Electromagnetic induction > AC electrical circuits > Problem 11.4.21

Statement

$11.4.21.$

Prove that in two parallel-connected inductors L1and L2, the sum of L1I1+
L2I2does not change. The direction of the currents is shown in the figure.
b. A capacitor of capacitance C, charged to a voltage V0, is discharged through
an inductor L1. What is the maximum current that can be obtained in the
inductor L2, if you close the key K at the moment when the inductor current
L1is maximum?

Solution

a) Conservation of$ L_1 I_1 + L_2 I_2$

The two coils are connected in parallel, so they share the same terminal voltage. The voltage across an ideal inductor is $V = L\frac{dI}{dt}$ Being in parallel:

$L_1 \frac{dI_1}{dt} = L_2 \frac{dI_2}{dt}$

Integrating over time from an initial instant to any later time, and assuming that initially both currents are zero, we obtain:

$L_1 I_1(t) + L_2 I_2(t) = \text{constant}$.

At the moment switch K is closed, the current in $L_1 $is maximum$ (I_0)$ and that in$ L_2$ is zero Therefore, the constant is $L_1 I_0$and the invariance is demonstrated:

$\boxed{L_1 I_1 + L_2 I_2 = L_1 I_0}$

b) Maximum current in $L_2$

The capacitor C charged to $V_0$ initially discharges only through$ L_1$ The frequency of that $L_1 $C circuit is $\omega = 1/\sqrt{L_1 C}$and the maximum current reached in L_1 (when the capacitor is fully discharged) is:

$I_0 = \frac{V_0}{\omega L_1} = V_0 \sqrt{\frac{C}{L_1}}$

Exactly at that instant, switch K is closed, connecting $L_2 $in parallel with $L_1 fFrom that moment on, the combination oscillates with a new frequency determined by the equivalent inductance of both coils in parallel:

$L_{\text{eq}} = \frac{L_1 L_2}{L_1 + L_2}$

$\omega' = \frac{1}{\sqrt{L_{\text{eq}} C}} = \frac{1}{\sqrt{\frac{L_1 L_2}{L_1 + L_2} C}}$

The total current$ I_1 - I_2 $$(the difference of currents at the common node) oscillates cosinusoidally with this frequency, starting from its maximum value I_0:

$I_1 - I_2 = I_0 \cos \omega' t$

Combining this equation with the conservation law from part (a):

$L_1 I_1 + L_2 I_2 = L_1 I_0$

we solve the system for I_2:

$I_2(t) = \frac{L_1}{L_1 + L_2} I_0 \bigl(1 + \cos \omega' t\bigr)$.

The maximum value of this current occurs when $\cos \omega' t = 1$:

$I_{2,\text{max}} = \frac{2 L_1}{L_1 + L_2} I_0
= \frac{2 L_1}{L_1 + L_2} \cdot V_0 \sqrt{\frac{C}{L_1}}
= 2 V_0 \sqrt{\frac{C}{L_1 + L_2}}$

$\boxed{I_{2,\text{max}} = 2 V_0 \sqrt{\frac{C}{L_1 + L_2}}}$

Answer

$\boxed{I_{2,\text{max}} = 2 V_0 \sqrt{\frac{C}{L_1 + L_2}}}$