Switch S in Fig. 30-61 is closed at time t = 0 s, initiating the buildup of current in the 12.2 mH inductor and the 14.8 Ω resistor. At what time (in ms) is the emf across the inductor equal to the potential difference across the resistor?
Figure: http://edugen.wileyplus.com/edugen/cours…
Figure: http://edugen.wileyplus.com/edugen/cours…
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Link doesn't show a diagram btw
First find the L/R time constant: R/L = 1213
Rt/L = 1213t
Since this is a series ckt, the current is the same in R and L
IL = Vs/R *(1-e^-1213t) Let Vs be the voltage supply.
VR = IL*R = Vs(1-e^-1213t) = Vs-Vs*e^-1213t
At t=0 VR =0, at t approaching infinity VR = Vs
VL = LdIL/dt = Vs*e^-1213t
At t=0 VL = Vs, at t approaching infinity VL=0
Set VL = VR
Vs[2*e^-1213t = Vs
e^-1213t =1/2
t = [-1/1213]*ln(1/2) = 0.57ms
First find the L/R time constant: R/L = 1213
Rt/L = 1213t
Since this is a series ckt, the current is the same in R and L
IL = Vs/R *(1-e^-1213t) Let Vs be the voltage supply.
VR = IL*R = Vs(1-e^-1213t) = Vs-Vs*e^-1213t
At t=0 VR =0, at t approaching infinity VR = Vs
VL = LdIL/dt = Vs*e^-1213t
At t=0 VL = Vs, at t approaching infinity VL=0
Set VL = VR
Vs[2*e^-1213t = Vs
e^-1213t =1/2
t = [-1/1213]*ln(1/2) = 0.57ms