Electromagnetic induction > AC electrical circuits > Problem 11.4.22

Statement

$11.4.22.$

a. At the moment when the current in the inductor L1was equal to I, the key
K was closed. How much heat will be released on the resistance R after the
key is closed?
b. With a closed key K, the current in the inductor L1is I1, and in the inductor
L2is I2. Determine within what limits the current in the inductors L1and L2
will change after opening the key K.

Solution

a) Heat dissipated in R when closing K

Initially: current I in $L_1 $ zero in $L_2$
Magnetic fluxes cannot change abruptly. Upon closing K, the following is conserved:

$L_1 I_1 + L_2 I_2 = L_1 I$

In the final state,$ I_1 = I_2 = I_0, so I_0 = \frac{L_1}{L_1+L_2}I$
The dissipated energy is the difference between the initial and final magnetic energy:

$W = \frac{1}{2}L_1 I^2 - \frac{1}{2}(L_1+L_2)I_0^2
= \frac{L_1 L_2}{2(L_1+L_2)} I^2$

b) Current limits after opening K

With K closed, currents are$ I_1 $in $L_1$ and$ I_2 $in $L_2$
Upon opening K, the total flux remains constant.
Since they are in series, $I_1' = I_2' = I'$
Therefore:

$(L_1 + L_2) I' = L_1 I_1 + L_2 I_2 \quad\Rightarrow\quad I' = \frac{L_1 I_1 + L_2 I_2}{L_1 + L_2}$

The current in L_1 varies between:

$I_1 \quad\text{and}\quad I_1 - \frac{2(I_1 - I_2)}{1 + L_1/L_2}$

The current in L_2 varies between:

$I_2 \quad\text{and}\quad I_2 + \frac{2(I_1 - I_2)}{1 + L_2/L_1}$

Answer

$\boxed{W = \frac{1}{2}L_1 I^2 - \frac{1}{2}(L_1+L_2)I_0^2
= \frac{L_1 L_2}{2(L_1+L_2)} I^2}$

$\boxed{I_1 \quad\text{and}\quad I_1 - \frac{2(I_1 - I_2)}{1 + L_1/L_2}}$

$\boxed{I_2 \quad\text{and}\quad I_2 + \frac{2(I_1 - I_2)}{1 + L_2/L_1}}$