Consider the current pulse i ( t ) shown in Figure a below. The current begins a

Question

Consider the current pulse i(t) shown in Figure a below. The current begins at zero, becomes 10.0 A between t = 0 and t = 200 ?s, and then is zero once again. This pulse is applied to the input of the partial circuit shown in Figure b below. (Let R = 276 ? and L = 12.3 mH.)

Determine the current in the inductor as a function of time. (Use the following variable as necessary: t. If the current in a range is always zero enter 0.)

Explanation / Answer

as input was 0 for t<0, the current=0

for 0<t<200 us,

current will get divided between 276 ohms and 12.3 mH

so if voltage across both of them is v(as they are in parallel)

then 10= (v/276)+i

as v=L*di/dt

substituting,

10=(12.3*0.001*di/dt)/276 + i

taking laplace transform and noting that i(0)=0,

we get

10/s=4.45*10^(-5)*s*I(s)+ I(s)

I(s)=10/(s*(4.45*10^(-5)*s+1))

taking inverse laplace transform,

we get

i(t)=10-10*exp(-2.247*10^4*t)

at t=200 us,i(t)=9.8883 A

for t>200 us,the current in inductor will get discharged via resistance.
so current=i(t)=9.8883*exp(-(t-200)/T) where t is in us.

where T=time constant=L/R=44.5 us

so i(t)=9.8883*exp(-(t-200)/(44.5))

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