Solving First-Order Diff. Equations in Electrical Eng.: RC and RL Circuits, Schemes and Mind Maps of Mathematical Analysis

Download Solving First-Order Diff. Equations in Electrical Eng.: RC and RL Circuits and more Schemes and Mind Maps Mathematical Analysis in PDF only on Docsity! ENGINEERING ANALYSIS Application of Ordinary Differential Equations Name : Amr Gamal Mahmoud Mohamed Directed by :prof .Eng /shazley abdo Series RL Circuit The RL circuit shown above has a resistor and an inductor connected in series. A constant voltage V is applied when the switch is closed. The (variable) voltage across the resistor is given by: VR=iR The (variable) voltage across the inductor is given by: VL=L dt/di Kirchhoff's voltage law says that the directed sum of the voltages around a circuit must be zero. This results in the following differential equation: Ri+ L di/dt=V Once the switch is closed, the current in the circuit is not constant. Instead, it will build up from zero to some steady state. Solving the DE for a Series RL Circuit The solution of the differential equation Ri+ L dt/di=V is: i=V/R(1−e−(R/L)t) Example 1 An RL circuit has an emf of 5 V, a resistance of 50 Ω, an inductance of 1 H, and no initial current.Find the current in the Kirchhoff’s current law (KCL) says the incoming currents are equal to the outgoing currents at a node. Use KCL at Node A of the sample circuit to get iN(t) = iR(t) =i(t). Substitute iR(t) into the KCL equation to give you The RL parallel circuit is a first-order circuit because it’s described by a first-order differential equation, where the unknown variable is the inductor current i(t). A circuit containing a single equivalent inductor and an equivalent resistor is a first- order circuit. Knowing the inductor current gives you the magnetic energy stored in an inductor. Calculate the zero-input response for an RL parallel circuit Here is how the RL parallel circuit is split up into two problems: the zero-input response and the zero-state response. Here, you’ll start by analyzing the zero-input response. To simplify matters, you set the input source (or forcing function) equal to 0: iN(t) = 0 amps. This means no input current for all time — a big, fat zero. The first-order differential equation reduces to For an input source of no current, the inductor current iZI is called a zero-input response. No external forces are acting on the circuit except for its initial state (or inductor current, in this case). The output is due to some initial inductor current I0 at time t = 0. You make a reasonable guess at the solution (the natural exponential function!) and substitute your guess into the RL first- order differential equation. Assume the inductor current and solution to be iZI(t) = Bekt This is a reasonable guess because the time derivative of an exponential is also an exponential. Like a good friend, the exponential function won’t let you down when solving these differential equations. You determine the constants B and k next. Substitute your guess iZI(t) = Bekt into the differential equation: Replacing iZI(t) with Bekt and doing some math gives you the following: You have the characteristic equation after factoring out Bekt: The characteristic equation gives you an algebraic problem to solve for the constant k: Use k = –R/L and the initial inductor current I0 at t = 0. This implies that B = I0, so the zero-input response iZI(t) gives you the following: The constant L/R is called the time constant. The time constant provides a measure of how long an inductor current takes to go to 0 or change from one state to another. circuit is influenced by the charging and discharging of the capacitor. Differential Equation for a Parallel RC Circuit: The voltage across a capacitor (vc(t)) is given by the equation: VC=1/c ∫idt Ohm's law for a resistor is: vR(t) = Ri(t) Kirchhoff's current law for the parallel circuit states that the current through the resistor is equal to the total current, so: i(t) = iR(t) + ic(t) Combining these equations, we get: Ri(t) + 1/c ∫idt=v(t) Ordinary Differential Equation (ODE): Differentiating both sides with respect to time gives a first-order linear ODE: R di/dt+i/c= v(t) Solution and Applications: The solution to this ODE involves finding the current (\(i(t)\)) as a function of time. The behavior of the circuit is characterized by the charging and discharging of the capacitor. Applications include: 1.Charging and Discharging Analysis:Understanding how the capacitor charges and discharges over time. 2. Time Constant:Determining the time constant (\(RC\)) of the circuit, which influences the rate at which the capacitor charges or discharges. 3. Filter Design: Parallel RC circuits are used in filters, and ODEs help optimize filter performance. 4.Transient Response: Analyzing the transient response of the circuit during changes in input voltage.