Question

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?

Answer 1

The inductance of the coil  and time is mathematically given as

• L=0.1035H
• t=12.8 ms

What is the inductance of the coil and How long after S1 is opened will the current reach 1.00% of its original value?

Generally, the equation for Current is  mathematically given as

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

Therefore

$L=\frac{-t R}{\ln \left(\frac{I}{I_{0}}\right)}$

Hence, the Maximum current is,

$\begin{aligned}I_{0} &=\frac{\varepsilon}{R}=\frac{6.30}{15} \\&I_{0} =0.42 \end{aligned}$

Thus,

$L &=\frac{-\left(0.280 * 10^{-3})(13.0)}{\ln \left(0.280}{0.42})}$

L=0.1035H

b)

In conclusion, Time taken after S1 is opened is,

$(0.01) I_{0}=I_{0} e^{\frac{-t}{\tau}}$

Time constant is,

$\tau=\frac{L}{R}\\\\\tau=\frac{0.042}{15}$

$\tau=0.0028s$

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

t=12.8 ms

Read more about inductance

brainly.com/question/18575018

#SPJ1