A 12.0 V dc battery having no appreciable internal resistance, a 154.0 resistor,

Question

A 12.0 V dc battery having no appreciable internal resistance, a 154.0 resistor, an 11.0 mHinductor, and an open switch are all connected in series.

A.) After the switch is closed, what is the time constant for this circuit ? = __________ us

B.) After the switch is closed, what is the maximum current that flows through it ? Imax = ________ A

C.) What is the current 73.9 s after the switch is closed ? i = _________ A

D.) After the switch is closed, what is the maximum energy stored in the inductor? Umax = ________ uJ

Explanation / Answer

1)
time constant T = L/R
T = 11*10^-3/154
T = 71.42us

2)
After a long time the inductor will appear to be a short circuit, and the maximum current is
I = V/R
I = 12/154
I = 0.078 A

3)
i = Io*(1 - e-t/T)
i = 0.078*(1 - e-71.42/71.42)
i = 0.05A

4)
W = L*I^2/2
W = 11*10^-3*(0.05)^2/2
W = 1.375x10^-5J

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