The figure shows the current as a function of | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) In a full-wave rectifier,the output current consists of a series of half-sinusoidal pulses. To find the average current $(I_{avg})$,we integrate t

Vedclass · Accepted Answer

$\frac{2I_0}{\pi}$

Vedclass · Answer

(D) In a full-wave rectifier,the output current consists of a series of half-sinusoidal pulses. To find the average current $(I_{avg})$,we integrate the current over one complete cycle (from $t=0$ to $t=T/2$) and divide by the time interval $(T/2)$. The current is given by $I = I_0 \sin(\omega t)$. $I_{avg} = \frac{1}{T/2} \int_{0}^{T/2} I_0 \sin(\omega t) dt$ $I_{avg} = \frac{2}{T} \cdot I_0 \left[ -\frac{\cos(\omega t)}{\omega} \right]_{0}^{T/2}$ Since $\omega = \frac{2\pi}{T}$,we have $\omega \cdot \frac{T}{2} = \pi$. $I_{avg} = \frac{2 I_0}{T \omega} [-\cos(\pi) + \cos(0)] = \frac{2 I_0}{T (2\pi/T)} [1 + 1] = \frac{2 I_0}{2\pi} \cdot 2 = \frac{2 I_0}{\pi}$.

The figure shows the current as a function of time at the output of a full-wave rectifier. The average value of this current is:

Similar Questions

The figure below shows a circuit and its input voltage $V_i$ as a function of time $t$. Assuming the diodes to be ideal,which of the following graphs depicts the output voltage $V_o$ as a function of time $t$?

If in a $p-n$ junction circuit,a square input signal of $5\,V$ is applied as shown,then the output across $R_L$ will be:

If the following input signal is sent through a $PN$-junction diode,then the output signal across $R_L$ will be:

The function of a rectifier is

Vedclass Test Series

Exam Paper Generator

Online Exam Module