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A sinusoidal current \(i = I\space cos\space \omega\)t has an rms value \(I_{rms} =\)2.10 A. (a) What is the current amplitude? (b) The current is passed through a full-wave rectifier circuit. What is the rectified average current? (c) Which is larger: \(I_{rms}\) or \(I_{rav}\)? Explain, using graphs of \(i^2\) and of the rectified current.

Short Answer

Expert verified
(a) 2.97 A, (b) 1.89 A, (c) \(I_{rms}\) is larger than \(I_{rav}\).

Step by step solution

01

Understanding RMS and Amplitude Relationship

The RMS (Root Mean Square) value for a sinusoidal current is related to the peak (amplitude) value by the formula: \[ I_{rms} = \frac{I_{peak}}{\sqrt{2}} \]Given \( I_{rms} = 2.10 \text{ A} \), we can find the amplitude \( I_{peak} \) using the rearranged formula: \[ I_{peak} = I_{rms} \times \sqrt{2} \] Substitute \( I_{rms} = 2.10 \text{ A} \) into this equation to find \( I_{peak} \).
02

Calculating the Current Amplitude

Plug the RMS value into the equation: \[ I_{peak} = 2.10 \times \sqrt{2} = 2.10 \times 1.414 = 2.97 \text{ A} \] Thus, the current amplitude is \( 2.97 \text{ A} \).
03

Understanding Rectified Average Current

A full-wave rectifier converts AC to DC by making all the negative parts of the waveform positive. The average current through a full-wave rectifier is given by: \[ I_{rav} = \frac{2 \times I_{peak}}{\pi} \] where \( I_{peak} \) is the current amplitude calculated in Step 2.
04

Calculating Rectified Average Current

Using the amplitude \( I_{peak} = 2.97 \text{ A} \), \[ I_{rav} = \frac{2 \times 2.97}{\pi} \approx \frac{5.94}{3.14159} \approx 1.89 \text{ A} \].Thus, the rectified average current is approximately \( 1.89 \text{ A} \).
05

Comparing RMS and Rectified Average Current

To compare \( I_{rms} \) and \( I_{rav} \), consider that ever for a full-wave rectified signal, the power (related to the square of \( i \)) is greater than just its rectified mean current, meaning the RMS value is typically larger due to handling both positive and negative parts around zero more effectively. Thus, \( I_{rms} = 2.10 \text{ A} \) is indeed larger than \( I_{rav} = 1.89 \text{ A} \).

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Key Concepts

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

RMS value
The RMS value, or Root Mean Square value, is a statistical measure of the magnitude of a varying quantity. For sinusoidal alternating currents, it provides a useful measure to compare the current's ability to produce heat in a resistor, similar to how a direct current (DC) would.

To find the RMS value of a sinusoidal current, the equation is given by \[ I_{rms} = \frac{I_{peak}}{\sqrt{2}} \] where \( I_{peak} \) is the peak or maximum current amplitude of the waveform.

In simple terms, the RMS value is approximately 0.707 times the peak value, meaning it essentially takes the effective value of the AC waveform and relates it to a DC value that would produce the same amount of heat in the load.
  • This makes it highly beneficial for electrical engineers and technicians to evaluate AC systems in terms of their DC equivalents.
In the problem, knowing the RMS value allows us to backtrack and find the peak value since the RMS value is given as 2.10 A.
rectified average current
Rectified average current, denoted often as \( I_{rav} \), is the average value of current during one cycle in a rectified waveform. A full-wave rectifier converts both halves of an AC waveform to a DC waveform by flipping the negative part to positive.

When dealing with full-wave rectification, the rectified average current can be calculated using the formula: \[ I_{rav} = \frac{2 \times I_{peak}}{\pi} \] Here, \( I_{peak} \) is the peak current amplitude that we calculate based on the given RMS value.
  • The factor of \( \frac{2}{\pi} \) accounts for the transformation of the sine wave into a constant DC voltage without polarity changes across the entire cycle.
  • This is useful when converting AC to DC as it helps us determine the average effective current for designing circuits and ensuring compatibility with DC components.
In this context, the \( I_{rav} \) was calculated to be approximately 1.89 A, thus showing how the peak value translates into an average over time.
full-wave rectifier circuit
A full-wave rectifier circuit is a type of diode rectifier circuit that efficiently converts an alternating current (AC) signal into a direct current (DC) signal by inverting all the negative cycles of the AC waveform to positive.

This rectification is usually performed using a configuration of diodes known as a bridge rectifier, which makes use of at least four diodes to guide the current in the same direction throughout the load.
  • Full-wave rectification is advantageous as it increases the output current and reduces the ripple effect since it makes the electrical current flow across the load for both parts of the cycle.
  • This process effectively reduces the gaps in the rectified output, providing a smoother and more continuous DC output compared to a half-wave rectifier.
For the given sinusoidal input in the exercise, employing a full-wave rectifier helps convert the magnitude effectively by producing a higher average current output which is easier to filter into a cleaner DC signal compared to its AC input.

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Most popular questions from this chapter

An \(L-R-C\) series circuit has \(R\) = 60.0 \(\Omega\), \(L\) = 0.800 H, and \(C\) = 3.00 \(\times\) 10\(^{-4}\) F. The ac source has voltage amplitude 90.0 V and angular frequency 120 rad/s. (a) What is the maximum energy stored in the inductor? (b) When the energy stored in the inductor is a maximum, how much energy is stored in the capacitor? (c) What is the maximum energy stored in the capacitor?

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A toroidal solenoid has 2900 closely wound turns, cross-sectional area 0.450 cm\(^2\), mean radius 9.00 cm, and resistance \(R\) = 2.80 \(\Omega\). Ignore the variation of the magnetic field across the cross section of the solenoid. What is the amplitude of the current in the solenoid if it is connected to an ac source that has voltage amplitude 24.0 V and frequency 495 Hz?

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