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Austin Feldman April 16, 2011 Brewer - Mat 275 Honors Project Calculating Current in an LRC Circuit using Undetermined Coefficients and Laplace Transforms 1. Current through an LRC circuit can be calculated in a wide variety of ways. Although using undetermined coefficients is not the easiest method, it creates a conceptual model of current that allows for a better understanding of the calculations behind the formulas. Below is a simple LRC circuit with their corresponding expressions; R for resistor, L for inductor, C for capacitor, and Vac for an AC source. KVL: Sum of voltages in a closed loop is equal to zero. βππ΄πΆ + ππ + ππΏ + ππΆ = 0 Note that the voltage source has a negative sign since the designated current will go from the negative terminal to the positive terminal. Also, we know that the voltage across a resistor is iR, the voltage across an inductor is Ldi/dt, the voltage across a capacitor is Q/C, and the voltage across the given voltage source is Vsin(Οt). Using these equations and substituting them in to the above equation we derive the following equation. πΏ( ππ π ) + π β π + = ππ ππ(ππ‘) ππ‘ πΆ It is also known that the derivative of charge, Q, is equal to current i. Taking the derivative of the above equation and substituting i in for dQ/dt we derive the following. 1 π2π ππ 1 πΏ ( 2 ) + π ( ) + β π = πππππ (ππ‘) ππ‘ ππ‘ πΆ It is important to note that this equation has the same mathematical form as the equation for a damped oscillator. Thus we can solve the equation by first solving for the general solution and then the particular solution. First we will solve for the general solution. 1 πΏπ 2 + π π + ( ) = 0 πΆ Using the quadratic equation, we can solve for r and see that we will get repeated roots. βπ ± βπ 2 β 4πΏ/πΆ βπ ± βπ 2 β 4πΏ/πΆ π= , 2πΏ 2πΏ From these repeated values, we can deduce that the general solution will be equal to the following. ππ = βπΉ±βπΉπ βππ³/πͺ ππ³ πͺπ π + βπΉ±βπΉπ βππ³/πͺ ππ³ πͺπ ππ If initial values are given, it is important to note that we cannot plug them in here. They can only be plugged into a combined equation of yc and yp. To find a particular solution, we will use the form shown below. ππ = π¨πππ(ππ) + π©πππ(ππ) Taking the first and second derivatives we derive the following. π¦πβ² = ππ΄πππ (ππ‘) β ππ΅π ππ(ππ‘) π¦πβ²β² = βπ2 π΄π ππ(ππ‘) β π2 π΅πππ (ππ‘) Plugging these values into the left side of the original equation, we can solve for A and B. We find the following. ππ πΉπ½ π¨= π (ππ π³ β πͺ)π + (ππΉ)π π©= π βππ½(ππ π³ β πͺ) π (ππ π³ β πͺ)π + (ππΉ)π 2 At this time, if initial values are given, we can combine the general and particular solution and solve for C1 and C2. From these calculations we can see that the method of undetermined coefficients can derive an accurate mathematical formula for the current in an LRC circuit. Studying LRC circuits using undetermined coefficients lets us not only have proof that the calculations often assumed are correct, but also helps us understand how the mathematical expressions are derived. However, it is not practical to use undetermined coefficients to solve everyday circuits. While there are numerous ways to derive to current in a circuit, one way in particular is simple and widely used; Laplace Transforms. 2. Laplace Transforms are often implemented on each component in LRC circuits to quickly derive a desired function from the circuit. A basic LRC circuit is shown below along with values assigned to each component of the circuit. In order to derive an expression for current for the circuit shown above, the circuit must be converted to what is known as the βs domainβ. To do this, the Laplace Transform must be taken for each component and the resulting circuit is shown. 3 Using Kirchhoffβs Voltage Loop rule, which states that the voltages in any independent loop will sum to equal zero, a value for the current in the circuit in the βs domainβ can be determined. 2 12 3 β πΌ(π ) + π β πΌ(π ) + β πΌ(π ) β =0 π π 2 12 πΌ(π ) β (π + 3 + ) = π π π 2 + 3π + 2 12 πΌ(π ) β ( )= π π πΌ(π ) = 12 π β( 2 ) π π + 3π + 2 πΌ(π ) = πΌ(π ) = π 2 12 + 3π + 2 12 (π + 1)(π + 2) Using separation of variables, values of A and B are able to be calculated. 12 π΄ π΅ πΌ(π ) = (π +1)(π +2) = π +1 + π +2 12 = π΄(π + 2) + π΅(π + 1) 12 = π΄π + π΅π + 2π΄ + π΅ s: 0=A+B β B=-A constant: 12=2A+B 12=2A-A β A=12 β B=-12 4 12 12 πΌ(π ) = ( β ) π +1 π +2 Now that an equation for the current in the s domain has been found, the current in the regular circuit can be calculated. Since the s domain was found by taking the Laplace Transform of each component, going from the s domain back to the circuit can be done by taking the inverse Laplace Transform. The calculations to derive the current in the circuit are shown below. π(π‘) = πΏβ1 πΌ(π ) = πΏβ1 ( 12 12 β ) = 12π βπ‘ β 12π β2π‘ π +1 π +2 By taking the Laplace Transform in the original circuit, an expression for the current in the so domain was able to be derived. From there, the inverse Laplace Transform was taken in order to derive an expression for the current in the initial circuit. This method is both accurate and faster when compared to using undetermined coefficients. This explains why Laplace Transformations are a more accepted way of analyzing circuits in general. 5