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NEET PHYSICSALTERNATING CURRENTMedium

Question

An AC source given by V=Vmsin⁡(ωt)V = V_m \sin(\omega t) is connected to a pure inductor LL in a circuit and ImI_m is the peak value of the AC current. The instantaneous power supplied to the inductor is:

A

VmIm2sin⁡(2ωt)\frac{V_m I_m}{2} \sin(2\omega t)

B

−VmIm2sin⁡(2ωt)-\frac{V_m I_m}{2} \sin(2\omega t)

C

VmImsin⁡2(ωt)V_m I_m \sin^2(\omega t)

D

−VmImsin⁡2(ωt)-V_m I_m \sin^2(\omega t)

Step-by-Step Solution

According to the sources, when an AC voltage V=Vmsin⁡(ωt)V = V_m \sin(\omega t) is applied to a pure inductor, the resulting current lags the voltage by one-quarter cycle (π/2\pi/2), which is expressed as i=Imsin⁡(ωt−π/2)=−Imcos⁡(ωt)i = I_m \sin(\omega t - \pi/2) = -I_m \cos(\omega t) . The instantaneous power (pLp_L) supplied to the inductor is the product of the instantaneous voltage and current: pL=Vi=(Vmsin⁡ωt)(−Imcos⁡ωt)=−VmImsin⁡ωtcos⁡ωtp_L = Vi = (V_m \sin \omega t)(-I_m \cos \omega t) = -V_m I_m \sin \omega t \cos \omega t. By applying the trigonometric identity sin⁡(2ωt)=2sin⁡ωtcos⁡ωt\sin(2\omega t) = 2 \sin \omega t \cos \omega t, the expression simplifies to pL=−VmIm2sin⁡(2ωt)p_L = -\frac{V_m I_m}{2} \sin(2\omega t) .

Exam Context & Concepts Covered

This question aligns with the NEET PHYSICS syllabus, specifically targeting concepts from ALTERNATING CURRENT. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

PHYSICSALTERNATING CURRENTsourcesinomegaconnectedinductorcircuit

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