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Question
Show that in an a.c. circuit containing a pure inductor, the voltage is ahead of current by π/2 in phase ?
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Solution
AC voltage applied to an inductor
Source, v = vm sin ωt
Using Kirchhoff’s loop rule,
`sum epsi(t) =0`
`v - (Ldi)/dt =0 `
`(di)/dt =v/L =v_m/L sin omegat`
Integrating di/dt with respect to time,
`int (di)/dt dt =v_m/L int sin(omega)dt`
`i=-(v_m)/omegaL cos (omegat) +`constant
`-cos omegat = sin(omegat -pi/2)`
`therefore i =i_msin (omegat - pi/2)`
`because`Where,`i_m = (v_m)/(omegaL)`is the amplitude of current
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