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The maximum charge on a capacitor in an LC circuit can be calculated using a simple formula. The formula is: \( Q_{\text{max}} = C \times V_{\text{max}} \)
Here, **Q** represents the charge, **C** is the capacitance, and **V** is the maximum voltage.
Given in our problem:

• Capacitance, \( C = 1.0 \text{ nF} = 1.0 \times 10^{-9} \text{ F} \)
• Maximum Voltage, \( V_{\text{max}} = 3.0 \text{ V} \)

Using these values, we can find the maximum charge by plugging them into the formula:
\( Q_{\text{max}} = (1.0 \times 10^{-9} \text{ F}) \times (3.0 \text{ V}) = 3.0 \times 10^{-9} \text{ C} \)
So, the maximum charge on the capacitor is \( 3.0 \times 10^{-9} \text{ C} \).

In an LC circuit, the maximum current can be determined by knowing the charge and the values of the inductance and capacitance. For maximum current, we use the formula: \( I_{\text{max}} = \frac{Q_{\text{max}}}{\text{sqrt}\big(\frac{L}{C}\big)}\)
Let's start with the given values:

• Inductance, \( L = 3.0 \text{ mH} = 3.0 \times 10^{-3} \text{ H} \)
• Capacitance, \( C = 1.0 \times 10^{-9} \text{ F} \)

First, calculate \( \text{sqrt}\big(\frac{L}{C}\big) \):
\[ \text{sqrt}\big(\frac{L}{C}\big) = \text{sqrt}\big(\frac{3.0 \times 10^{-3} \text{ H}}{1.0 \times 10^{-9} \text{ F}}\big) = \text{sqrt}(3.0 \times 10^6) \ \text{sqrt}(3.0 \times 10^6) \ \text{approx } 1.73 \times 10^3 \text{ s}^{-1} \]
Now, the maximum current can be computed as:
\( I_{\text{max}} = \frac{3.0 \times 10^{-9} \text{ C}}{1.73 \times 10^3} \ \text{approx } 1.73 \times 10^{-6} \text{ A} \)
Thus, the maximum current in the circuit is \( 1.73 \times 10^{-6} \text{ A} \).

The energy stored in the magnetic field of the coil at its maximum can be calculated using the inductance and the maximum current. The energy in the magnetic field, symbolized as **U**, is given by:
\( U = \frac{1}{2} L I_{\text{max}}^2 \)
Given:
\( L = 3.0 \times 10^{-3} \text{ H} \)
\( I_{\text{max}} = 1.73 \times 10^{-6} \text{ A} \)
First, square the value of \(I_{\text{max}}\):
\[ (1.73 \times 10^{-6})^2 \ \text{approx } 2.99 \times 10^{-12} \]
Now, multiply with \( L \) and divide by 2:
\[ U = \frac{1}{2} \times 3.0 \times 10^{-3} \text{ H} \times 2.99 \times 10^{-12} \]
\[ U \ \text{approx } \frac{1}{2} \times 3.0 \times 10^{-3} \times 2.99 \times 10^{-12} \]
\[ U \ \text{approx } 4.5 \times 10^{-15} \text{ J} \]
Therefore, the maximum energy stored in the magnetic field of the coil is \( 4.5 \times 10^{-15} \text{ J} \).