Subfield: Electromagnetism


Question: Initially, a switch $S$ is unshorted in the circuit shown in the figure on the right, a capacitor of capacitance $2 C$ carries the electric charge $q_{0}$, a capacitor of capacitance $C$ is uncharged, and there are no electric currents in both coils of inductance $L$ and $2 L$, respectively. The capacitor starts to discharge and at the moment when the current in the coils reaches its maximum value, the switch $S$ is instantly shorted. Find the maximum current $I_{\max }$ through the switch $S$ thereafter.

<img_4374>

Solution:
At the moment when the current in the coils is a maximum, the total voltage across the coils is equal to zero, so the capacitor voltages must be equal in magnitude and opposite in polarity. Let $U$ be a voltage on the capacitors at the time moment just mentioned and $I_{0}$ be that maximum current. According to the law of charge conservation

thus,

$$
q_{0}=2 C U+C U
\tag{C1.1}
$$

$$
U=\frac{q_{0}}{3 C}
\tag{C1.2}
$$

Then, from the energy conservation law

$$
\frac{q_{0}^{2}}{2 \cdot 2 C}=\frac{L I_{0}^{2}}{2}+\frac{2 L I_{0}^{2}}{2}+\frac{C U^{2}}{2}+\frac{2 C U^{2}}{2}
\tag{C1.3}
$$

the maximum current is found as

$$
I_{0}=\frac{q_{0}}{3 \sqrt{2 L C}}
\tag{C1.4}
$$

After the key $K$ is shortened there will be independent oscillations in both circuits with the frequency

$$
\omega=\frac{1}{\sqrt{2 L C}}
\tag{C1.5}
$$

and their amplitudes are obtained from the corresponding energy conservation laws written as

$$
\frac{2 C U^{2}}{2}+\frac{L I_{0}^{2}}{2}=\frac{L J_{1}^{2}}{2}
\tag{C1.6}
$$
$$
\frac{C U^{2}}{2}+\frac{2 L I_{0}^{2}}{2}=\frac{2 L J_{2}^{2}}{2} .
\tag{C1.7}
$$

Hence, the corresponding amplitudes are found as

$$
J_{1}  =\sqrt{5} I_{0},
\tag{C1.8}
$$
$$
J_{2}  =\sqrt{2} I_{0} .
\tag{C1.9}
$$

Choose the positive directions of the currents in the circuits as shown in the figure on the right. Then, the current flowing through the key is written as follows

$$
I=I_{1}-I_{2}
\tag{C1.10}
$$

The currents depend on time as

$$
I_{1}(t)=A \cos \omega t+B \sin \omega t
\tag{C1.11}
$$
$$
I_{2}(t)=D \cos \omega t+F \sin \omega t
\tag{C1.12}
$$

<img_4418>

The constants $A, B, D, F$ can be determined from the initial values of the currents and their amplitudes by putting down the following set of equations

$$
I_{1}(0)=A=I_{0},
\tag{C1.13}
$$
$$
A^{2}+B^{2}=J_{1}^{2}, 
\tag{C1.14}
$$
$$
I_{2}(0)=D=I_{0}, 
\tag{C1.15}
$$
$$
D^{2}+F^{2}=J_{2}^{2}
\tag{C1.16}
$$

Solving (C1.13)-(C1.16) it is found that

$$
B=2 I_{0}, 
\tag{C1.17}
$$
$$
F=-I_{0},
\tag{C1.18}
$$

The sign in $F$ is chosen negative, since at the time moment of the key shortening the current in the coil $2 L$ decreases.

Thus, the dependence of the currents on time takes the following form

$$
I_{1}(t)=I_{0}(\cos \omega t+2 \sin \omega t),
\tag{C1.19}
$$
$$
I_{2}(t)=I_{0}(\cos \omega t-\sin \omega t) .
\tag{C1.20}
$$

In accordance with (C1.10), the current in the key is dependent on time according to

$$
I(t)=I_{1}(t)-I_{2}(t)=3 I_{0} \sin \omega t .
\tag{C1.21}
$$

Hence, the amplitude of the current in the key is obtained as

$$
I_{\max }=3 I_{0}=\omega q_{0}=\frac{q_{0}}{\sqrt{2 L C}}
\tag{C1.22}
$$
Instead of determining the coefficients $A, B, D, F$ the vector diagram shown in the figure on the right can be used. The segment $A C$ represents the current sought and its projection on the current axis is zero at the time of the key shortening. The current $I_{1}$ in the coil of inductance $L$ grows at the same time moment because the capacitor $2 C$ continues to discharge, thus, this current is depicted in the figure by the segment $O A$. The current $I_{2}$ in the coil of inductance $2 L$ decreases at the time of the key shortening since it continues to charge the capacitor $2 C$, that is why this current is depicted in the figure by the segment $O C$.

It is known for above that $O B=I_{0}, O A=\sqrt{5} I_{0}, O C=\sqrt{2} I_{0}$. Hence, it is found from the Pythagorean theorem that

$$
A B=\sqrt{O A^{2}-O B^{2}}=2 I_{0}
\tag{C2.1}
$$
$$
B C=\sqrt{O C^{2}-O B^{2}}=I_{0}
\tag{C2.2}
$$

<img_4426>

Thus, the current sought is found as

$$
I_{\max }=A C=A B+B C=3 I_{0}=\omega q_{0}=\frac{q_{0}}{\sqrt{2 L C}}
\tag{C2.3}
$$

Method 3. Heuristic approach

It is clear that the current through the key performs harmonic oscillations with the frequency

$$
\omega=\frac{1}{\sqrt{2 L C}}
\tag{C3.1}
$$

and it is equal to zero at the time of the key shortening, i.e.

$$
I(t)=I_{\max } \sin \omega t
\tag{C3.2}
$$

Since the current is equal to zero at the time of the key shortening, then the current amplitude is equal to the current derivative at this time moment divided by the oscillation frequency. Let us find that current derivative. Let the capacitor of capacitance $2 C$ have the charge $q_{1}$. Then the charge on the capacitor of capacitance $C$ is found from the charge conservation law as

$$
q_{2}=q_{0}-q_{1} .
\tag{C3.3}
$$

After shortening the key the rate of current change in the coil of inductance $L$ is obtained as

$$
\dot{I}_{1}=\frac{q_{1}}{2 L C}
\tag{C3.4}
$$

whereas in the coil of inductance $2 L$ it is equal to

$$
\dot{I}_{2}=-\frac{q_{0}-q_{1}}{2 L C}
\tag{C3.5}
$$

Since the voltage polarity on the capacitors are opposite, then the current derivative with respect to time finally takes the form

$$
\dot{I}=\dot{I}_{1}-\dot{I}_{2}=\frac{q_{0}}{2 L C}=\omega^{2} q_{0}
\tag{C3.6}
$$

Note that this derivative is independent of the time of the key shortening!

Hence, the maximum current is found as

$$
I_{\max }=\frac{\dot{I}}{\omega}=\omega q_{0}=\frac{q_{0}}{\sqrt{2 L C}}
\tag{C3.7}
$$

and it is independent of the time of the key shortening!

Final answer: $I_{\max }=\frac{q_{0}}{\sqrt{2 L C}}$