Chapter 19: Problem 2583

The output current versus time curve of a rectifier is shown in the figure. The average value of the output-current is (A) 0 (B) $\left(\mathrm{I}_{0} / 2\right)$ (C) $\left(2 \mathrm{I}_{0} / \pi\right)$ (D) I $_{0}$

Short Answer

Expert verified

The average value of the output current for the given rectifier is (C) $\left(2 \mathrm{I}_{0} / \pi\right)$.

Step by step solution

Identify the waveform

The question describes the output-current curve of a rectifier. The provided options suggest that the curve is that of a half-wave rectifier (HW). The output of a half-wave rectifier has a sinusoidal positive half-cycle, and the negative half-cycle is zero. Therefore, the current waveform is given by, $$ i(t) = \begin{cases} I_0 \sin(\omega t), & 0\leq\omega t\leq\pi \\ 0, & \pi\leq\omega t\leq2\pi \end{cases} $$ where $I_0$ is the peak current, $\omega$ is the angular frequency, and $t$ is time.

Calculate the average output current

To determine the average value ($I_{avg}$) of the output current, we will calculate the area under the curve over one complete time period and divide it by the time period ($T$). Since the curve is symmetric and the output is zero during the negative half-cycle, we can focus on the positive half-cycle for the calculation. The average output current can be found by integrating the waveform over one time period $(0\leq\omega t\leq\pi)$ and dividing by the time period: $$ I_{avg} = \dfrac{1}{T} \int_0^T i(t) dt $$

Perform the integration

First, we will integrate the waveform for the positive half-cycle (from $0$ to $\pi$) and then divide by the time period $T=\dfrac{2\pi}{\omega}$. This will give us the average output current: $$ I_{avg} = \dfrac{1}{T} \int_{0}^{\pi / \omega} I_0 \sin (\omega t) dt $$ Applying the integral of $\sin(\omega t)$: $$ I_{avg} = \dfrac{1}{T} \left[-\dfrac{I_0}{\omega} \cos(\omega t)\right]_{0}^{\pi / \omega} $$ Now, substitute the limits of integration: $$ I_{avg} = \dfrac{1}{T}\left(-\dfrac{I_0}{\omega} \cos(\pi) - \left(-\dfrac{I_0}{\omega} \cos(0)\right)\right) $$

Simplify the expression

Now, we will simplify the expression and substitute the time period $T$: $$ I_{avg} = \dfrac{1}{\frac{2 \pi}{\omega}}\left(-\dfrac{I_0}{\omega}(-1) - \left(-\dfrac{I_0}{\omega}\right)\right) $$ Simplify further: $$ I_{avg} = \dfrac{\omega}{2\pi}\left(\dfrac{2I_0}{\omega}\right) $$ Finally, the average output current: $$ I_{avg} = \dfrac{2I_0}{\pi} $$ With the calculated average value of the output current, the correct option is (C) $\left(2 \mathrm{I}_{0} / \pi\right)$.

Unlock Step-by-Step Solutions & Ace Your Exams!

Full Textbook Solutions
Get detailed explanations and key concepts
Unlimited Al creation
Al flashcards, explanations, exams and more...
Ads-free access
To over 500 millions flashcards
Money-back guarantee
We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

Recommended explanations on English Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

The output current versus time curve of a rectifier is shown in the figure. The average value of the output-current is (A) 0 (B) \(\left(\mathrm{I}_{0} / 2\right)\) (C) \(\left(2 \mathrm{I}_{0} / \pi\right)\) (D) I \(_{0}\)

Short Answer

Step by step solution

Identify the waveform

Calculate the average output current

Perform the integration

Simplify the expression

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on English Textbooks

Prosody

English Grammar

Semiotics

Pragmatics

Listening and Speaking

Blog

Study anywhere. Anytime. Across all devices.