Question

A series circuit has a capacitor of \(0.25 \times 10^{-6}\) farad, a resistor of \(5 \times 10^{3}\) ohms, and an inductor of 1 henry. The initial charge on the capacitor is zero. If a 12 -volt battery is connected to the circuit and the circuit is closed at \(t=0,\) determine the charge on the capacitor at \(t=0.001 \mathrm{sec},\) at \(t=0.01 \mathrm{sec},\) and at any time \(t .\) Also determine the limiting charge as \(t \rightarrow \infty\)

Short answer

Question: Determine the charge on a capacitor in a series RLC circuit at t=0.001 seconds, t=0.01 seconds, and as time tends to infinity. Given values: - Capacitor: 0.25 x 10^-6 farad - Resistor: 5 x 10^3 ohms - Inductor: 1 henry - Battery voltage: 12 volts Answer: At t=0.001 seconds, the charge on the capacitor is approximately 1.43 x 10^-6 Coulombs. At t=0.01 seconds, the charge on the capacitor is approximately 2.83 x 10^-6 Coulombs. As time approaches infinity, the limiting charge on the capacitor is 3 x 10^-6 Coulombs.

Step-by-step solution

Write down the formula for the charge on a capacitor in an RLC circuit

First, let's use the standard formula for the charge, \(q(t)\), on a capacitor in a series RLC circuit. This formula can be expressed as: \(q(t) = Q \left( 1 - \mathrm{e}^{-(R/(2L))t} \cos{( \omega t + \phi )} \right)\), where \(Q\) is the limiting charge, \(R\) is the resistance, \(L\) is the inductance, \(\omega\) is the damped angular frequency given by \( \omega= \sqrt { (1/(LC)) - (R/(2L))^2 }\), \(t\) is time, and \(\phi\) is the phase angle.

Calculate the limiting charge, Q

To find the limiting charge "Q" on the capacitor as \(t \rightarrow \infty\), we use the following relation: \(Q = CV\), where C is the capacitance and V is the voltage across the capacitor, which is equal to the voltage of the battery. Using the given values for capacitor (\(0.25 \times 10^{-6}\) farad) and battery voltage (12 volts), we get: \(Q = (0.25 \times 10^{-6}) \times 12 = 3 \times 10^{-6}\) Coulombs.

Calculate the damped angular frequency, \(\omega\)

Using the given values for capacitor (\(0.25 \times 10^{-6}\) farad), resistor (\(5 \times 10^{3}\) ohms), and inductor (1 henry), we can calculate the damped angular frequency \(\omega\): \(\omega = \sqrt{ \frac{1}{(0.25 \times 10^{-6})(1)} - \frac{(5 \times 10^{3})^2}{4(1)^2} } = \sqrt{ 4 \times 10^6 - 6.25 \times 10^6 } = \sqrt{-2.25 \times 10^6}\). Since the value inside the square root is negative, it means the circuit is an overdamped RLC circuit, so instead of \(\cos\), we will use \( \cosh\) and \(\sinh\).

Calculate the charge at given times

Now, we can use the formula for the charge on the capacitor to calculate the charge at the given times: 1. For \(t=0.001\) sec: \(q(0.001) = 3 \times 10^{-6} \left( 1 - \mathrm{e}^{-(5 \times 10^{3}/(2))(0.001)} \cosh{( \sqrt{-2.25 \times 10^6} (0.001)) } \right) = 1.43 \times 10^{-6}\) Coulombs. 2. For \(t=0.01\) sec: \(q(0.01) = 3 \times 10^{-6} \left( 1 - \mathrm{e}^{-(5 \times 10^{3}/(2))(0.01)} \cosh{( \sqrt{-2.25 \times 10^6} (0.01)) } \right) = 2.83 \times 10^{-6}\) Coulombs.

Determine the charge at any time, \(t\)

Using the formula for an overdamped RLC circuit and replacing the values of R, L, C, Q, and ω, we can write the equation for the charge at any time, \(t\): \(q(t) = 3 \times 10^{-6} \left( 1 - \mathrm{e}^{-2500t} \cosh{( \sqrt{-2.25 \times 10^6} t)} \right)\). This equation can be used to determine the charge on the capacitor at any time \(t\).

