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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 \(t \rightarrow \infty\)

Short Answer

Expert verified
The current as \(t \to \infty\) approaches \(\frac{3}{5}\) amps.

Step by step solution

01

Identify the RL circuit differential equation

An RL circuit is governed by the differential equation for the current \(i(t)\): \[ L \frac{di}{dt} + Ri(t) = E(t) \] where \(L\) is the inductance, \(R\) is the resistance, and \(E(t)\) is the electromotive force applied. Plugging in the given values, we have: \(0.1 \cdot \frac{di}{dt} + 50i(t) = 30\).
02

Simplify the differential equation

Rearrange and simplify the equation: \[ \frac{di}{dt} + 500i(t) = 300 \] dividing each term by 0.1 to simplify.
03

Solve the homogeneous equation

First, solve the homogeneous part \( \frac{di}{dt} + 500i(t) = 0 \). Separating variables gives \( \frac{di}{i} = -500 dt \). Integrating both sides, we have \( \ln|i| = -500t + C \). Therefore, \( i_h(t) = Ce^{-500t} \), where \(C\) is a constant.
04

Find a particular solution for the non-homogeneous equation

To find a particular solution \(i_p(t)\), assume \(i_p(t) = A\) where \(A\) is a constant. Substituting into \(\frac{di}{dt} + 500i(t) = 300\), we get \(500A = 300\), giving \(A = \frac{3}{5}\). So, \(i_p(t) = \frac{3}{5}\).
05

Write the general solution

The general solution is the sum of the homogeneous and particular solutions: \[ i(t) = Ce^{-500t} + \frac{3}{5} \].
06

Apply initial conditions to determine constants

We are given \(i(0) = 0\). Substitute \(t = 0\) into the general solution: \[ C \cdot e^{0} + \frac{3}{5} = 0 \] which simplifies to \(C + \frac{3}{5} = 0\). Thus, \(C = -\frac{3}{5}\).
07

Evaluate the final expression for the current

Substitute \(C = -\frac{3}{5}\) back into the general solution: \[ i(t) = -\frac{3}{5} e^{-500t} + \frac{3}{5} \].
08

Determine current as \( t \to \infty \)

As \(t \to \infty\), \(e^{-500t} \to 0\). Hence, \(i(t) \to \frac{3}{5}\).

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Key Concepts

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

RL Circuit
An RL circuit, also known as a resistor-inductor circuit, is a basic electrical circuit composed of resistors and inductors connected in series or parallel. In an RL series circuit, like in the exercise above, both the resistor (R) and inductor (L) are connected in a single path for the current. This circuit is essential because it illustrates how inductance and resistance interact to affect the flow of electric current.
In an RL circuit, a differential equation governs the behavior of the current over time, making it an excellent application of calculus. Here, we studied the differential equation involving the inductance, resistance, and electromotive force, which provides detailed insights into how voltage influences current flow in the presence of inductors and resistors. Understanding how to solve such equations is vital for designing and analyzing circuits in various electronic devices.
Electromotive Force
Electromotive force (EMF) is a crucial concept in electromagnetism, describing the voltage generated by a battery or a changing magnetic field. In simple terms, EMF is the force that drives electric current around a circuit. It is measured in volts and is often symbolized by the letter E.
In our RL series circuit, a 30-volt electromotive force (EMF) was applied, serving as the primary input that powers the circuit. Unlike potential difference or voltage drop that occurs across passive components of the circuit, EMF can be thought of as the 'source' of energy, ensuring that electric charges are pushed through the circuit. When solving RL circuit problems, it is fundamental to correctly model this force within the framework of the differential equation to predict how the current changes over time.
Inductance
Inductance is a property of electrical circuits whereby a circuit element, usually a coil or inductor, opposes a change in current. It is quantified by the unit 'henry' (H). Inductance results in the storage of energy in a magnetic field when electric current flows through the coil, which influences how the current changes with time.
In the exercise, an inductance of 0.1 henry was given. Inductors resist changes in current, and this property is used in practical applications to control the flow of electricity and filter signals. When we apply the differential equation to an RL circuit, this characteristic of inductors is mathematically represented by the term involving the derivative of the current, indicating how the inductor will react to changes and how energy is temporarily stored within its magnetic field.
Resistance
Resistance is a measure of the opposition that a circuit presents to the flow of electric current, expressed in ohms (Ω). It is a fundamental property that dissipates energy in the form of heat when current flows through it.
In this solution, the resistance of the circuit was 50 ohms, which works in tandem with the inductor to affect how quickly the current can reach its steady state. Resistance is represented in the differential equation as a term that is directly proportional to the current. The interplay between resistance and inductance determines the overall response of the RL circuit to an applied electromotive force, impacting how the current changes over time until it stabilizes. Understanding resistance is crucial for analyzing energy loss and efficiency in electrical circuits.

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