Question

In the following figure (Figure 1), R = 15.3 and the battery emf is 6.50 V . With switch S, open, switch Si is closed. After several minutes, S, is opened and S2 is closed. Part A At 2.18 ms after Si is opened, the current has decayed to 0.340 A. Calculate the inductance of the coil. Express your answer to three significant figures and include the appropriate units. 0 НА ? L = Value Units Submit Request Answer Part B Figure < 1 of 1 How long after Sy is opened will the current reach 1.00% of its original value? Express your answer to three significant figures and include the appropriate units. 7 НА ? 1 = Value Units 0000 L Submit Request Answer R Provide Feedback S2

Answer 1

Part A.

Initially when switch S1 is closed, and S2 is open, then after several minutes inductor will be fully charged. Now given that S2 is closed and S1 is opened, So current will start decaying in the Inductor.

Current decya in inductor is given by:

i = i0*e^(-tR/L)

Here given that: i0 = maximum current = E0/R = 6.50/15.3 = 0.425 Amp

Now after t = 2.18 ms, current in inductor is 0.340 Amp, So

e^(-tR/L) = i/i0

-tR/L = ln (i/i0)

L = -t*R/ln (i/i0)

R = resistance = 15.3 ohm, So

L = -2.18*10^-3*15.3/ln (0.340/0.425) = 0.14947 H

In three significant figures

L = 0.149 H = Inductance of the coil

Part B.

Now we need time when i = 1.00% of i0 = 0.01*i0, So

-tR/L = ln (i/i0)

t = -(L/R)*ln (i/i0)

t = -(0.14947/15.3)*ln (0.01*i0/i0)

t = 0.044989 sec

In three significant figures

t = 0.0450 sec = 45.0*10^-3 s = 45.0 ms

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