Answer
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Hint:A full wave rectifier’s duty is to rectify the negative element of the voltage given as the input, into a positive voltage which is further converted to a DC voltage using configuration of a diode. This major point is to be remembered while solving the given question under given conditions.
Complete answer:
Let us consider an AC current wave with a peak value of ${{I}_{0}}$. Then when it passes through a full wave rectifier, it converts the AC wave to DC wave, and all the negative components are eliminated. In both the cases we can say that there is no variation in the peak value ${{I}_{0}}$.
Now we have to find the average value of the output direct current that is produced from the rectifier. The average value of the direct output current in a full wave rectifier can be found by calculating the average value of the current over a given cycle.
That is, $2{{I}_{0}}/\pi $
Thus, we can say that option D is the correct answer.
Note:The given method is the easiest way to solve the given problem. We can also find the answer using integration. That is, the average value of the output direct current can be found from the area under the graph. Hence, the area of the curve after rectification can be integrated so as to find the required value in the question. This method applies the direct formula to find the answer.
Complete answer:
Let us consider an AC current wave with a peak value of ${{I}_{0}}$. Then when it passes through a full wave rectifier, it converts the AC wave to DC wave, and all the negative components are eliminated. In both the cases we can say that there is no variation in the peak value ${{I}_{0}}$.
Now we have to find the average value of the output direct current that is produced from the rectifier. The average value of the direct output current in a full wave rectifier can be found by calculating the average value of the current over a given cycle.
That is, $2{{I}_{0}}/\pi $
Thus, we can say that option D is the correct answer.
Note:The given method is the easiest way to solve the given problem. We can also find the answer using integration. That is, the average value of the output direct current can be found from the area under the graph. Hence, the area of the curve after rectification can be integrated so as to find the required value in the question. This method applies the direct formula to find the answer.
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