Circuits with Capacitor and Inductor


 Elmer Lamb
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1 Circuits with Capacitor and Inductor We have discussed so far circuits only with resistors. While analyzing it, we came across with the set of algebraic equations. Hereafter we will analyze circuits with inductors and capacitors. When we apply KCL or KVL to the circuits (in time domain) containing inductors and capacitors, it will produce differential equations. Capacitor and Inductor do not allow either voltage or current to jump instantaneously 1 from one state to another state. 1 It will be possible If infinite current or infinite voltage is available.
2 C Circuit Let us consider this circuit. The capacitor has an initial charge. (V c (0 ) = V 0 ) t = 0 V s V c (t) C By KCL at t = 0, C dv c dt V c V s = 0 C dv c V c = V s dt It is a first order differential equation.
3 Let us follow the three simple steps to solve 2 the differential equations. 1. Find the homogeneous solution. 2. Find the particular solution. 3. Find the total solution which is sum of above two. 4. Find the constants using initial conditions. To find the homogeneous solution, set the forcing function (V s ) to zero. C dv ch V ch = 0 dt Let us assume the solution. Plug into the equation, V ch = Ae st CAse st Ae st = 0 2 Solving differential equations is simply a guess work
4 Cs 1 = 0 s = 1 C V ch = Ae (t/c) Let τ = C. where τ is a time constant. Particular solution: V ch = Ae (t/τ) C dv cp dt V cp = V s Guess any solution that satisfies this equation. In this case, it is very simple. V cp = V s
5 Total solution: V c = V ch V cp V c = Ae (t/c) V s To find the constant, If V 0 < V s, V c V s V c (t) V c (0 ) = V c (0 ) = V 0 At t = 0, V c (0 ) = A V s V 0 t (s) A = V 0 V s The total solution is V c = (V 0 V )e (t/c) V s The first component of equation is called transient (natural) response. The second component is steady state (forced) response.
6 Example  C Circuit Let us consider this circuit. The capacitor has a zero initial charge. (V c (0 ) = 0V ) t = 0 1 Ω 3 V V c (t) 1 F τ = C = 1 sec. V c = 3(1 e t ) V V c &i 3 V V c (t) i = C dv c dt = 3e t A i(t) t (s)
7 C Circuit Let us consider this case. t = 0 V s V c (t) C Before t = 0, the capacitor has been charged to V s 2 volts. For t > 0 V c (0) = V 2 V c (t) C C dv c dt C dv c dt V c = 0 V c = 0
8 Since there is no forcing function, the solution V c (t) will contain only homogeneous solution. V c (t) = Ae (t/c) To find A, we have to use an initial condition. At t = 0, V c (0) = A V c (t) = V c (0)e (t/c) V c V c (0) V c (t) t (s)
9 L Circuit Let us consider the L circuit. The current in the inductor before t = 0 is I 0. t = 0 V s v L (t) i L L After t = 0, V s = i L L di L dt L di L dt i L = V s It is a first order differential equation.
10 Let us follow the same steps to solve this equation. Homogeneous solution: Set the forcing function to zero. L di LH dt i LH = 0 Assume i LH = Ae st and substitute in the above equation. L Asest Ae st = 0 L s 1 = 0 s = L i LH = Ae (t/l) Let τ = L. where τ is a time constant. i LH = Ae (t/τ)
11 Particular solution: di LP L i LP = V s dt Guess any solution that satisfies it. Total solution: I LP = V s i L = i LH i LP i L (t) = Ae (t/τ) V s To find constant A, let us use an initial condition. At t = 0, i L (0) = A V s A = I 0 V s
12 i L (t) = (I 0 V s )e(t/τ) V s where τ = L. Let I = V s be the current at steady state. i L (t) = (I 0 I )e (t/τ) I First term is a natural (transient) response. Second term is a steady state (forced) response. i L I I 0 If I 0 < I, i L (t) t (s) The voltage across the inductor is v L (t) = L di L dt = (I I 0 )e (t/τ)
13 L Circuit Let us consider this case. t = 0 I L i L Before t = 0, the inductor current is i L (0) = I. After t = 0, L di L dt i L = 0 Since it is a homogeneous first order differential equation, i L (t) will contain only natural (transient) response.
