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

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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

A first-order RL parallel circuit has one resistor (or network of resistors) and a single inductor. First-order circuits can be analyzed using first-order differential equations. By analyzing a first-order circuit, you can understand its timing and delays. Analyzing such a parallel RL circuit, like the one shown here, follows the same process as analyzing an RC series circuit. So if you are familiar with that procedure, this should be a breeze. Start with the simple RL parallel circuit Because the resistor and inductor are connected in parallel in the example, they must have the same voltage v(t). The resistor current iR(t) is based on Ohm’s law: The element constraint for an inductor is given as where i(t) is the inductor current and L is the inductance. Now substitute v(t) = L di(t)/dt into Ohm’s law because you have the same voltage across the resistor and inductor:

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.

Replacing iZI(t) with Be kt and doing some math gives you the following: You have the characteristic equation after factoring out Be kt : The characteristic equation gives you an algebraic problem to solve for the constant k: Use k = – R/L and the initial inductor current I 0 at t = 0. This implies that B = I 0 , 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.

To analyze the RL parallel circuit further, you must calculate the circuit’s zero-state response, and then add that result to the zero- input response to find the total response for the circuit. Series RC Circuit In this section we see how to solve the differential equation arising from a circuit consisting of a resistor and a capacitor. (See the related section Series RL Circuit in the previous section.) In an RC circuit, the capacitor stores energy between a pair of plates. When voltage is applied to the capacitor, the charge builds up in the capacitor and the current drops off to zero. Case 1: Constant Voltage The voltage across the resistor and capacitor are as follows: VR=Ri and VC=1/c ∫idt Kirchhoff's voltage law says the total voltages must be zero. So applying this law to a series RC circuit results in the equation: Ri+1/c ∫idt=V One way to solve this equation is to turn it into a differential equation , by differentiating throughout with respect to t: R di/dt+i/c= 0

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.