The circuit shown has the following component values: E = 11 V, R1 = 68 k, R2 =

Question

The circuit shown has the following component values: E = 11 V, R1 = 68 k, R2 = 1.5 k, and L = 15 mH.

a. Determine the time constant () of the circuit after the switch closes. ms
b. After the switch closes, determine the steady-state current through the inductor. mA
c. Assume the switch has been closed for a long period of time, determine the energy stored in the inductor. J
d. Assume the switch has been closed for a long period of time. Determine the amount of voltage that appears across the inductor at the instant the switch opens (pay careful attention to the polarity). v

Explanation / Answer

a) (68000+1500)*i+0.015*di/dt=0==>=0.015H/(68000+1500)=0.00022ms

b) at steady state there is no back emf==>i coil=11V/1500=7.33mA=Io==>iL=Io*e^(-/216E-9)=0A

c) WL=1/2*L*(iL)2=0 joules for t=

d) vcoil+vR1=0==>vcoil=-vR1==>-7.33mA*1500=-11V

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