Question

Derive Eq. 8.43. [Hint: Use the method of Section 7.2.4, building the two currents up from zero to their final values.]

Short answer

The equation 8.43 is derived as$\mathit{W}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}{\mathit{L}}_{\mathbf{a}}{\mathit{l}}_{\mathbf{a}}\mathbf{+}\frac{\mathbf{1}}{\mathbf{2}}{\mathit{L}}_{\mathbf{b}}{\mathit{l}}_{\mathbf{b}}\mathbf{+}\mathit{M}{\mathit{L}}_{\mathbf{a}}{\mathit{l}}_{\mathbf{b}}$ .

Step-by-step solution

Expression for the power delivered to the two loops:

Write the expression for the power delivered to the two loops.

$\mathit{W}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}\mathit{L}{\mathit{l}}^{\mathbf{2}}$

Here, L is the inductance, and I is the current.

Derive equation 8.43 as W=12LaIa2+12LbIb2+MIaIb

Write the expression for the total energy stored in the pair of coils.

$\frac{\mathbf{d}\mathbf{W}}{\mathbf{d}\mathbf{t}}\mathbf{=}\mathbf{-}{\mathit{\epsilon }}_{\mathbf{a}}{\mathit{l}}_{\mathbf{a}}\mathbf{-}{\mathit{\epsilon }}_{\mathbf{b}}{\mathit{l}}_{\mathbf{b}}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{\left(}\mathbf{1}\mathbf{\right)}$

Where,

$$\begin{gathered} {\mathit{\epsilon }}_{\mathbf{a}}\mathbf{=}\mathbf{-}{\mathit{L}}_{\mathbf{a}}\frac{\mathbf{d}\mathbf{l}}{\mathbf{d}\mathbf{t}}\mathbf{-}\mathit{M}\frac{\mathbf{d}{\mathbf{l}}_{\mathbf{a}}}{\mathbf{d}\mathbf{t}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{-}{\mathit{L}}_{\mathbf{b}}\frac{\mathbf{d}{\mathbf{l}}_{\mathbf{b}}}{\mathbf{d}\mathbf{t}}\mathbf{-}\mathit{M}\frac{\mathbf{d}{\mathbf{l}}_{\mathbf{b}}}{\mathbf{d}\mathbf{t}} \end{gathered}$$

Substitute the known values in equation (1).

$$\begin{gathered} \frac{\mathbf{d}\mathbf{W}}{\mathbf{d}\mathbf{t}}\mathbf{=}\mathbf{-}\left(-L_a\frac{d{l}_a}{dt}-M\frac{d{l}_a}{dt}\right)\mathbf{-}\left(-L_a\frac{dl}{dt}-M\frac{d{l}_b}{dt}\right) \\ \frac{\mathbf{d}\mathbf{W}}{\mathbf{d}\mathbf{t}}\mathbf{=}\mathbf{-}\left(-L_a{l}_a\frac{d{l}_a}{dt}-M{l}_a\frac{d{l}_b}{dt}\right)\mathbf{+}\left(L_b\frac{dl}{dt}+M\frac{d{l}_b}{dt}\right) \\ \frac{\mathbf{d}\mathbf{W}}{\mathbf{d}\mathbf{t}}\mathbf{=}\frac{\mathbf{d}}{\mathbf{d}\mathbf{t}}\left(-\frac{1}{2}{L}_a{l}_a^2+\frac{1}{2}{L}_a{l}_b^2+M{l}_a{l}_b\right) \end{gathered}$$

Integrate with respect to t.

$\mathit{W}\mathbf{=}\mathbf{-}\frac{\mathbf{1}}{\mathbf{2}}{\mathit{L}}_{\mathbf{a}}{\mathit{l}}_{\mathbf{a}}^{\mathbf{2}}\mathbf{+}\frac{\mathbf{1}}{\mathbf{2}}{\mathit{L}}_{\mathbf{b}}{\mathit{l}}_{\mathbf{b}}^{\mathbf{2}}\mathbf{+}\mathit{M}{\mathit{l}}_{\mathbf{a}}{\mathit{l}}_{\mathbf{b}}$

Therefore, equation 8.43 is derived as$\mathit{W}\mathbf{=}\mathbf{-}\frac{\mathbf{1}}{\mathbf{2}}{\mathit{L}}_{\mathbf{a}}{\mathit{l}}_{\mathbf{a}}^{\mathbf{2}}\mathbf{+}\frac{\mathbf{1}}{\mathbf{2}}{\mathit{L}}_{\mathbf{b}}{\mathit{l}}_{\mathbf{b}}^{\mathbf{2}}\mathbf{+}\mathit{M}{\mathit{l}}_{\mathbf{a}}{\mathit{l}}_{\mathbf{b}}$ .