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A direct current of 5A is superimposed on an alternating current I=10sinωt flowing through a wire, The effective value of the resulting current will be
(A) (152)A
(B) 53A
(C) 55A
(D) 15A

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
IR=Idc+Iac
According to the question, Idc=5A and Iac=10sinωt. Substituting these in the above equation, we get the resultant current as
IR=5+10sinωt
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
IR2=(5+10sinωt)2
We know that (a+b)2=a2+2ab+b2. So we write the above equation as
IR2=52+2(5)(10sinωt)+(10sinωt)2
IR2=25+100sinωt+100sin2ωt (1)
Now, we know that the mean of a function is given by
M=1T0Tf(t)dt
The period of the function IR2 is clearly equal to 2πω. So we integrate it from 0 to 2πω to get its mean as
M=12π/ω02πωIR2dt
M=ω2π02πωIR2dt
Putting (1) above, we get
M=ω2π02πω(25+100sinωt+100sin2ωt)dt
M=ω2π(02πω25dt+02πω100sinωtdt+02πω100sin2ωtdt)
We know that the average value of the sinusoidal functions is equal to zero. So we can put 02πω100sinωtdt=0 in the above equation to get
M=ω2π(02πω25dt+02πω100sin2ωtdt)
We know that sin2θ=1cos2θ2. So we write the above equation asM=ω2π(02πω25dt+02πω100(1cos2ωt2)dt)
M=ω2π(2502πωdt+5002πω(1cos2ωt)dt)
M=ω2π(2502πωdt+5002πωdt02πωcos2ωtdt)
We know that dt=t. So we get
M=ω2π(25[t]02πω+5025[t]02πω5002πωcos2ωtdt)
M=ω2π(25(2πω)+50(2πω)02πωcos2ωtdt)
Since cos2ωt is also sinusoidal, its average over a cycle will also be zero, that is, 02πωcos2ωtdt=0.
M=ω2π(25(2πω)+50(2πω))
M=(25+50)=75A2
Now, we take the square root of this mean value to get the final RMS value of the resulting current as
RMS=M
RMS=75A=53A
Thus the effective value of the resulting current is equal to 53A.

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 0 to 2π.