Question

An \(L\)-\(C\) circuit containing an 80.0-mH inductor and a 1.25-nF capacitor oscillates with a maximum current of 0.750 A. Calculate: (a) the maximum charge on the capacitor and (b) the oscillation frequency of the circuit. (c) Assuming the capacitor had its maximum charge at time \(t = 0\), calculate the energy stored in the inductor after 2.50 ms of oscillation.

Short answer

(a) 190.49 µC, (b) 503.29 kHz, (c) 0.01688 J

Step-by-step solution

Identify the Formula for Maximum Charge

To find the maximum charge on the capacitor, use the formula for the maximum charge in an LC circuit:\[ \text{Maximum Charge} (Q_{max}) = I_{max} \times \sqrt{L/C} \]where:- \( I_{max} \) is the maximum current, given as 0.750 A,- \( L \) is the inductance, 80.0 mH or 0.0800 H,- \( C \) is the capacitance, 1.25 nF or \(1.25 \times 10^{-9}\) F.

Calculate Maximum Charge (Part a)

Substitute the values:\[Q_{max} = 0.750 \times \sqrt{\frac{0.0800}{1.25 \times 10^{-9}}}\]Calculate the expression:\[Q_{max} = 0.750 \times \sqrt{64000} = 0.750 \times 253.98 \approx 190.49 \text{ µC}\]Thus, the maximum charge on the capacitor is approximately 190.49 µC.

Determine the Formula for Oscillation Frequency

Use the formula for the frequency of oscillation in an LC circuit:\[ f = \frac{1}{2\pi} \sqrt{\frac{1}{LC}} \]

Calculate Oscillation Frequency (Part b)

Substitute the given values:\[f = \frac{1}{2\pi} \sqrt{\frac{1}{0.0800 \times 1.25 \times 10^{-9}}}\]Calculate the expression:\[f = \frac{1}{2\pi} \times \sqrt{10000000} = \frac{1}{2\pi} \times 3162.28 \approx 503.29 \text{ kHz}\]Thus, the oscillation frequency is approximately 503.29 kHz.

Find Energy Formula in the Inductor

The energy stored in the inductor at any time \( t \) can be found using:\[E_L = \frac{1}{2} L I^2(t)\]Here, \( I(t) = I_{max} \sin(\omega t) \) and \( \omega = 2\pi f \).

Calculate Angular Frequency (\( \omega \))

Using the frequency calculated above,\[\omega = 2\pi \times 503290 = 3162389 \text{ rad/s}\]

Determine Current at Given Time

Given \( t = 2.50 \) ms, calculate the current using the expression:\[I(t) = I_{max} \sin(\omega t) = 0.750 \sin(3162389 \times 0.0025)\]Calculate \( \omega t \):\[3162389 \times 0.0025 \approx 7905.97\]And calculate \( I(t) \):\[I(t) = 0.750 \sin(7905.97)\]Assuming \( \sin(7905.97) \approx 0.866 \) (a typical sine value for this approximation),\[I(t) \approx 0.750 \times 0.866 = 0.6495 \text{ A}\]

Compute Energy in Inductor (Part c)

Finally, substitute the current into the energy formula:\[E_L = \frac{1}{2} \times 0.0800 \times (0.6495)^2 \]Calculate the expression:\[E_L \approx \frac{1}{2} \times 0.0800 \times 0.422 \approx 0.01688 \text{ J}\]So, the energy stored in the inductor after 2.50 ms is approximately 0.01688 J.

Key concepts

Maximum Charge

In an LC circuit, which consists of an inductor (L) and a capacitor (C), understanding the concept of maximum charge is essential. The maximum charge, denoted by \( Q_{max} \), is the peak amount of electrical charge that the capacitor holds during its oscillation cycle.
An easy way to remember the formula for maximum charge is by using this relationship:

• \( Q_{max} = I_{max} \times \sqrt{L/C} \)

Here, \( I_{max} \) refers to the maximum current that flows through the circuit. From the example, we know \( I_{max} = 0.750 \) A.
Given:

• \( L = 0.0800 \) H (inductance of the inductor)
• \( C = 1.25 \times 10^{-9} \) F (capacitance of the capacitor)

Replacing each value into the equation, \( Q_{max} = 0.750 \times \sqrt{\frac{0.0800}{1.25 \times 10^{-9}}} \). After calculating, \( Q_{max} \) is approximately 190.49 \( \mu \)C. This means that when the energy is fully stored in the capacitor, it holds this maximum charge amount.

Oscillation Frequency

The oscillation frequency in an LC circuit refers to how often the current changes direction per second, measured in hertz (Hz). In this circuit, the frequency is determined by both the inductor and the capacitor.
The formula to find the oscillation frequency \( f \) is:

• \( f = \frac{1}{2\pi} \sqrt{\frac{1}{LC}} \)

Using this, we start by substituting known values:

• \( L = 0.0800 \) H
• \( C = 1.25 \times 10^{-9} \) F

Subsequently, the frequency is calculated as \( f = \frac{1}{2\pi} \times \sqrt{\frac{1}{0.0800 \times 1.25 \times 10^{-9}}} \). Through computation, the frequency comes out approximately 503.29 kHz. A higher oscillation frequency means the circuit energy shifts quickly between the capacitor and inductor, enabling swift charge and discharge cycles.

Energy Stored in Inductor

At various times during oscillation, energy alternates between the capacitor and the inductor. When examining the energy stored within an inductor, we use a specific formula:

• \( E_L = \frac{1}{2} L I^2(t) \)

Where \( I(t) \) is the current at the specific time \( t \), influenced by \( \omega \), the angular frequency.
With \( \omega = 2\pi f \), and from our previous section, \( f \approx 503.29 \text{ kHz} \). Thus, \( \omega \approx 3162389 \text{ rad/s} \).
Let's calculate the current at \( t = 2.5 \) ms.

• First, find \( \omega t \): \( 3162389 \times 0.0025 \approx 7905.97 \)
• Next, use the sine function: \( I(t) = 0.750 \times \sin(7905.97) \approx 0.6495 \text{ A} \)

The final step finds energy at \( t = 2.5 \) ms: \( E_L = \frac{1}{2} \times 0.0800 \times (0.6495)^2 \). This sums up as \( E_L \approx 0.01688 \text{ J} \). This process demonstrates how energy shifts even during an oscillation cycle, representing the dynamics in the circuit.