Question

A sinusoidal current $\mathit{i}\mathbf{=}\mathit{l}\mathbf{}\mathit{c}\mathit{o}\mathit{s}\mathit{\omega }\mathit{t}$has an rms value${\mathbf{l}}_{\mathbf{rms}}\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{10}$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:${\mathbf{l}}_{\mathbf{rms}}$or${\mathbf{l}}_{\mathbf{avg}}$? Explain, using graphs ofand of the rectified current.

Short answer

The maximum value of the current is 2.97 A. The rectified average current is 1.89 A and the rms current is more than the rectified average current.

Step-by-step solution

Step-1: Formulas used  

The maximum value of the current${\mathbf{l}}_{\mathbf{max}}$is the amplitude of the current wave

${\mathbf{l}}_{\mathbf{max}}\mathbf{=}{\sqrt{\mathbf{2}\mathbf{l}}}_{\mathbf{rms}}$,where is the root mean square current.

The rectified average current${\mathbf{l}}_{\mathbf{avg}}$represents the constant current that produced from using diodes to make the total charges flow similar.

${\mathbf{l}}_{\mathbf{avg}}\mathbf{=}\frac{\mathbf{2}}{\mathbf{\tau \tau }}{\mathbf{l}}_{\mathbf{max}}$

Step-2: Calculations for the current amplitude

$$\begin{gathered} {\mathrm{l}}_{\max }=\sqrt{2}\left(2.1\mathrm{A}\right) \\ =2.97\mathrm{A} \end{gathered}$$

Step-3: Calculations for rectified average current

$$\begin{gathered} {\mathrm{l}}_{\mathrm{avg}}=\frac{2}{\mathrm{\pi }}\left(2.1\mathrm{A}\right) \\ =1.89\mathrm{A} \end{gathered}$$

Step-4: Analysing graphs of i2 and of the rectified current.

The root mean square current is larger than the rectified current ${\mathrm{l}}_{\mathrm{rms}}>{\mathrm{l}}_{\mathrm{avg}}$. And it always ${\mathrm{l}}_{\mathrm{rms}}>{\mathrm{l}}_{\mathrm{avg}}$because the average of $i^2$is more than the average of I.

Therefore the maximum value of the current is 2.97A. The rectified average current is 1.89A and the rms current is more than the rectified average current.