The maximum charge on a capacitor in an LC circuit, often denoted as $Q_{max}$ (or simply $Q$), is found by analyzing the energy conservation within the ideal circuit. In an ideal LC circuit, energy continuously oscillates between the electric field stored in the capacitor and the magnetic field stored in the inductor.
How to Find the Maximum Charge ($Q_{max}$)
The maximum charge is typically determined by the initial conditions of the circuit or by knowing the maximum current achieved during oscillation.
1. From Initial Capacitor Voltage
The most direct way to find the maximum charge is if you know the initial voltage ($V_0$) to which the capacitor was charged before being connected to the inductor. At the moment of connection (and subsequently, at every peak charge cycle in an ideal circuit), the charge on the capacitor will reach this initial value.
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Formula: $Q_{max} = C \cdot V_0$
Where:
- $Q_{max}$ is the maximum charge in Coulombs (C).
- $C$ is the capacitance in Farads (F).
- $V_0$ is the initial voltage across the capacitor in Volts (V).
This formula is derived from the definition of capacitance: $Q = C \cdot V$. When the capacitor is fully charged to $V_0$, it holds its maximum possible charge for that specific energy input.
2. From Maximum Current
Alternatively, the maximum charge can be determined using the principle of energy conservation. In an oscillating LC circuit, the total energy ($U_{total}$) remains constant. This total energy is fully stored in the capacitor's electric field when the charge is maximum, and fully stored in the inductor's magnetic field when the current is maximum.
- Maximum Electric Energy: $U{E,max} = \frac{1}{2} \frac{Q{max}^2}{C}$
- Maximum Magnetic Energy: $U{B,max} = \frac{1}{2} L I{max}^2$
Since $U{E,max} = U{B,max} = U_{total}$, we can set the two energy expressions equal:
$\frac{1}{2} \frac{Q{max}^2}{C} = \frac{1}{2} L I{max}^2$
Solving for $Q_{max}$:
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Formula: $Q{max} = I{max} \sqrt{LC}$
Where:
- $Q_{max}$ is the maximum charge in Coulombs (C).
- $I_{max}$ is the maximum current in Amperes (A).
- $L$ is the inductance in Henrys (H).
- $C$ is the capacitance in Farads (F).
Summary of Key Formulas
| Parameter to Find | Given Information | Formula |
|---|---|---|
| Maximum Charge ($Q_{max}$) | Initial Voltage ($V_0$) | $Q_{max} = C \cdot V_0$ |
| Maximum Charge ($Q_{max}$) | Maximum Current ($I_{max}$) | $Q{max} = I{max} \sqrt{LC}$ |
Energy Distribution: A Deeper Look
Understanding how energy distributes within an LC circuit provides further insight into its behavior. The maximum charge ($Q_{max}$) signifies the point where all energy is purely electrical.
When the energy is stored equally between the electric field of the capacitor and the magnetic field of the inductor, the charge ($q$) on the capacitor is not at its maximum. If $Q_{max}$ represents the maximum charge (as denoted by 'Q' in various contexts), then the charge on the capacitor when energy is equally split is:
- Charge at Equal Energy Split: $q = \frac{Q_{max}}{\sqrt{2}}$
This relationship arises because, at this specific instant, the electric potential energy is half of the total maximum energy:
$\frac{1}{2} \frac{q^2}{C} = \frac{1}{2} \left( \frac{1}{2} \frac{Q{max}^2}{C} \right)$
$q^2 = \frac{1}{2} Q{max}^2$
$q = \frac{Q_{max}}{\sqrt{2}}$
Practical Considerations
- Ideal vs. Real Circuits: These formulas apply to ideal LC circuits, which assume no energy loss due to resistance. In real-world circuits, some resistance is always present, leading to damped oscillations where the maximum charge (and energy) decreases over time.
- Initial Energy Input: The maximum charge is fundamentally determined by the total energy initially supplied to the circuit. This energy might come from a charged capacitor, a current flowing through an inductor, or an external power source.
Example:
Suppose you have an LC circuit with a 5 µF capacitor and a 2 mH inductor.
If the capacitor is initially charged to 12 V:
$Q_{max} = C \cdot V_0 = (5 \times 10^{-6} \text{ F}) \cdot (12 \text{ V}) = 60 \times 10^{-6} \text{ C} = 60 \text{ µC}$
If, instead, you know the maximum current achieved in the circuit is 0.5 A:
First, calculate $\sqrt{LC} = \sqrt{(2 \times 10^{-3} \text{ H}) \cdot (5 \times 10^{-6} \text{ F})} = \sqrt{10 \times 10^{-9}} = \sqrt{10^{-8}} = 10^{-4} \text{ s}$
$Q{max} = I{max} \sqrt{LC} = (0.5 \text{ A}) \cdot (10^{-4} \text{ s}) = 50 \times 10^{-6} \text{ C} = 50 \text{ µC}$
