Question

A 6.40-nF capacitor is charged to 24.0 V and then disconnected from the battery in the circuit and connected in series with a coil that has \(L =\) 0.0660 H and negligible resistance. After the circuit has been completed, there are current oscillations. (a) At an instant when the charge of the capacitor is 0.0800 \(\mu\)C, how much energy is stored in the capacitor and in the inductor, and what is the current in the inductor? (b) At the instant when the charge on the capacitor is 0.0800 \(\mu\)C, what are the voltages across the capacitor and across the inductor, and what is the rate at which current in the inductor is changing?

Short answer

(a) Energy: Capacitor = 0.50 µJ, Inductor = 1.34 µJ, Current = 0.202 A. (b) Voltages: Capacitor = 12.5 V, Inductor = -12.5 V, Rate of current change = 189.39 A/s.

Step-by-step solution

Calculate Initial Energy Stored in Capacitor

The energy stored in the capacitor when it is fully charged is given by the formula \( E = \frac{1}{2} C V^2 \). Substituting the given values: \( C = 6.40 \times 10^{-9} \text{ F} \) and \( V = 24.0 \text{ V} \), we calculate:\[E = \frac{1}{2} (6.40 \times 10^{-9}) (24.0)^2 = 1.8432 \times 10^{-6} \text{ J}.\]

Energy in Capacitor at 0.0800 µC Charge

When the charge on the capacitor is 0.0800 \( \mu\text{C} \), the potential energy in the capacitor is given by \( U_{C} = \frac{1}{2} \frac{Q^2}{C} \). Using \( Q = 0.0800 \times 10^{-6} \text{ C} \) and \( C = 6.40 \times 10^{-9} \text{ F} \):\[U_{C} = \frac{1}{2} \frac{(0.0800 \times 10^{-6})^2}{6.40 \times 10^{-9}} = 5.0 \times 10^{-7} \text{ J}.\]

Energy in Inductor

The total energy is conserved, so the energy in the inductor \( U_L \) is the initial energy minus the current energy in the capacitor. Thus,\[U_L = 1.8432 \times 10^{-6} - 5.0 \times 10^{-7} = 1.3432 \times 10^{-6} \text{ J}.\]

Calculate Current in Inductor

The energy in the inductor is given by \( U_L = \frac{1}{2} L I^2 \). Solving for \( I \) and substituting \( U_L = 1.3432 \times 10^{-6} \text{ J} \) and \( L = 0.0660 \text{ H} \), we find:\[I = \sqrt{\frac{2 \times 1.3432 \times 10^{-6}}{0.0660}} = 0.202 \text{ A}.\]

Voltage Across Capacitor

The voltage across the capacitor \( V_C \) can be found using \( V_C = \frac{Q}{C} \). Substituting \( Q = 0.0800 \times 10^{-6} \text{ C} \) and \( C = 6.40 \times 10^{-9} \text{ F} \):\[V_C = \frac{0.0800 \times 10^{-6}}{6.40 \times 10^{-9}} = 12.5 \text{ V}.\]

Voltage Across Inductor

The voltage across the inductor \( V_L \) is equal to the negative of the rate of change of current \( L \frac{dI}{dt} \). Since the energy in the inductor equals the work done on moving charges, it remains balanced with the capacitor voltage. Assuming resonance where the total voltage around the loop is zero, at this instant, \( V_L = -V_C \), hence, \( V_L = -12.5 \text{ V} \).

Rate of Change of Current in Inductor

Using the relation \( V_L = -L \frac{dI}{dt} \), where \( V_L = -12.5 \text{ V} \), calculate \( \frac{dI}{dt} \):\[12.5 = 0.0660 \frac{dI}{dt} \\frac{dI}{dt} = \frac{12.5}{0.0660} = 189.39 \text{ A/s}.\]

Key concepts

Energy Conservation in Circuits

In circuits, especially LC circuits like the one in the exercise, energy conservation is a key principle. This principle states that the total energy within the system remains constant over time, although it can change forms. For a closed LC circuit, energy oscillates between the capacitor and the inductor.

When a capacitor is charged and then connected to an inductor, the energy stored in the capacitor (\(E = \frac{1}{2} C V^2\)) begins to transfer to the inductor. Initially, all of the energy resides in the capacitor. As the circuit oscillates, this energy converts back and forth.

• The energy stored in the capacitor is in the form of electric potential energy.
• The energy in the inductor is stored as magnetic energy.

Understanding this energy conservation helps us solve problems by knowing that the sum of energies in the capacitor and inductor at any point remains equal to the initial energy the capacitor had when it was fully charged.

Capacitor Energy

A capacitor stores energy in its electric field created by the separation of charges. When a capacitor is charged to a voltage, it holds electric potential energy.

The energy \(U_C\) stored in a capacitor when it holds charge \(Q\) and has capacitance \(C\) is given by:\[U_C = \frac{1}{2} \frac{Q^2}{C}\]This formula allows us to calculate the energy at any point during the oscillation process. In the exercise, for a capacitor with charge \(Q = 0.0800 \ \mu\text{C}\) and capacitance \(C = 6.40 \times 10^{-9}\ \text{F}\), we can determine the exact amount of energy stored at that particular moment.

• This energy decreases as the charge transfers from the capacitor to the inductor.
• It is a crucial part of understanding resonant oscillations, as it signifies how potential energy is transformed into kinetic energy.

Inductor Energy

Inductors are coils of wire that store energy in their magnetic fields when a current passes through them. The energy \(U_L\) in an inductor with inductance \(L\) and a current \(I\) is given by:\[U_L = \frac{1}{2} L I^2\]This relation is vital in solving LC circuit problems as it helps determine the energy dynamics between the capacitor and the inductor.

In our exercise, this concept explains how once the circuit is completed, the initial energy stored in the capacitor begins to appear as energy in the inductor. The current through the inductor starts small, grows as the capacitor discharges, and reaches a maximum when the capacitor is momentarily empty.

• The exchange between electric and magnetic energy is continuous.
• This energy conservation allows us to find other quantities like the current in the inductor.

Resonant Oscillations

Resonant oscillations occur in an LC circuit when energy alternates between the capacitor's electric field and the inductor's magnetic field. This happens at a natural frequency determined by the circuit's components.

The frequency of these oscillations is given by:\[f = \frac{1}{2\pi\sqrt{LC}}\]At this resonant frequency, the energy alternates most efficiently between the capacitor and the inductor, without losses in a perfect scenario.

• This efficient energy exchange leads to maximum current oscillation in the circuit.
• Understanding resonance is crucial for electronic devices, ensuring they operate correctly without excessive energy losses.

Recognizing these oscillations in the exercise helps solve for currents and voltage changes, ensuring the preservation of total circuit energy.