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www.GOALias.blogspot.comAlternating Current2d q dq qL + R + = v sin2mω t(7.28)dtdtCThis is like the equation for a forced, damped oscillator, [see Eq. {14.37(b)}in Class XI Physics Textbook]. Let us assume a solutionq = q msin (ω t + θ)[7.29(a)]so that d q= qmω cos( ωt+ θ)[7.29(b)]dt2d q2and =− q sin( )2 mω ωt+ θ[7.29(c)]dtSubstituting these values in Eq. (7.28), we get[ cos( ) ( )sin( )]qmω R ωt + θ + XC − XLωt+ θ = vmsin ω t(7.30)where we have used the relation X c= 1/ωC, X L= ω L. Multiplying and2dividing Eq. (7.30) by ( ) 2Z = R + X − X , we havecLR( XCXL)q ω ⎡m Z cos( t ) sin( t )Zω θ −Zω θ ⎤⎢ + + + ⎥⎣⎦= v sin ω t (7.31)mNow, letandRZ= cosφ( XC− XL)= sin φZX − Xso that φRSubstituting this in Eq. (7.31) and simplifying, we get:−1C L= tan(7.32)q ω Z cos( ωt + θ − φ) = v sin ωt(7.33)mComparing the two sides of this equation, we see thatwherev = q ω Z = i Zm m mmmmi = q ω[7.33(a)]ππand θ − φ = − or θ =− + φ[7.33(b)]22Therefore, the current in the circuit isd qi = = qmω cos( ωt+ θ)dt= i mcos(ωt + θ)or i = i msin(ωt + φ ) (7.34)vmvmwhere im= =Z 2 2R + ( XC− XL)−1X C − X Land φ = tanR[7.34(a)]247