AB. \[0\le t\le 15\] \[i(t)=\SQRT{5}\,{{t}^{2}}c\] \[{{i}_{rms}}=\sqrt{<{{i}^{2}}>}=\sqrt{\frac{5\int\limits_{0}^{1}{\,{{t}^{4}}DT}}{\int\limits_{0}^{1}{dt}}}=\,1\,mA\]

"> AB. \[0\le t\le 15\] \[i(t)=\SQRT{5}\,{{t}^{2}}c\] \[{{i}_{rms}}=\sqrt{<{{i}^{2}}>}=\sqrt{\frac{5\int\limits_{0}^{1}{\,{{t}^{4}}DT}}{\int\limits_{0}^{1}{dt}}}=\,1\,mA\]

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Figure shows one cycle of an alternating current with the segments AB, BC, CD, DE being symmetrical and parabolic. The root mean square value of this current over one cycle is x mA, find x.  

Joint Entrance Exam - Main (JEE Main) Physics in Joint Entrance Exam - Main (JEE Main) . 10 months ago

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[a] RMS value over one cycle=RMS value over AB. \[0\le t\le 15\] \[i(t)=\SQRT{5}\,{{t}^{2}}c\] \[{{i}_{rms}}=\sqrt{<{{i}^{2}}>}=\sqrt{\frac{5\int\limits_{0}^{1}{\,{{t}^{4}}DT}}{\int\limits_{0}^{1}{dt}}}=\,1\,mA\]

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