To solve the problem involving the L-C circuit, we need to analyze the behavior of the inductor and capacitor over time. The circuit consists of an inductor with an inductance of L = 0.09 H and a capacitor with a capacitance of C = 4 × 10^-4 F. The initial charge on the capacitor is given as Q₀ = 5 × 10^-6 C, and the initial current in the inductor is zero. We will find the maximum current in the inductor and the charge on the capacitor when the current in the inductor reaches half of its maximum value.
Understanding the L-C Circuit Dynamics
In an L-C circuit, energy oscillates between the inductor and the capacitor. The maximum current in the inductor occurs when the capacitor is fully discharged, and the maximum charge on the capacitor occurs when the inductor's current is zero. The total energy in the system can be expressed as:
- Energy in the capacitor: \( U_C = \frac{1}{2} C V^2 \)
- Energy in the inductor: \( U_L = \frac{1}{2} L I^2 \)
Initially, all the energy is stored in the capacitor:
Calculating Maximum Current
The voltage across the capacitor can be calculated using the initial charge:
Using the formula \( V = \frac{Q}{C} \), we have:
V = \( \frac{5 \times 10^{-6} \text{ C}}{4 \times 10^{-4} \text{ F}} = 0.0125 \text{ V} \)
The initial energy stored in the capacitor is:
U₀ = \( \frac{1}{2} C V^2 = \frac{1}{2} \times 4 \times 10^{-4} \times (0.0125)^2 \)
Calculating this gives:
U₀ = \( \frac{1}{2} \times 4 \times 10^{-4} \times 1.5625 \times 10^{-4} = 3.125 \times 10^{-8} \text{ J} \)
At maximum current, all this energy is stored in the inductor:
Setting \( U_L = U₀ \):
\( \frac{1}{2} L I_{max}^2 = 3.125 \times 10^{-8} \)
Substituting L = 0.09 H:
\( \frac{1}{2} \times 0.09 \times I_{max}^2 = 3.125 \times 10^{-8} \)
Solving for \( I_{max} \):
0.045 I_{max}^2 = 3.125 \times 10^{-8}
Thus, \( I_{max}^2 = \frac{3.125 \times 10^{-8}}{0.045} \approx 6.944 \times 10^{-7} \)
Taking the square root:
\( I_{max} \approx 8.33 \times 10^{-4} \text{ A} \) or 0.833 mA.
Finding Charge When Current is Half of Maximum
When the current in the inductor is half of its maximum value, we have:
\( I = \frac{1}{2} I_{max} = \frac{1}{2} \times 8.33 \times 10^{-4} \approx 4.165 \times 10^{-4} \text{ A} \)
At this point, we can use the conservation of energy principle. The total energy remains constant:
\( U_{total} = U_C + U_L \)
Where \( U_L = \frac{1}{2} L I^2 \) and \( U_C = \frac{1}{2} C V^2 \). We know \( U_{total} = 3.125 \times 10^{-8} \text{ J} \).
Calculating \( U_L \):
\( U_L = \frac{1}{2} \times 0.09 \times (4.165 \times 10^{-4})^2 \approx 7.5 \times 10^{-9} \text{ J} \)
Now, substituting into the total energy equation:
\( 3.125 \times 10^{-8} = U_C + 7.5 \times 10^{-9} \)
Thus, \( U_C = 3.125 \times 10^{-8} - 7.5 \times 10^{-9} = 2.375 \times 10^{-8} \text{ J} \)
Now, we can find the voltage across the capacitor:
\( U_C = \frac{1}{2} C V^2 \Rightarrow V^2 = \frac{2 U_C}{C} \)
Substituting the values:
\( V^2 = \frac{2 \times 2.375 \times 10^{-8}}{4 \times 10^{-4}} \approx 0.11875 \)
Taking the square root gives:
\( V \approx 0.345 \text{ V} \)
Finally, we can find the charge on the capacitor at this voltage:
\( Q = C \times V = 4 \times 10^{-4} \times 0.345 \approx 1.38 \times 10^{-4} \text{ C} \)
Summary of Results
To summarize:
- Maximum current in the inductor: \( I_{max} \approx 0.833 \text{ mA} \)
- Charge on the capacitor when the current is half its maximum: \( Q \approx 1.38 \times 10^{-4} \text{ C} \)
This analysis illustrates the oscillatory nature of energy transfer in an L-C circuit, showcasing how the energy shifts between the inductor and capacitor while maintaining the total energy constant. If you have any further questions or need clarification on any part of this process, feel free to ask!
