EE101: RLC Circuits (with DC sources)
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1 EE101: ircuits (with D sources) M. B. Patil mbpatil@ee.iitb.ac.in Department of Electrical Engineering Indian Institute of Technology Bombay
2 Series circuit i V V V 0 V
3 Series circuit i V V V 0 V KV: V + V + V = V 0 i + di dt + 1 Z i dt = V 0
4 Series circuit i V V V 0 V KV: V + V + V = V 0 i + di dt + 1 Z Differentiating w. r. t. t, we get, di dt + d2 i dt i = 0. i dt = V 0
5 Series circuit i V V V 0 V KV: V + V + V = V 0 i + di dt + 1 Z Differentiating w. r. t. t, we get, di dt + d2 i dt i = 0. i.e., d2 i dt 2 + di dt + 1 i = 0, a second-order ODE with constant coefficients. i dt = V 0
6 Parallel circuit i i i I 0 V
7 Parallel circuit i i i I 0 V K: i + i + i = I 0 1 V + 1 Z V dt + dv dt = I 0
8 Parallel circuit i i i I 0 V K: i + i + i = I 0 1 V + 1 Differentiating w. r. t. t, we get, 1 dv dt + 1 V + d2 V dt 2 = 0. Z V dt + dv dt = I 0
9 Parallel circuit i i i I 0 V K: i + i + i = I 0 1 V + 1 Differentiating w. r. t. t, we get, 1 dv dt + 1 V + d2 V dt 2 = 0. i.e., d2 V dt dv dt + 1 V = 0, a second-order ODE with constant coefficients. Z V dt + dv dt = I 0
10 Series/Parallel circuits i V V i i i I 0 V V 0 V
11 Series/Parallel circuits i V V i i i I 0 V V 0 V * A series circuit driven by a constant current source is trivial to analyze.
12 Series/Parallel circuits i V V i i i I 0 V V 0 V * A series circuit driven by a constant current source is trivial to analyze. Since the current through each element is known, the voltage can be found in a straightforward manner. V = i, V = di dt, V = 1 Z i dt.
13 Series/Parallel circuits i V V i i i I 0 V V 0 V * A series circuit driven by a constant current source is trivial to analyze. Since the current through each element is known, the voltage can be found in a straightforward manner. V = i, V = di dt, V = 1 Z i dt. * A parallel circuit driven by a constant voltage source is trivial to analyze.
14 Series/Parallel circuits i V V i i i I 0 V V 0 V * A series circuit driven by a constant current source is trivial to analyze. Since the current through each element is known, the voltage can be found in a straightforward manner. V = i, V = di dt, V = 1 Z i dt. * A parallel circuit driven by a constant voltage source is trivial to analyze. Since the voltage across each element is known, the current can be found in a straightforward manner. i = V /, i = dv dt, i = 1 Z V dt.
15 Series/Parallel circuits i V V i i i I 0 V V 0 V * A series circuit driven by a constant current source is trivial to analyze. Since the current through each element is known, the voltage can be found in a straightforward manner. V = i, V = di dt, V = 1 Z i dt. * A parallel circuit driven by a constant voltage source is trivial to analyze. Since the voltage across each element is known, the current can be found in a straightforward manner. i = V /, i = dv dt, i = 1 Z V dt. * The above equations hold even if the applied voltage or current is not constant, and the variables of interest can still be easily obtained without solving a differential equation.
16 Series/Parallel circuits A general circuit (with one inductor and one capacitor) also leads to a second-order ODE. As an example, consider the following circuit:
17 Series/Parallel circuits A general circuit (with one inductor and one capacitor) also leads to a second-order ODE. As an example, consider the following circuit: i 1 V 0 V 2
18 Series/Parallel circuits A general circuit (with one inductor and one capacitor) also leads to a second-order ODE. As an example, consider the following circuit: i 1 V 0 V 2 V 0 = 1 i + di dt + V (1) i = dv dt + 1 V 2 (2)
19 Series/Parallel circuits A general circuit (with one inductor and one capacitor) also leads to a second-order ODE. As an example, consider the following circuit: i 1 V 0 V 2 Substituting (2) in (1), we get V 0 = 1 i + di dt + V (1) i = dv dt + 1 V 2 (2) V 0 = 1 ˆV + V / 2 + ˆV + V / 2 + V, (3) V [] + V [ 1 + / 2 ] + V [1 + 1 / 2 ] = V 0. (4)
20 General solution onsider the second-order ODE with constant coefficients, d 2 y dt 2 + a dy dt + b y = K (constant).
21 General solution onsider the second-order ODE with constant coefficients, d 2 y dt 2 The general solution y(t) can be written as, + a dy dt + b y = K (constant). y(t) = y (h) (t) + y (p) (t), where y (h) (t) is the solution of the homogeneous equation, d 2 y dt 2 and y (p) (t) is a particular solution. + a dy dt + b y = 0,
22 General solution onsider the second-order ODE with constant coefficients, d 2 y dt 2 The general solution y(t) can be written as, + a dy dt + b y = K (constant). y(t) = y (h) (t) + y (p) (t), where y (h) (t) is the solution of the homogeneous equation, d 2 y dt 2 and y (p) (t) is a particular solution. + a dy dt + b y = 0, Since K = constant, a particular solution is simply y (p) (t) = K/b.
