In the following figure ( 6754630015.jpg ) , R = 14.8 ? and the battery emf is 6
ID: 2244858 • Letter: I
Question
In the following figure (6754630015.jpg) , R = 14.8? and the battery emf is 6.40V . With switch S2 open, switch S1is closed. After several minutes, S1 is opened and S2 is closed.
A. At 2.15
ms after S1 is opened, the current has decayed to 0.299A . Calculate the inductance of the coil.
B. How long after
S1 is opened will the current reach 1.00% of its original value?
In the following figure (6754630015.jpg) , R = 14.8? and the battery emf is 6.40V . With switch S2 open, switch S1is closed. After several minutes, S1 is opened and S2 is closed. At 2.15 ms after S1 is opened, the current has decayed to 0.299A . Calculate the inductance of the coil. How long after S1 is opened will the current reach 1.00% of its original value?Explanation / Answer
a)
Io=E/R =6.4/14.8 =0.43 A
I=Ioe^(-t/T)
0.299 =0.432 e^(-2.15m/T)
-2.15m/T =ln(0.69)=-0.369
T=5.83 ms
T=L/R =5.83*10^-3
L=14.8*5.83*10^-3
L=0.086 H or 86.24 mH
b)
given
I=0.01Io
0.01Io =Ioe^(-t/5.83*10^-3)
ln(0.01) =-t/5.83*10^-3
-4.605 =-t/5.83*10^-3
t=0.02685 s or 26.85 ms
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