60. (a) The energy delivered by the battery is the integral of Eq. 28-14 (where we use Eq. 31-43 for thecurrent):Zt0Pbatterydt=Zt0E2R³1−e−Rt/L´dt=E2R·t+LR³e−Rt/L−1´¸=(10.0V)26.70 Ω·2.00 s +(5.50 H)(e−(6.70 Ω)(2.00 s)/5.50 H−1)6.70 Ω¸=18.7J.(b) The energy stored in the magnetic Feld is given by Eq. 31-51:UB=12Li2(t)=12LµER¶2(1−
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