Problem 86

Question

Peak alternating current Suppose that at any given time $t$ (in seconds) the current $i$ (in amperes) in an alternating current circuit is $i=2 \cos t+2 \sin t .$ What is the peak current for this circuit (largest magnitude)?

Step-by-Step Solution

Verified

Answer

Peak current is $2\sqrt{2}$ amperes.

1Step 1: Understanding the Problem

We need to find the largest magnitude of the current given by the equation $ i = 2 \cos t + 2 \sin t $. This is known as the peak current in the circuit.

2Step 2: Express the Current in a Single Trigonometric Function

Use the trigonometric identity for expressing a sum of cosine and sine with the same frequency: $ i(t) = A \cos(t - \phi) $, where $ A $ is the amplitude. This is done using the identity: $ A = \sqrt{a^2 + b^2} $, where $ a $ and $ b $ are the coefficients of $ \cos $ and $ \sin $ respectively.

3Step 3: Calculate the Amplitude

For the given function $ i(t) = 2 \cos t + 2 \sin t $, the coefficients are 2 and 2. Calculate the amplitude as follows:\[A = \sqrt{2^2 + 2^2} = \sqrt{4 + 4} = \sqrt{8} = 2\sqrt{2}.\]

4Step 4: Find the Peak Current Value

With the amplitude $ A = 2\sqrt{2} $, the peak current, or maximum magnitude, of the current in the circuit is $ 2\sqrt{2} $ amperes.

Key Concepts

Alternating CurrentTrigonometric IdentitiesAmplitude

Alternating Current

Alternating current (often abbreviated as AC) is a type of electrical current where the direction of flow reverses periodically. It contrasts with direct current (DC), where the flow of charge is always in one direction.
This reversal of direction is characterized by the sinusoidal (wave-like) behavior of AC, which makes it versatile for power transmission.

AC can be easily transformed to different voltages. This is useful for efficient transmission over long distances.
It is the type of current that powers most homes and businesses, providing the electricity for everything from lighting to heating devices.
AC's behavior is mathematically described using trigonometric functions like sine and cosine.

Understanding AC is essential for analyzing circuits, as it helps in calculating the current at various times using these trigonometric expressions.

Trigonometric Identities

Trigonometric identities involve relationships between trigonometric functions that hold true for every value of the involved variables. They are crucial for simplifying complex equations and making calculations manageable in electrical engineering, particularly with AC circuits.

The identity $ \sin^2 x + \cos^2 x = 1 $ is probably the best known and is used in various derivations and proofs.
For AC circuits, an important identity is expressing a combination of sine and cosine functions as a single cosine function: $ A \cos(t - \phi) $, where $ A $ is the amplitude calculated as $ \sqrt{a^2 + b^2} $.
This transformation provides not only ease of calculation but also a direct interpretation of the signal characteristics, like amplitude and phase.

The efficient use of these identities simplifies finding peaks in AC signals, revealing the maximum current or voltage possible, which is critical in the design and safety of electrical systems.

Amplitude

Amplitude in the context of alternating current is a measure of the maximum value of the current or voltage. It represents the peak of the sinusoidal wave and is crucial for understanding how powerful an AC signal is.

In the equation of an AC signal like $ i(t) = 2 \cos t + 2 \sin t $, the amplitude $ A $ can be calculated using the formula $ A = \sqrt{a^2 + b^2} $, where $ a $ and $ b $ are the coefficients of the cosine and sine terms.
For the given example, this calculation results in the amplitude $ A = 2\sqrt{2} $.
The amplitude reflects the maximum current that can flow through the circuit and is measured in amperes for current or volts for voltage.

Amplitude is a key parameter in the analysis and design of AC circuits, as it helps to determine the load conditions and potential energy outputs of the electrical systems involved.

Problem 85

Problem 86

Other exercises in this chapter

Problem 85

The height of a body moving vertically is given by $$s=-\frac{1}{2} g t^{2}+v_{0} t+s_{0}, \quad g>0$$ with $s$ in meters and $t$ in seconds. Find the body'

View solution

Problem 85

Right, or wrong? Say which for each formula and give a brief reason for each answer. a. $\int(2 x+1)^{2} d x=\frac{(2 x+1)^{3}}{3}+C$ b. \(\int 3(2 x+1)^{2} d

View solution

Problem 86

Right, or wrong? Say which for each formula and give a brief reason for each answer. a. $\int \sqrt{2 x+1} d x=\sqrt{x^{2}+x+C}$ b. \(\int \sqrt{2 x+1} d x=\s

View solution

Problem 87

Then find the extreme values of the function on the interval and say where they occur. $$f(x)=|x-2|+|x+3|, \quad-5 \leq x \leq 5$$

View solution