14 The solution i L (t) = Ae (t/τ) where τ = L. To fins A, i L (0) = A A = I i L (t) = Ie (t/τ) i L I i L (t) t (s)
15 Example L Circuit Find the time constant of the following circuit. t = 0 V s L i L τ = L Th where Th is the Thevenin resistance from the inductor terminals after voltage source is shorted. Th = 2 τ = 2L
16 LC Circuit Let us consider this circuit. t = 0 L i L V s V c C After t = 0, Since i L = C dv c dt, V s = i L L di L dt V c V s = C dv c dt LC d 2 V c dt 2 V c LC d 2 V c dt 2 C dv c V c = V s dt It is a second order linear constant coefficient differential equation.
17 Let us use the same steps to solve this. Homogeneous solution: To obtain this, make V s = 0. LC d 2 V ch dt 2 C dv ch dt V ch = 0 Assume, V ch = Ae st and substitute. LCs 2 Ae st CsAe st Ae st = 0 s 2 L s 1 LC = 0 It is called a characteristics equation. The roots of the characteristics equation are s 1,2 = L ± ( L 2 ) 2 4 LC
18 Let α = 2L and ω 0 = 1 LC. ( s 1,2 = ) 2 2L ± 1 2L LC s 1,2 = α ± α 2 ω0 2 Case 1: Over Damping (α > ω 0 ). The two roots are real. Let s 1 = α 1 and s 2 = α 2. V ch = A 1 e α 1t A 2 e α 2t Case 2: Critical Damping (α = ω 0 ). The two roots are real and equal. Let s 1, s 2 = α. V ch = e αt (A 1 t A 2 )
19 Case 3: Under Damping (α < ω 0 ). The two roots will be complex. s 1,2 = α ± j ω0 2 α2 Let ω d = ω 2 0 α2. s 1,2 = α ± jω d s 1 = α jω d, s 2 = α jω d V ch = A 1 e (αjω d )t A 2 e (αjω d )t A 1 and A 2 must be complex conjugate of each other in order to make V c (t) real. V ch = e αt (B 1 cos(ω d t) B 2 sin(ω d t))
20 Particular Solution: LC d 2 V cp dt 2 C dv cp dt V cp = V s Guess any solution that satisfies the above equation. Total solution: 1. Over Damping: 2. Critical Damping: 3. Under Damping: V cp = V s V c (t) = A 1 e α1t A 2 e α2t V s V c (t) = e αt (A 1 t A 2 ) V s V c (t) = e αt (B 1 cos(ω d t) B 2 sin(ω d t)) V s To find constants: Use initial conditions. i.e., V c (0) and i L (0).
21 V c (t) Under Damping V s Critical Damping Over Damping t (s) Figure: Plot of V c (t) for different cases
22 LC Circuit Let us consider this circuit. t = 0 L i L V s V c C After t = 0, Since i L = C dv c dt, V s = L di L dt V c V s = LC d 2 V c dt 2 V c LC d 2 V c dt 2 V c = V s
23 Homogeneous solution: To obtain this, make V s = 0. LC d 2 V ch dt 2 V ch = 0 Assume, V ch = Ae st and substitute. LCs 2 Ae st Ae st = 0 s 2 1 LC = 0 The roots of the characteristics equation are Since ω 0 = 1 LC, 1 1 s 1,2 = j, j LC LC s 1,2 = jω 0, jω 0 V ch = A 1 e (jω 0)t A 2 e (jω 0)t
24 In order V c (t) to be real, A 1 and A 2 must be complex conjugates of each other. V ch = B 1 cos(ω 0 t) B 2 sin(ω 0 t) Particular solution: Total solution: V cp = V s V c (t) = B 1 cos(ω 0 t) B 2 sin(ω 0 t) V s To find constants: Use initial conditions. Since there is no esistor in the circuit, the response will oscillate forever. It is an undamped circuit.
25 V c (t)&i L (t) 2V s V c (t) V s i L (t) t (s) Figure: Plots of V c (t) and i L (t) for LC circuit
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