A 8.15 H inductor with negligible resistance is placed in series with a 13.3 V b

Question

A 8.15 H inductor with negligible resistance is placed in series with a 13.3 V battery, a 3.00 0 resistor, and a switch. The switch is closed at timet0 seconds (a) Calculate the initial current at t 0 seconds. ) Calculate the current after we wait a very long time Le. as time approaches infinity) Number (c) Calculate the current at a time of 2.17s Number (d) Determine how long it takes for the current to reach half of its maximum. Number Previous Give Up&View; Solution Check Answer NextExit Hint

Explanation / Answer

a)initial current = 0 A {inductor act as an open circuit}

b)at infinity time inductor behaves as a conducting wire

I1 = 13.3/3 = 4.43A

c)I = I1*{1-exp[- (t*R)/L]}
I = 2.17*{1-exp[ - (4.43*3)/8.15]}
I = 1.745 A

d)0.5 = 1- exp[ - (t*3)/8.15]

e^-3t/8.15 = 0.5

-3t/8.15 = ln(0.5)

3t = 0.693*8.15

t =1.88 secs

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