23 General solution onsider the second-order ODE with constant coefficients, d 2 y dt 2 The general solution y(t) can be written as, + a dy dt + b y = K (constant). y(t) = y (h) (t) + y (p) (t), where y (h) (t) is the solution of the homogeneous equation, d 2 y dt 2 and y (p) (t) is a particular solution. + a dy dt + b y = 0, Since K = constant, a particular solution is simply y (p) (t) = K/b. In the context of circuits, y (p) (t) is the steady-state value of the variable of interest, i.e., y (p) = lim t y(t), which can be often found by inspection.
24 General solution For the homogeneous equation, d 2 y dt 2 + a dy dt + b y = 0, we first find the roots of the associated characteristic equation, r 2 + a r + b = 0. et the roots be r 1 and r 2. We have the following possibilities:
25 General solution For the homogeneous equation, d 2 y dt 2 + a dy dt + b y = 0, we first find the roots of the associated characteristic equation, r 2 + a r + b = 0. et the roots be r 1 and r 2. We have the following possibilities: * r 1, r 2 are real, r 1 r 2 ( overdamped ) y (h) (t) = 1 exp(r 1 t) + 2 exp(r 2 t).
26 General solution For the homogeneous equation, d 2 y dt 2 + a dy dt + b y = 0, we first find the roots of the associated characteristic equation, r 2 + a r + b = 0. et the roots be r 1 and r 2. We have the following possibilities: * r 1, r 2 are real, r 1 r 2 ( overdamped ) y (h) (t) = 1 exp(r 1 t) + 2 exp(r 2 t). * r 1, r 2 are complex, r 1,2 = α ± jω ( underdamped ) y (h) (t) = exp(αt) [ 1 cos(ωt) + 2 sin(ωt)].
27 General solution For the homogeneous equation, d 2 y dt 2 + a dy dt + b y = 0, we first find the roots of the associated characteristic equation, r 2 + a r + b = 0. et the roots be r 1 and r 2. We have the following possibilities: * r 1, r 2 are real, r 1 r 2 ( overdamped ) y (h) (t) = 1 exp(r 1 t) + 2 exp(r 2 t). * r 1, r 2 are complex, r 1,2 = α ± jω ( underdamped ) y (h) (t) = exp(αt) [ 1 cos(ωt) + 2 sin(ωt)]. * r 1 = r 2 = α ( critically damped ) y (h) (t) = exp(αt) [ 1 t + 2 ].
28 Parallel circuit I 0 i i i V =10 Ω =1 µf =0.44 mh I 0 = 100 ma
29 Parallel circuit I 0 i i i V =10 Ω =1 µf =0.44 mh I 0 = 100 ma i (0 ) = 0 A i (0 + ) = 0 A. V (0 ) = 0 V V (0 + ) = 0 V.
30 Parallel circuit I 0 i i i V =10 Ω =1 µf =0.44 mh I 0 = 100 ma i (0 ) = 0 A i (0 + ) = 0 A. V (0 ) = 0 V V (0 + ) = 0 V. d 2 V dt dv dt + 1 V = 0 (as derived earlier)
31 Parallel circuit I 0 i i i V =10 Ω =1 µf =0.44 mh I 0 = 100 ma i (0 ) = 0 A i (0 + ) = 0 A. V (0 ) = 0 V V (0 + ) = 0 V. d 2 V dt dv dt + 1 V = 0 (as derived earlier) The roots of the characteristic equation are (show this): r 1 = s 1, r 2 = s 1.
32 Parallel circuit I 0 i i i V =10 Ω =1 µf =0.44 mh I 0 = 100 ma i (0 ) = 0 A i (0 + ) = 0 A. V (0 ) = 0 V V (0 + ) = 0 V. d 2 V dt dv dt + 1 V = 0 (as derived earlier) The roots of the characteristic equation are (show this): r 1 = s 1, r 2 = s 1. The general expression for V (t) is, V (t) = A exp(r 1 t) + B exp(r 2 t) + V ( ),
33 Parallel circuit I 0 i i i V =10 Ω =1 µf =0.44 mh I 0 = 100 ma i (0 ) = 0 A i (0 + ) = 0 A. V (0 ) = 0 V V (0 + ) = 0 V. d 2 V dt dv dt + 1 V = 0 (as derived earlier) The roots of the characteristic equation are (show this): r 1 = s 1, r 2 = s 1. The general expression for V (t) is, V (t) = A exp(r 1 t) + B exp(r 2 t) + V ( ), i.e., V (t) = A exp( t/τ 1 ) + B exp( t/τ 2 ) + V ( ), where τ 1 = 1/r 1 = 15.4 µs, τ 2 = 1/r 1 = 28.6 µs.
