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RC circuit with voltage source K. Shum Apr, 2010. Objective: Model an RC circuit by linear differential equation The voltage source is sin(t) (measured in volt). The switched is closed at t=0. The initial voltage across the capacity is -2 V. Suppose that R=1 Ohm, C=3 H and =1. Describe the voltage across the resistor VR(t) after the switch is closed. (To help “debugging”, we will work with symbol R, C and during derivation, and plug in numerical values at the end) Let the voltage of the capacitor be Vc(t) and the current (clockwise) be I(t). From KVL, ( )= ( )+ ( ) Using the relationship V = I/C for capacitor and Ohm’s law V= IR, we get ( ) = ( ) / + ′( ) We write the differential equation in standard form 1 ( )+ ( )= ( ) (*) Note that the coefficient of I(t) is 1/RC. (RC is usually called the time constant) We solve the differential equation in (*) by integrating factor. That is, we want to find a function f(t) so that the derivative of f(t) is f(t)/RC. From our knowledge of exponential function, we see that et/RC is such a function. Multiply both sides of (*) by et/RC, / 1 / / ( )+ ( )= ( ) The left hand side is now the derivative of e / I(t) / / ( ) = ( ) . (**) Integrate both sides of (**), / / ( ) =∫ ( ) . (***) Using the integral = ( + )+ + from calculus, (I use the symbol K for the integration constant), we integrate the right hand side of (***), and get / 1 ( )= sin + cos + . 1 +( ) Here, K is an arbitrary real number. The voltage across the resistor is 1 / R I(t) = sin + cos + . 1 + ( ) 2 Substitute R,C and by their numerical values, the general solution is 9 1 / ( )= sin + cos + . 10 3 The sine and cosine in the above equation can be combined and written in an amplitude-and-phase form. ( )= / cos( − (tan 3)) + General solution The constant K can be any real number. The phase tan 3 is approximately equal to 1.25. From the initial condition (0) = 2, we solve for the constant , which turns out to be 1.7 in this case. The answer to the original problem is 9 / ( )= cos( − (tan 3)) + 1.7 . 10 The solution with initial voltage 2 is plotted in the following figure. 2.5 2 1.5 1 VR(t) 0.5 0 -0.5 -1 0 5 10 15 20 25 30 t We observe an initial transient period of about 15 seconds. After 15 seconds, the voltage oscillates with amplitude (9/10) = 0.95. 3 Discussion: We plot below several solutions corresponding to different initial conditions. 5 4 3 2 1 VR 0 -1 -2 -3 -4 0 5 10 15 20 25 30 t Steady state We see that after about 15 seconds, the five sample solutions, each for a particular initial condition, are visually indistinguishable in the graph. In the general solution, the first term is called the stead-state solution. ( )= / cos( − (tan 3)) + Steady-state solution The entire solution is called the transient-state solution. It describes how the system transits from rest to steady state. The second term K exp(-t/3) is responsible for the transient effect. Its rate of decay is dictated by the time constant 3 seconds. Also note that the steady-state solution does not depend on the initial condition. The initial condition only affects the constant K, which is attached to the second term only. 4