Problem 32 - Advanced Engineering Mathematics

Question

A 200-volt electromotive force is applied to an \(R C\)-series circuit in which the resistance is 1000 ohms and the capacitance is \(5 \times 10^{-6}\) farad. Find the charge \(q(t)\) on the capacitor if \(i(0)=0.4\). Determine the charge and current at \(t=0.005 \mathrm{~s}\). Determine the charge as \(t \rightarrow \infty\).

Step-by-Step Solution

Answer

\( q(t) \approx 0.0002642 \, C \) and \( i(t) \approx 0.147 \, A \) at \( t = 0.005 \, s \). As \( t \to \infty \), \( q(t) = 10^{-3} \, C \).

Step 1: Write Down the Known Values

The given problem states that a 200 V electromotive force E is applied to an RC-series circuit where the resistance R is 1000 ohms, and the capacitance C is \( 5 \times 10^{-6} \) farads. Initial current \( i(0) = 0.4 \), and we need to find the charge \( q(t) \) on the capacitor.

Step 2: Establish the Differential Equation

For an \( RC \) circuit with a constant voltage, the differential equation governing the charge \( q(t) \) is \( R \frac{dq}{dt} + \frac{q}{C} = E \). Substituting the known values, we get \( 1000 \frac{dq}{dt} + \frac{q}{5\times10^{-6}} = 200 \).

Step 3: Solve the Homogeneous Equation

Firstly, solve the homogeneous equation \( 1000 \frac{dq}{dt} + \frac{q}{5\times10^{-6}} = 0 \). This simplifies to \( \frac{dq}{dt} = -\frac{q}{5\times10^{-3}} \). Solving this gives the general solution \( q_h(t) = Ce^{-200t} \).

Step 4: Solve for Particular Solution

Assume a particular solution of the form \( q_p = A \), where \( A \) is a constant. Substitute into the non-homogeneous equation to find \( A \, , 1000 \cdot 0 + \frac{A}{5 \times 10^{-6}} = 200 \Rightarrow A = 10^{-3} \). Therefore, the particular solution is \( q_p(t) = 10^{-3} \).

Step 5: Find the General Solution

Combine the homogeneous and particular solutions to find the general solution: \( q(t) = Ce^{-200t} + 10^{-3} \).

Step 6: Apply Initial Condition to Find Constant C

Use the initial condition \( i(0) = 0.4 \) which implies \( \frac{dq}{dt}(0) = 0.4 \). Differentiating \( q(t) = Ce^{-200t} + 10^{-3} \) results in \( \frac{dq}{dt} = -200Ce^{-200t} \). Substituting the initial condition gives \( -200C = 0.4 \), leading to \( C = -0.002 \).

Step 7: Write the Final Solution for q(t)

With the constant \( C \) known, the final solution for charge over time is \( q(t) = -0.002e^{-200t} + 10^{-3} \).

Step 8: Calculate Charge at t=0.005 s

Substitute \( t = 0.005 \) into the charge equation: \( q(0.005) = -0.002e^{-200 \times 0.005} + 10^{-3} \). Calculate \( e^{-1} \approx 0.3679 \). Hence, \( q(0.005) = -0.002 \times 0.3679 + 10^{-3} \approx 0.0002642 \) Coulombs.

Step 9: Find the Current at t=0.005 s

Use the derivative of \( q(t) \) to find current: \( i(t) = \frac{dq}{dt} = -200(-0.002)e^{-200t} = 0.4e^{-200t} \). Substitute \( t = 0.005 \) gives \( i(0.005) = 0.4e^{-1} \approx 0.147 \, A \).

Step 10: Determine q(t) as t Approaches Infinity

As \( t \to \infty \), the exponential term \( e^{-200t} \) approaches zero. Therefore, \( q(t) = \lim_{t \to \infty}(-0.002e^{-200t} + 10^{-3}) = 10^{-3} \).

Key Concepts

Differential Equations

In electronics, analyzing circuits often involves the use of differential equations to model the relationship between changing quantities over time. An RC circuit, which consists of a resistor (R) and a capacitor (C) in series, is no exception. This system can be described by a first-order linear differential equation. This relationship expresses how the electric charge, current, and voltage change over time due to applied forces like an electromotive force (EMF).

For an RC circuit subjected to a constant voltage, the differential equation that governs the charge, \( q(t) \), on the capacitor is given by:

• \( R \frac{dq}{dt} + \frac{q}{C} = E \)

where:\

• \( R \) is the resistance,
• \( C \) is the capacitance,
• \( E \) is the electromotive force (in volts).

The term \( \frac{dq}{dt} \) is the rate of change of charge over time, representing the current through the circuit. Solving this differential equation involves finding both the homogeneous solution, representing natural circuit behavior, and a particular solution that accounts for the constant voltage.

Initial Conditions

Initial conditions play a crucial role in solving differential equations, as they allow us to find specific solutions that fit real-world situations. In an RC circuit, the initial condition might be the initial current or initial charge on the capacitor at the start of the observation, often time \( t = 0 \).

In the provided problem, the initial condition given is \( i(0) = 0.4 \). This information is used to determine the constants of integration when solving the differential equation for charge \( q(t) \). In this context, knowing the current at \( t = 0 \) helps determine the integration constant for the homogeneous solution of the equation:

• \( \frac{dq}{dt}(0) = -200Ce^{-200 \times 0} = 0.4 \)

From this, the constant \( C \) is found to be \( -0.002 \). Initial conditions ensure that the model accurately reflects the circuit's performance starting from a known state.

Charge on Capacitor

The charge on a capacitor in an RC circuit is a fundamental concept that represents the amount of electric energy stored. Understanding how charge evolves over time under the influence of a constant voltage is key to analyzing and predicting circuit behavior.

Using the general solution to the differential equation, \( q(t) = Ce^{-200t} + 10^{-3} \), the charge on the capacitor can be determined at any time \( t \). By recognizing how the exponential term influences \( q(t) \), we see:

• As time increases, \( e^{-200t} \) approaches zero, highlighting that the charge stabilizes.
• For \( t = 0.005 \) s, we substitute to find \( q(0.005) = -0.002 \times e^{-1} + 10^{-3} \approx 0.0002642 \) Coulombs.
• As \( t \rightarrow \infty \), \( q(t) \rightarrow 10^{-3} \), the system reaches a steady state where the capacitor is fully charged to \( 10^{-3} \) Coulombs.

The exponential decay of the initial charge term \( Ce^{-200t} \) illustrates how quickly the system approaches its steady state, reflecting the dynamics of charging within the circuit.