
A direct current of is superimposed on an alternating current flowing through a wire, The effective value of the resulting current will be
(A)
(B)
(C)
(D)
Answer
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Hint We can obtain the equation of the resulting current through the wire by adding the direct current to the alternating current. The effective value of current is equal to the root mean square value of the current. So we need to take the square, calculate its mean by integrating it over a time period, and take the square root to obtain the RMS value which will be the required effective current.
Complete step-by-step solution:
Since the direct current is superimposed over the alternating current, the resulting current will be equal to the sum of the direct and the alternating current. So we have
According to the question, and . Substituting these in the above equation, we get the resultant current as
Now, we know that the effective value of the current is nothing but the RMS, or the root mean square value of the current. By the definition of RMS, we have to take the square, then take the mean, and finally take the square root of the current. So on squaring both sides of the above equation, we get
We know that . So we write the above equation as
(1)
Now, we know that the mean of a function is given by
The period of the function is clearly equal to . So we integrate it from to to get its mean as
Putting (1) above, we get
We know that the average value of the sinusoidal functions is equal to zero. So we can put in the above equation to get
We know that . So we write the above equation as
We know that . So we get
Since is also sinusoidal, its average over a cycle will also be zero, that is, .
Now, we take the square root of this mean value to get the final RMS value of the resulting current as
Thus the effective value of the resulting current is equal to .
Hence, the correct answer is option B.
Note: We can integrate the square of the current between any time interval which makes a complete time period. But it will be convenient to set the lower limit of the integral equal to zero. So we chose the period from to .
Complete step-by-step solution:
Since the direct current is superimposed over the alternating current, the resulting current will be equal to the sum of the direct and the alternating current. So we have
According to the question,
Now, we know that the effective value of the current is nothing but the RMS, or the root mean square value of the current. By the definition of RMS, we have to take the square, then take the mean, and finally take the square root of the current. So on squaring both sides of the above equation, we get
We know that
Now, we know that the mean of a function is given by
The period of the function
Putting (1) above, we get
We know that the average value of the sinusoidal functions is equal to zero. So we can put
We know that
We know that
Since
Now, we take the square root of this mean value to get the final RMS value of the resulting current as
Thus the effective value of the resulting current is equal to
Hence, the correct answer is option B.
Note: We can integrate the square of the current between any time interval which makes a complete time period. But it will be convenient to set the lower limit of the integral equal to zero. So we chose the period from
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