2bat||40 V8.0 10 A s .0.050 HLdidtLε===×(c) This circuit becomes equivalent to that analyzed in §30-9 when we replace the parallel set of 20000 Ωresistors with R= 10000 Ω. Now, with τL= L/R= 5 ×10–6s, we have t/L= 3/5, and we apply Eq. 30-41: ()353bat11.810A.ieR−−=−≈×(d) The rate of change of the current is figured from the loop rule (and Eq. 30-35): bat||0.LiR−−=Using the values from part (c), we obtain |L| ≈22 V. Then, 2bat22 V4.4 10 A s .0.050 HLdidtL≈×(e) As t→∞, the circuit reaches a steady state condition, so that dibat/dt= 0 and L= 0. The loop rule then leads to 3batbat40 V4.010A.10000Liεε−−−=
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This note was uploaded on 06/03/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.