Question

A coil is connected in series with a 8.41 kΩ resistor. An ideal 68.6 V battery is applied across the two devices, and the current reaches a value of 1.64 mA after 6.26 ms. (a) Find the inductance of the coil. (b) How much energy is stored in the coil at this same moment?

Answer 1

Answer:

The inductance and stored energy are 234 H and $3.15\times10^{-4}\ J$

Explanation:

Given that,

Resistance R= 8.41 kΩ

Voltage V= 68.6 V

Current I = 1.64 mA

Time t = 6.26 ms

We need to calculate the maximum current of the coil

Using formula of maximum current

$I_{max}=\dfrac{V}{R}$

$I_{max}=\dfrac{68.6}{8.41\times10^{3}}$

$I_{max}=8.15\times10^{-3}\ A$

We need to calculate the inductance of the coil

$I_{f}=I(1-e^{\frac{-t}{\tau}})$

$t=-\dfrac{L}{R}In(1-\dfrac{I_{f}}{I_{max}})$

$6.26\times10^{-3}=-\dfrac{L}{8.41\times10^{3}}In(1-\dfrac{1.64\times10^{-3}}{8.15\times10^{-3}})$

$L=\dfrac{{6.26\times10^{-3}\times8.41\times10^{3}}}{ln(1-\dfrac{1.64\times10^{-3}}{8.15\times10^{-3}})}$

$L=234\ H$

(b). We need to calculate the stored energy

Using formula of stored energy

$U=\dfrac{1}{2}LI^2$

$U=\dfrac{1}{2}\times234\times(1.64\times10^{-3})^2$

$U=3.15\times10^{-4}\ J$

Hence, The inductance and stored energy are 234 H and $3.15\times10^{-4}\ J$