Alternating Current

JEE Main level, single correct, medium

40 of 190

Question 40 of 190

An ACAC current is given by I=I1sinωt+I2cosωt.I={I}_{1}sin\omega t+{I}_{2}cos\omega t.
A hot wire ammeter will give a reading:

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Answer: (B) I12+I222\sqrt{\dfrac{{I}_{1}^{2}+{I}_{2}^{2}}{2}}

I=I1sinωt+I2cosωtI={I}_{1}sin\omega t+{I}_{2}cos\omega t
∴I0=I12+I22∴{I}_{0}=\sqrt{{I}_{1}^{2}+{I}_{2}^{2}}
∴Irms=I02=I12+I222∴{I}_{rms}=\dfrac{{I}_{0}}{\sqrt{2}}=\sqrt{\dfrac{{I}_{1}^{2}+{I}_{2}^{2}}{2}}

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