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Physics Question on Alternating current

A direct current of 5 A5\, A is superposed on an alternating current I=10 sin⁡ωtI = 10 \,\sin \omega t flowing through the wire. The effective value of the resulting current will be

A

(15/2)A(15/2) A

B

53A5\sqrt{3 A}

C

55A 5\sqrt{5A}

D

15 A15\, A

Answer

53A5\sqrt{3 A}

Explanation

Solution

Total carrent, l=(5+10 sin⁡ωt)l=(5+10\, \sin \omega t)
⇒Ieff=[∫0TI2dt∫0Tdt]1/2\Rightarrow I_{eff}=\left[\frac{\int_{0}^{T} I^{2} d t}{\int_{0}^{T} d t}\right]^{1 / 2}
=[1T∫0T(5+10 sin⁡ωt)2dt]1/2=\left[\frac{1}{T} \int_{0}^{T}(5+10\, \sin \omega t)^{2} d t\right]^{1 /2}
=[1T∫0T(25+100sin⁡ωt+100sin⁡2ωt)]1/2=\left[\frac{1}{T} \int_{0}^{T}\left(25+100 \sin \omega t+100 \sin ^{2} \omega t\right)\right]^{1 /2}
But 1T∫0Tsin⁡ωt.dt=0\frac{1}{T} \int_{0}^{T} \sin \omega t . d t=0 and
1T∫0Tsin⁡2ωt.dt=12\frac{1}{T} \int_{0}^{T} \sin ^{2} \omega t . d t=\frac{1}{2}
So, Ieff=[25+12×100]1/2I_{eff}=\left[25+\frac{1}{2} \times 100\right]^{1/2}
=53A=5 \sqrt{3 A}