JEE MAIN - Physics (2021 - 17th March Morning Shift - No. 13)
An AC current is given by I = I1 sin$$\omega$$t + I2 cos$$\omega$$t. A hot wire ammeter will give a reading :
$${{{I_1} + {I_2}} \over {\sqrt 2 }}$$
$$\sqrt {{{I_1^2 - I_2^2} \over 2}} $$
$$\sqrt {{{I_1^2 + I_2^2} \over 2}} $$
$${{{I_1} + {I_2}} \over {2\sqrt 2 }}$$
Explanation
$${I_{RMS}} = \sqrt {{{\int {{I^2}dt} } \over {\int {dt} }}} $$
$$I_{RMS}^2 = \int\limits_0^T {{{{{({I_1}\sin \omega t + {I_2}\cos \omega t)}^2}dt} \over T}} $$
$$ = {1 \over T}\int\limits_0^T {(I_1^2{{\sin }^2}\omega t + I_2^2{{\cos }^2}\omega t + 2{I_1}{I_2}\sin \omega t\cos \omega t)dt} $$
$$ = {{I_1^2} \over 2} + {{I_2^2} \over 2} + 0$$
$${I_{RMS}} = \sqrt {{{I_1^2 + I_2^2} \over 2}} $$
$$I_{RMS}^2 = \int\limits_0^T {{{{{({I_1}\sin \omega t + {I_2}\cos \omega t)}^2}dt} \over T}} $$
$$ = {1 \over T}\int\limits_0^T {(I_1^2{{\sin }^2}\omega t + I_2^2{{\cos }^2}\omega t + 2{I_1}{I_2}\sin \omega t\cos \omega t)dt} $$
$$ = {{I_1^2} \over 2} + {{I_2^2} \over 2} + 0$$
$${I_{RMS}} = \sqrt {{{I_1^2 + I_2^2} \over 2}} $$
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