(a) The circuit consists of one generator across one inductor; therefore, \\varepsilon _{m}=V_{L} . The current amplitude is I_m=\\dfrac{\\varepsilon _{m}}{X_{L}}=\\dfrac{\\varepsilon _{m}}{\\omega _{d}L}=\\dfrac{25.0V}{\\left ( 377rad/s \\right )\\left ( 12.7H \\right )}=5.22 \ imes 10^{-3}A .(b) I = I_m sin(\\omega _d t - \\phi) --(i)Current will be maximum when sin(\\omega _d t - \\phi) = 1 or, \\omega _d t - \\phi = \\dfrac{\\pi}{2} At t = 0 potential difference equals zero as sin(\\omega _d t) = 0 \\phi = -\\dfrac{\\pi}{2} This is the phase angle between current and potential difference.Now, I = I_m sin(\\omega_d t + \\dfrac{\\pi}{2}) or, I = I_m(cos \\omega _d t) So, maximum current occurs when cos(\\omega _d t) = 1 That gives t = 0 At this instant potential difference is zero: \\varepsilon = 0 (c) Consider equation \\xi =\\xi _{m}\\sin \\omega _{d}t with \\varepsilon =-\\dfrac{\\varepsilon _{m}}{2} . In order to satisfy this equation, we require \\sin \\left ( \\omega _{d}t \\right )=-1/2 . Now we note that the problem states that ε is increasing in magnitude, which (since it is already negative) means that it is becoming more negative. Thus, differentiating equation with respect to time (and demanding the result be negative) we must also require \\cos \\left ( \\omega _{d}t \\right )< 0 . These conditions imply that \\omega t must equal \\left ( 2n\\pi -5\\pi /6 \\right ) [ n = integer]. Consequently, equation i=I\\sin \\left ( \\omega _{d}t+\\phi\\right) yields (for all values of n ) i=I\\sin \\left ( 2n\\pi -\\frac{5\\pi}{6}-\\frac{\\pi}{2} \\right )=\\left ( 5.22 \ imes 10^{-3}A \\right )\\left ( \\frac{\\sqrt{3}}{2} \\right )=4.51 \ imes 10^{-3}A .