34 Parallel circuit As t, V = di dt = 0 V V ( ) = 0 V.
35 Parallel circuit As t, V = di dt = 0 V V ( ) = 0 V. V (t) = A exp( t/τ 1) + B exp( t/τ 2),
36 Parallel circuit As t, V = di dt = 0 V V ( ) = 0 V. V (t) = A exp( t/τ 1) + B exp( t/τ 2), Since V (0 + ) = 0 V, we have, A + B = 0. (1)
37 Parallel circuit As t, V = di dt = 0 V V ( ) = 0 V. V (t) = A exp( t/τ 1) + B exp( t/τ 2), Since V (0 + ) = 0 V, we have, A + B = 0. (1) Our other initial condition is i (0 + ) = 0 A, which can be used to obtain dv dt (0+ ).
38 Parallel circuit As t, V = di dt = 0 V V ( ) = 0 V. V (t) = A exp( t/τ 1) + B exp( t/τ 2), Since V (0 + ) = 0 V, we have, A + B = 0. (1) Our other initial condition is i (0 + ) = 0 A, which can be used to obtain dv dt (0+ ). i (0 + ) = I 0 1 V (0+ ) dv dt (0+ ) = 0 A, which gives
39 Parallel circuit As t, V = di dt = 0 V V ( ) = 0 V. V (t) = A exp( t/τ 1) + B exp( t/τ 2), Since V (0 + ) = 0 V, we have, A + B = 0. (1) Our other initial condition is i (0 + ) = 0 A, which can be used to obtain dv dt (0+ ). i (0 + ) = I 0 1 V (0+ ) dv dt (0+ ) = 0 A, which gives (A/τ 1) + (B/τ 2) = I 0/. (2)
40 Parallel circuit As t, V = di dt = 0 V V ( ) = 0 V. V (t) = A exp( t/τ 1) + B exp( t/τ 2), Since V (0 + ) = 0 V, we have, A + B = 0. (1) Our other initial condition is i (0 + ) = 0 A, which can be used to obtain dv dt (0+ ). i (0 + ) = I 0 1 V (0+ ) dv dt (0+ ) = 0 A, which gives From (1) and (2), we get the values of A and B, and (SEQUE file: ee101 rlc 1.sqproj) (A/τ 1) + (B/τ 2) = I 0/. (2) V (t) = 3.3 [exp( t/τ 1) exp( t/τ 2)] V. (3)
41 Parallel circuit As t, V = di dt = 0 V V ( ) = 0 V. V (t) = A exp( t/τ 1) + B exp( t/τ 2), Since V (0 + ) = 0 V, we have, A + B = 0. (1) Our other initial condition is i (0 + ) = 0 A, which can be used to obtain dv dt (0+ ). i (0 + ) = I 0 1 V (0+ ) dv dt (0+ ) = 0 A, which gives From (1) and (2), we get the values of A and B, and (SEQUE file: ee101 rlc 1.sqproj) (A/τ 1) + (B/τ 2) = I 0/. (2) V (t) = 3.3 [exp( t/τ 1) exp( t/τ 2)] V. (3) I 0 i i i V =10 Ω =1 µf =0.44 mh I 0 = 100 ma V (Volts) i (ma) i (ma) i (ma) time (ms) time (ms)
42 Series circuit: home work i V V 0 V 5 V t=0 V s V =1 mh =1 µf
43 Series circuit: home work i V V 0 V 5 V t=0 V s V =1 mh =1 µf (a) Show that the condition for critically damped response is = 63.2 Ω.
44 Series circuit: home work i V V 0 V 5 V t=0 V s V =1 mh =1 µf (a) Show that the condition for critically damped response is = 63.2 Ω. (b) For = 20 Ω, derive expressions for i(t) and V (t) for t > 0 (Assume that V (0 ) = 0 V and i (0 ) = 0 A). Plot them versus time.
45 Series circuit: home work i V V 0 V 5 V t=0 V s V =1 mh =1 µf (a) Show that the condition for critically damped response is = 63.2 Ω. (b) For = 20 Ω, derive expressions for i(t) and V (t) for t > 0 (Assume that V (0 ) = 0 V and i (0 ) = 0 A). Plot them versus time. (c) epeat (b) for = 100 Ω.
46 Series circuit: home work i V V 0 V 5 V t=0 V s V =1 mh =1 µf (a) Show that the condition for critically damped response is = 63.2 Ω. (b) For = 20 Ω, derive expressions for i(t) and V (t) for t > 0 (Assume that V (0 ) = 0 V and i (0 ) = 0 A). Plot them versus time. (c) epeat (b) for = 100 Ω. (d) ompare your results with the following plots. (SEQUE file: ee101 rlc 2.sqproj)
47 Series circuit: home work i V V 0 V 5 V t=0 V s V =1 mh =1 µf (a) Show that the condition for critically damped response is = 63.2 Ω. (b) For = 20 Ω, derive expressions for i(t) and V (t) for t > 0 (Assume that V (0 ) = 0 V and i (0 ) = 0 A). Plot them versus time. (c) epeat (b) for = 100 Ω. (d) ompare your results with the following plots. (SEQUE file: ee101 rlc 2.sqproj) 8 4 = 20 Ω V 5 = 100 Ω V 0 4 V V 0 V V time (ms) time (ms)
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