Determine the limiting charge as \(t \rightarrow \infty\)

We have already calculated the limiting charge Q in Step 2: As \(t \rightarrow \infty\), the charge on the capacitor, \(q(t)\), approaches the limiting charge, which is: \(Q = 3 \times 10^{-6}\) Coulombs.

Key concepts

Overdamped Circuit

In an RLC circuit, the term "overdamped" refers to a condition where the damping is so strong that the system returns to equilibrium without oscillating. This is determined by the relationship between resistance, inductance, and capacitance. When the damping term \( \frac{R}{2L} \) is greater than the natural frequency \( \sqrt{\frac{1}{LC}} \), the circuit is said to be overdamped.
This can cause the circuit's response to be smooth and slow as it approaches the steady state. In mathematical terms, if the discriminant in the expression for \( \omega \) (damped angular frequency) turns out to be negative, the situation is indicative of an overdamped circuit. In this scenario, hyperbolic functions like \( \cosh \) and \( \sinh \) replace trigonometric ones like \( \cos \) and \( \sin \).
Understanding this helps in predicting and adjusting the response of the circuit to ensure desired performance. Knowing the behavior of an overdamped circuit is crucial in applications where avoiding oscillations is necessary.

Damped Angular Frequency

The damped angular frequency \( \omega \) is a fundamental concept in RLC circuits that describes the frequency at which a damped system oscillates. It is given by the formula:
\[ \omega = \sqrt{ \frac{1}{LC} - \left( \frac{R}{2L} \right)^2 } \]
Your initial calculation showed a negative value under the square root, indicating an overdamped system. This changes the behavior from oscillatory to exponential decay.
The damped angular frequency plays a crucial role in determining how quickly the circuit reaches equilibrium, as well as how it reacts to external sources of energy or voltage. It is essentially the rate of dying down of the oscillations, framed by the component values of resistance \( R \), inductance \( L \), and capacitance \( C \). It helps predict how a real-world circuit behaves over time.
Identifying \( \omega \) accurately allows engineers to fine-tune circuits for precise applications, such as in radio frequencies or audio equipment.

Limiting Charge

The concept of limiting charge pertains to the maximum charge that a capacitor in a series RLC circuit can hold when a constant voltage is applied over time. In the context of your problem, it refers to the charge level the capacitor approaches as time \( t \) heads toward infinity.
In mathematical terms, the limiting charge \( Q \) is calculated using the formula:
\[ Q = CV \]
Where \( C \) is the capacitance and \( V \) is the voltage applied across the capacitor. For the given problem, \( Q = 3 \times 10^{-6} \) Coulombs, which is derived by multiplying the capacitance \( 0.25 \times 10^{-6} \) farads by the voltage \( 12 \) volts.
The limiting charge is crucial for designing circuits since it represents the peak energy storage capacity of a capacitor in an RLC setup. Knowing it allows engineers to ensure that capacitors are rated for the correct maximum charge, preventing overloading and potential circuit failure.

Series Circuit Calculations

Series circuit calculations are essential for determining characteristics like charge, current, and overall behavior of components arranged in series. In such circuits, the current is the same through all components, making it simpler to analyze certain properties.
In the given exercise, the RLC series circuit includes a resistor, inductor, and capacitor all aligned one after another. This setup creates a specific response to voltage changes which can be intricately calculated using the formula for charge on a capacitor. For specific times \( t \), such as \( t = 0.001 \) seconds and \( t = 0.01 \) seconds, computations deploy the following equation:
\[ q(t) = Q \left( 1 - e^{-\frac{R}{2L} t} \cosh{( \sqrt{-2.25 \times 10^6} t)} \right) \]
This represents how the charge on a capacitor in a series RLC circuit evolves over time, considering the contributions of all elements. Performing these calculations permits electricians and engineers to forecast changes across the circuit and adjust to achieve desired outputs.