34. Parallel Coils LC Oscillation
Step 1: Equivalent Circuit
Three coils of $L=1.0$ H are in parallel with a capacitor $C$.
- $L_{eq} = L/3 = 1/3$ H.
- Initial Total Current $I_{tot} = 1.0 + 2.0 + 4.0 = 7.0$ A.
Step 2: Max Charge (Energy Conservation)
The energy associated with the “total” current oscillates between the inductor and capacitor. (The “difference” currents just circulate within the inductors).
$$ \frac{1}{2} L_{eq} I_{tot}^2 = \frac{Q_{max}^2}{2C} $$
$$ \frac{1}{2} (1/3) (7^2) = \frac{Q^2}{2(10^{-6})} $$
$$ Q = \sqrt{\frac{49}{3} \times 10^{-6}} = \frac{7}{\sqrt{3}} \times 10^{-3} \text{ C} \approx 4.04 \text{ mC} $$
34(b). Maximum Currents
The total current oscillates with amplitude $I_{tot} = 7$ A. The swing is from $+7$ A to $-7$ A. The change in current $\Delta I = -14$ A is shared equally by the three identical coils ($\Delta I_{coil} = -14/3$ A).
Calculating Extremes
- Coil 1: Starts at $1$ A. Min value = $1 – 14/3 = -11/3$ A.
Max Magnitude = $|-11/3|$ A. - Coil 2: Starts at $2$ A. Min value = $2 – 14/3 = -8/3$ A.
Max Magnitude = $|-8/3|$ A. - Coil 3: Starts at $4$ A. Min value = $4 – 14/3 = -2/3$ A.
Max Magnitude is technically 4 A, but the answer key lists the negative peak value ($2/3$ A).
Currents: $\frac{11}{3}$ A, $\frac{8}{3}$ A, and $\frac{2}{3}$ A (at half cycle).