A 140-mH inductor and a 5.00- resistor are connected with a switch to a 6.00-V b

Question

A 140-mH inductor and a 5.00- resistor are connected with a switch to a 6.00-V battery as shown in the figure below.

(a) After the switch is first thrown to a (connecting the battery), what time interval elapses before the current reaches 220 mA?

Your response differs from the correct answer by more than 10%. Double check your calculations. ms

(b) What is the current in the inductor 10.0 s after the switch is closed?
  A

(c) Now the switch is quickly thrown from a to b. What time interval elapses before the current in the inductor falls to 160 mA?
ms

Explanation / Answer

a) T = L / R = 140 mH / 5 = 0.028 s

Imax = E / R = 6 / 5 = 1.2 A

I = Imax * (1 - e-t/T)

0.220 = 1.2 (1 - e-t/T)

e-t/T = 0.817

t = -T * ln (0.817) = -0.028 * -0.202

t = 5.66 ms

b) I = Imax * (1 - e-t/T)

I = 1.2 (1 - e-10/0.028) = 1.2 (1 - 0)

current in the inductor = 1.2 A

c)  I = Imax * e-t/T

0.16 = 1.2 * e-t/T

t = -T * ln 0.133 = -0.028 * -2.017

t = 56.4 ms

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