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) $\left(2 \mathrm{I}_{0} / \pi\right)$.

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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

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