n Fig. 30.11, R = 15.0 Ω and the battery emf is 6.30 V. With switch S2 open, switch S1 is closed. After several minutes, S1 is opened and S2 is closed. (a) At 2.00 ms after S1 is opened, the current has decayed to 0.280 A. Calculate the inductance of the coil. (b) How long after S1 is opened will the current reach 1.00% of its original value?

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okpalawalter8 okpalawalter8 · Answer 1 · 2022-07-25T23:24:14+02:00

The inductance of the coil and time is mathematically given as

Generally, the equation for Current is mathematically given as

[tex]\begin{aligned}I(t) &=I_{0} e^{\frac{-t}{\tau}} \\&I(t) =I_{0} e^{\frac{-R t}{L}}\end{aligned}[/tex]

Therefore

[tex]L=\frac{-t R}{\ln \left(\frac{I}{I_{0}}\right)}[/tex]

Hence, the Maximum current is,

[tex]\begin{aligned}I_{0} &=\frac{\varepsilon}{R}=\frac{6.30}{15} \\&I_{0} =0.42 \end{aligned}[/tex]

Thus,

[tex]L &=\frac{-\left(0.280 * 10^{-3})(13.0)}{\ln \left(0.280}{0.42})} \\[/tex]

L=0.1035H

b)

In conclusion, Time taken after S1 is opened is,

[tex](0.01) I_{0}=I_{0} e^{\frac{-t}{\tau}}[/tex]

Time constant is,

[tex]\tau=\frac{L}{R}\\\\\tau=\frac{0.042}{15} \\[/tex]

[tex]\tau=0.0028s[/tex]

[tex](0.01) I_{o} &=I_{o} e^{\frac{-t}{(0.00028)}} \\\\t=-\ln (0.01)(0.0028) \\\\t=-(-4.605)(0.0033) \\\\t=0.01289s[/tex]

t=12.8 ms