}{6}\right) \] Solve for \( t \): \[ t = -\frac{\ln\left(\frac{1}{6}\right)}{2} \] \[ t \approx 0.896 \text{ seconds} \]

Graph the equation

To graph the equation \[ I(t) = 1.2\left[1 - e^{-2t}\right] \] select a range of \( t \) values and calculate the corresponding \( I \) values. Plot these points on a graph with time \( t \) on the x-axis and current \( I \) on the y-axis.

Key concepts

RL circuit equations

An RL circuit includes a resistor (R) and an inductor (L) in series with a voltage source (E). The main equation describing the current in such a circuit over time is:
\[ I = \frac{E}{R} \left[ 1 - e^{-(R/L)t} \right] \]
Here's what each term represents:

• \(I\) is the current in amperes.
• \(E\) is the electromotive force (EMF) in volts.
• \(R\) is the resistance in ohms.
• \(L\) is the inductance in henrys.
• \(t\) is the time in seconds.

This equation shows how the current changes with time due to the interplay of resistance and inductance. Initially, the current is zero, and it gradually increases, approaching a maximum value of \(\frac{E}{R}\). The rate at which the current rises is influenced by the values of \(R\) and \(L\).

exponential functions

Exponential functions play a key role in RL circuits. The function \( e^{-(R/L)t} \) in our equation, known as an exponential decay function, dictates how quickly the current grows.

• For large values of \(t\), \( e^{-(R/L)t} \) approaches zero, making the current \(I\) nearly equal to \(\frac{E}{R}\).
• For small values of \(t\), the term \(1 - e^{-(R/L)t}\) is small, resulting in a lower current.

Exponential functions are thus critical in predicting the timing of electrical phenomena within the circuit. These functions are balanced by the circuit's resistance and inductance, determining the speed of the current change.

natural logarithms

Natural logarithms are essential for solving RL circuit equations. To find the exact time \(t\) when the current reaches a certain value, you often need to invert the exponential function using natural logarithms (\(\ln\)).
For example, to solve for the time when the current \(I\) is 0.5 amperes, we set up the following equation: \( 0.5 = 1.2 \left[ 1 - e^{-2t} \right] \).
We isolate the exponential part: \( e^{-2t} = \frac{7}{12} \).
Taking the natural logarithm of both sides, we get: \( -2t = \ln \left( \frac{7}{12} \right) \).
Solving for \(t\): \( t = -\frac{ \ln \left( \frac{7}{12} \right)}{2} \).
This use of natural logarithms allows us to convert multiplicative relationships into additive ones, making the equations easier to manage.

graphing electrical equations

To understand the behavior of an RL circuit over time, graphing the current equation is very helpful. The equation \[ I(t) = 1.2 \left[ 1 - e^{-2t} \right] \] can be plotted to visualize how the current evolves as time progresses.
Here's how to do it:

• Choose a range of time values (\(t\)) — for instance, from 0 to 5 seconds.
• Calculate the current (\(I\)) for each time value using the equation.
• Plot these points on a graph with time \(t\) on the x-axis and current \(I\) on the y-axis.

The shape of the resulting curve will illustrate an exponential growth, asymptotically approaching the maximum current value of 1.2 amperes (since \(\frac{E}{R} = 1.2\)). Graphing these equations is crucial for visualizing and understanding how current changes over time in the circuit.