Question

A circuit has R = 16.6 Ω and the battery emf is 6.50 V . With switch S2 open, switch S1 is closed. After several minutes, S1 is opened and S2 is closed.

1) At 2.14 ms after S1 is opened, the current has decayed to 0.235 A . Calculate the inductance of the coil.

2) How long after S1 is opened will the current reach 1.00% of its original value?
-----------------------------------------
I know that inductance (L) of a coil is mu(N^2)A / length but I'm not sure if this formula applies here. I don't really know where to start.

Answer 1

1. VR = 6.50 Volts(t = 0).
VR = I*R = 0.235 * 16.6 = 3.901 Volts(t = 2.14 mS).

6.5/e^(t/T) = 3.90,
e^(t/T) = 6.50/3.90 = 1.666,
t/T = 0.5104,
T = t/0.5104 = 0.00214/0.5104 = 0.004193.

T = L/R.
0.004193 = L/16.6,
L = 0.0696 henrys = 69.6 mH.

Answer 2

2. I max = E/R = 6.5/116.6 = 0.392A.
I min = 0.01 * 0.392 = 0.00392A.

Imax/e^(t/T) = Imin.
e^(t/T) = Imax/Imin = 0.392/0.00392 = 99.89,
t/T = 4.60.

T = L/R = 0.0696/16.6 = 0.004193 = L-R time constant.

t/0.004193 = 4.60.
t = ?

Answer 3

post