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Chapter 3: Problem 1

An RL circuit with a $$5 - \Omega$$ resistor and a $$0.05 - \mathrm { H }$$ inductor carries a current of 1 A at $$t = 0$$, at which time a voltage source $$E ( t ) = 5 \cos 120 t \mathrm { V }$$ is added. Determine the subsequent inductor current and voltage.

Short Answer

Expert verified
The subsequent current and voltage across the inductor can be found by solving the differential equation given by Kirchhoff's voltage law and applying the initial condition. The subsequent current is given by the general solution \(i = C e^{-5t/0.05} + A cos(120t + \phi)\), and the voltage is \(L\frac{di}{dt}\). The constants in the expressions will be determined by the initial conditions and requirements of the voltage source.

Step by step solution

01

Setting up the differential equation using Kirchhoff's voltage law (KVL)

KVL states that the sum of the electrical potential differences (voltage) around any closed loop is zero. We have a voltage source \(E(t)\) and the voltage across an inductor \(L\) and a resistor \(R\) in one loop. For an inductor, the voltage can be represented as \(Ldi/dt\) where \(di/dt\) is rate of change of current. Therefore, \(E(t) - L\frac{di}{dt} - Ri = 0\) where \(R\) is the resistance and \(i\) is the current.
02

Substituting given parameters

Substitute given parameters into the equation designed in Step 1. Here given \(E(t) = 5 \cos 120t\), \(R= 5\Omega\), \(L = 0.05 H\), we have \(5cos120t - 0.05 \frac{di}{dt} - 5i = 0\) or equivalently, \(0.05\frac{di}{dt} + 5i = 5cos120t\).
03

Solve the differential equation

This is a linear first order differential equation. We first need to solve the associated homogeneous equation \(0.05\frac{di}{dt} + 5i = 0\), which has a solution \(i_h = Ce^{-5t/0.05}\) for any constant \(C\). To solve the inhomogeneous equation, we assume a particular solution of the form \(i_p = A cos(120t + \phi)\). By substituting \(i_p\) back into the equation, we can solve for \(A\) and \(\phi\). The general solution is \(i = i_h + i_p\).
04

Applying the initial condition

We apply the initial condition \(i(0) = 1A\) to find the final solution to the differential equation. This will give us the subsequent current through the circuit.
05

Find the voltage across the inductor

Finally, the voltage across the inductor is given by \(L\frac{di}{dt}\). We can differentiate the current expression obtained in Step 4 and multiply it by the inductance \(L = 0.05 H\) to find the voltage.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Kirchhoff's voltage law
Kirchhoff's Voltage Law (KVL) is an essential principle in circuit analysis. It states that the sum of the voltages around any closed loop in a circuit must equal zero. This is because both potential gains and drops must cancel each other out to conserve energy. In the context of the RL circuit:
  • "voltage gain" comes from a source like a battery or AC generator.
  • "voltage drop" occurs across resistors and inductors.

In our problem, we apply KVL to the loop consisting of a voltage source, a resistor (\( R \)), and an inductor (\( L \)). The voltage source is given by \( E(t) = 5\cos(120t) \). The drops across the resistor and inductor are represented by \( Ri \) and \( L\frac{di}{dt} \), respectively. Applying KVL gives:\[E(t) - L\frac{di}{dt} - Ri = 0\]
This equation balances the voltage gain and drops, providing a foundation for determining how the current changes over time.
linear first order differential equation
A linear first order differential equation is a type of equation prevalent in circuit analysis and physiology. It is characterized by the formula \( a\frac{dy}{dt} + by = f(t) \). Here, \( y \) is the function to be solved, \( a \) and \( b \) are constants, and \( f(t) \) is some given function. For our RL circuit problem:
  • \( L = 0.05 \) and is associated with the derivative term (inductance effect).
  • \( R = 5 \) is linked to the current term (resistance effect).

The equation to solve was simplified to:\[0.05\frac{di}{dt} + 5i = 5\cos(120t)\]
To solve this, we handle both the homogeneous part, \( 0.05\frac{di}{dt} + 5i = 0 \), and the particular solution by assuming a certain form, typically based on the right-side expression. By combining these, we get the complete solution for how current \( i(t) \) behaves over time through the circuit.
initial condition
The term "initial condition" refers to a known value of the function or its derivatives at a specific point, usually at the start of our time measurement. For a realistic model of an electrical circuit, initial conditions are crucial because they account for the state of the system before any external influence, like a voltage source, comes into play.
  • In the given RL circuit problem, the initial current is specified as \( i(0) = 1 \) A.
  • This initial condition helps us determine any unknown constants in the general solution of the differential equation.

The process involves substituting \( t = 0 \) and \( i(0) = 1 \) into the general solution obtained from the differential equation, allowing us to solve for the constant \( C \). This step adapts the theoretical findings of the equation to the starting physical conditions of the circuit.

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