"An AC current is given by I = I1 sin$\\omega$t + I2 cos$\\omega$t. A hot wire ammeter will give a reading :"

"${I_{RMS}} = \\sqrt {{{\\int {{I^2}dt} } \\over {\\int {dt} }}}$ $$I_{RMS}^2 = \\int\\limits_0^T {{{{{({I_1}\\sin \\omega t + {I_2}\\cos \\omega t)}^2}dt} \\over T}} $$ $$ = {1 \\over T}\\int\\limits_0^T {(I_1^2{{\\sin }^2}\\omega t + I_2^2{{\\cos }^2}\\omega t + 2{I_1}{I_2}\\sin \\omega t\\cos \\omega t)dt} $$ $= {{I_1^2} \\over 2} + {{I_2^2} \\over 2} + 0$ ${I_{RMS}} = \\sqrt {{{I_1^2 + I_2^2} \\over 2}}$"

Medium MCQ +4 / -1 PYQ · JEE Mains 2021

An AC current is given by I = I₁ sin$\omega$t + I₂ cos$\omega$t. A hot wire ammeter will give a reading :

A ${{{I_1} + {I_2}} \over {\sqrt 2 }}$
B $\sqrt {{{I_1^2 - I_2^2} \over 2}}$
C $\sqrt {{{I_1^2 + I_2^2} \over 2}}$ Correct answer
D ${{{I_1} + {I_2}} \over {2\sqrt 2 }}$

Solution

${I_{RMS}} = \sqrt {{{\int {{I^2}dt} } \over {\int {dt} }}}$ $$I_{RMS}^2 = \int\limits_0^T {{{{{({I_1}\sin \omega t + {I_2}\cos \omega t)}^2}dt} \over T}} $$ $$ = {1 \over T}\int\limits_0^T {(I_1^2{{\sin }^2}\omega t + I_2^2{{\cos }^2}\omega t + 2{I_1}{I_2}\sin \omega t\cos \omega t)dt} $$ $= {{I_1^2} \over 2} + {{I_2^2} \over 2} + 0$ ${I_{RMS}} = \sqrt {{{I_1^2 + I_2^2} \over 2}}$

About this question

Subject: Physics · Chapter: Alternating Current · Topic: AC Circuits: R, L, C

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An AC current is given by I = I1 sin$\omega$t + I2 cos$\omega$t. A hot wire ammeter will give a reading :

Solution

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An AC current is given by I = I₁ sin$\omega$t + I₂ cos$\omega$t. A hot wire ammeter will give a reading :