258 InductanceP32.69Left-hand loop:ε−+−=IIR IR21 220bg.Outside loop:−=IIR LdIdt210.Eliminating I2gives′−=IRLdIdt0.This is of the same form as Equation 32.6, so its solution is of the sameform as Equation 32.7:FIG. P32.69ItReRtLafej=′′−−′1.But ′ =+RRR12and ′ =+R2, so′′=++=RRRRRRR121.ThusRea f=−−′11.P32.70When switch is closed, steady current I0=1.20 A . When the switch isopened after being closed a long time, the current in the right loop isIIe=−02soeIIRt L=0andRtLII=FHGIKJln0.Therefore,LRt====20100015012000956956ln..ln .a fsA 0.250 AH mHΩ.FIG. P32.70P32.71(a)While steady-state conditions exist, a 9.00 mA flows clockwise around the right loop of thecircuit. Immediately after the switch is opened, a 9.00 mA current will flow around the outerloop of the circuit. Applying Kirchhoff’s loop rule to this loop gives:
This is the end of the preview.
access the rest of the document.