Question

The flux linkage through a certain coil of $\mathit{R}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{75}\mathit{\Omega }$resistance would be ${\mathbf{\varphi }}_{\mathbf{B}}\mathbf{=}\mathbf{26}\mathbf{mWb}$if there were a current ofin it. (a) Calculate the inductance of $i=5.5A$the coil. (b) If a 6.0Videal battery were suddenly connected across the coil, how long would it take for the current to rise from 0 to 2.5 A?

Short answer

1. The inductance of the coil is L =$4.7\times {10}^{-3}H$
2. If a 6.0 V ideal battery were suddenly connected across the coil, how long would it take for the current to rise from 0 to 2.5 A is$t=2.3\times {10}^{-3}s$

Step-by-step solution

Given

$$\begin{gathered} R=0.75\Omega \\ {\varphi }_B=26mWb=26\times {10}^{-3}Wb \\ i=5.5A \end{gathered}$$

Understanding the concept

An inductor is a device that can be used to produce a magnetic field in aspecified region. If a current iis established through each of the N windings of an inductor, a magneticflux${\mathit{\varphi }}_{\mathbf{B}}$links those windings. The inductance Lof the inductor is given by equations 30-28. After that using the equation 30-41 for finding the total time required for current rises from 0 to 2.5.

Formula

$$\begin{gathered} L=\frac{{\Phi }_B}{i} \\ i=\frac{\epsilon }{R}\left(1-e^{-\frac{Rt}{L}}\right) \end{gathered}$$

(a) Calculate the inductance of the coil.

By using the equation 30-28 to find the inductance in the coil is

$$\begin{gathered} L=\frac{{\Phi }_B}{i} \\ \mathrm{So}, \\ L=\frac{(26\times {10}^{-3})}{5.5} \\ L=4.7\times {10}^{-3}\mathrm{H} \end{gathered}$$

(b) If a 6.0 V ideal battery were suddenly connected across the coil, calculate how long would it take for the current to rise from 0 to 2.5 A

Ifa constantis introduced into a single loop circuitcontaining a resistance Rand an inductance L, the current rises to an equilibrium value of$\epsilon /R$

So the rise of current is

$i=\frac{\epsilon }{R}\left(1-e^{-\frac{Rt}{L}}\right)$

Rearrange this equation to find the time

$$\begin{gathered} i\frac{\epsilon }{R}=\left(1-e^{-\frac{Rt}{L}}\right) \\ {e}^{-\frac{Rt}{L}}=1-i\frac{R}{\epsilon } \end{gathered}$$

Taking natural log of both side

$$\begin{gathered} -\frac{Rt}{L}=\ln \left(1-i\frac{R}{\epsilon }\right) \\ t=-\frac{L}{R}\ln \left(1-i\frac{R}{\epsilon }\right) \end{gathered}$$

By substituting the value

$$\begin{gathered} t=-\frac{4.7\times {10}^{-3}}{0.75}\ln \left(1-2.5\frac{0.75}{6.0}\right) \\ t=-6.27\times {10}^{-3}\ln (0.6875) \\ t=2.3\times {10}^{-3}s \end{gathered}$$