Problem 29 - Advanced Engineering Mathematics

Question

A 30-volt electromotive force is applied to an \(L R\) -series circuit in which the inductance is \(0.1\) henry and the resistance is 50 ohms. Find the current \(i(t)\) if \(i(0)=0\). Determine the current as \(t \rightarrow \infty\).

Step-by-Step Solution

Answer

The current \(i(t) = -0.6e^{-500t} + 0.6\), and as \(t\rightarrow \infty\), \(i(t)\rightarrow 0.6\) amperes.

Step 1: Write the Differential Equation

For an \(LR\)-series circuit with an applied electromotive force \(E(t)\), inductance \(L\), and resistance \(R\), the differential equation is given by \(L\frac{di}{dt} + Ri = E(t)\). In this problem, \(E(t) = 30\), \(L = 0.1\text{ henrys}\), and \(R = 50 \text{ ohms}\). Thus, we have: \[0.1\frac{di}{dt} + 50i = 30.\]

Step 2: Simplify the Differential Equation

Divide through by 0.1 to simplify, yielding: \[ \frac{di}{dt} + 500i = 300.\] This is a first-order linear differential equation.

Step 3: Solve the Homogeneous Equation

The homogeneous equation is \(\frac{di}{dt} + 500i = 0\). Solving gives: \[i_h(t) = Ce^{-500t},\] where \(C\) is a constant.

Step 4: Find the Particular Solution

For the particular solution of \(\frac{di}{dt} + 500i = 300\), let \(i_p(t) = A\), where \(A\) is a constant. Substituting into the equation gives \(500A = 300\), so \(A = 0.6\). Thus, \(i_p(t) = 0.6\).

Step 5: Form the General Solution

The general solution is the sum of the homogeneous and particular solutions: \[i(t) = i_h(t) + i_p(t) = Ce^{-500t} + 0.6.\]

Step 6: Use Initial Condition to Solve for C

We know \(i(0) = 0\). Substitute in the general solution: \[0 = Ce^{0} + 0.6\Rightarrow C + 0.6 = 0.\] Thus, \(C = -0.6\).

Step 7: Write the Specific Solution

Combine the values of \(C\) from the previous step into the general solution: \[i(t) = -0.6e^{-500t} + 0.6.\] This is the specific solution for \(i(t)\) given \(i(0) = 0\).

Step 8: Determine the Current as \(t \to \infty\)

As \(t\rightarrow \infty\), the term \(-0.6e^{-500t} \rightarrow 0\), thus \[i(t) \rightarrow 0.6.\] Therefore, as time goes to infinity, the current stabilizes at \(0.6\) amperes.

Key Concepts

LR circuit

An LR circuit is one that contains both an inductor (L) and a resistor (R). In these circuits, a key aspect to consider is how the inductance and resistance interact when a voltage is applied.

**Key components:**

• **Inductor (L)**: This component stores energy as a magnetic field. Inductance is measured in henrys (H). In our problem, the inductance is given as 0.1 henrys.
• **Resistor (R)**: This component resists the flow of electric current, which causes a drop in the electrical voltage. Resistance is measured in ohms (Ω), and we have a resistance of 50 ohms in the problem.
• **Voltage Source**: The electromotive force applied is 30 volts, providing the energy to drive current through the circuit.

An LR circuit is modeled by a first-order linear differential equation that quantifies the relationship between these elements: \(L\frac{di}{dt} + Ri = E(t)\).
This relationship depicts how quickly the current changes over time and how it stabilizes eventually, as seen through the terms involving both L and R.

initial conditions

Initial conditions in differential equations specify the value of the function or its derivatives at a specific point, usually at the starting point, to find a unique solution. In the context of this problem, the initial condition provided is \(i(0) = 0\).

This signifies that at time \(t = 0\), the current in the circuit starts at 0 amperes. By incorporating initial conditions into our calculations, we can determine the constant in our general solution, making it specific to the problem we're solving.

**Why initial conditions matter:**

• They allow for the determination of a particular solution from a general one, by establishing a unique function that fits the physical scenario.
• They ensure the solution accurately describes the behavior of the circuit as it evolves over time.
• Using \(i(0) = 0\) allows us to find that \(C = -0.6\) in the general solution \(i(t) = Ce^{-500t} + 0.6\).

Without this initial condition, we would not be able to specify the exact path of the current.

particular solution

Within the scope of differential equations, a particular solution covers the non-homogeneous part of a linear differential equation. For our scenario, the equation is given by \(\frac{di}{dt} + 500i = 300\), owing to the applied voltage.

To find the particular solution, you assume a constant current, \(i_p(t)\). In this step-by-step solution, we determined this constant as \(i_p(t) = 0.6\) by setting up the equation for this steady state, leading to \(500A = 300\) and concluding with \(A = 0.6\).

**The role of the particular solution:**

• It accounts for the external force or input in the system, the 30 volts here.
• Provides the constant term in the general solution.
• Brings together the homogeneous solution to give a complete description of the circuit's current over time.

The particular solution reflects how the circuit behaves in the presence of a consistent driving force, showing the steady state current eventually reached.

homogeneous equation

The homogeneous equation comes from removing the external force in the differential equation, leaving \(\frac{di}{dt} + 500i = 0\) in this case. By solving this, it reflects the natural response of the circuit without any external driving voltage.

**Solving the homogeneous equation:**

• The solution results in \(i_h(t) = Ce^{-500t}\), where \(C\) is an arbitrary constant determined by initial conditions.
• This component diminishes over time, representing transient phenomena or the part of the current that fades as the system approaches stability.

**Importance of the homogeneous solution:**

• It highlights how the system returns to equilibrium in absence of external inputs.
• Demonstrates the effect of internal properties, like resistance and inductance, on the system's decay towards steady state.

The homogeneous part is crucial to fully understanding the transitional dynamics within an LR circuit, and forms half of the solution when combined with the particular solution.