DeCarlo Ch18b Solutions

# Linear Circuit Analysis: The Time Domain and Phasor Approach

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5/31/01 Mag Crt Probs P18-1 © R. A. DeCarlo & P.M. Lin C HAPTER 18 P ROBLEM S OLUTIONS S OLUTION P ROBLEM 18.34. There is a correction to this problem: set M = 3 H. (a) The stored energy at t = 0 is: W (0) = 0.5 L 1 i 1 2 (0) + 0.5 L 2 i 2 2 (0) + Mi 1 (0) i 2 (0) = 8 J »L1 = 10; L2 = 2; M = 3; i10 = 1; i20 = -3; »W0 = 0.5*L1*i10^2 + 0.5*L2*i20^2 + M*i10*i20 W0 = 5 (b) Writing two differential mesh equations we obtain L 1 di 1 dt + M di 2 dt + R 1 i 1 = 10 di 1 dt + 3 di 2 dt + i 1 = 0 and L 2 di 2 dt + M di 1 dt + R 2 i 2 = 2 di 2 dt + 3 di 1 dt + i 2 = 0 Taking the Laplace transform of these equations yields 10 sI 1 - 10 i 1 (0) + 3 sI 2 - 3 i 2 (0) + I 1 = (10 s + 1) I 1 + 3 sI 2 - 10 i 1 (0) - 3 i 2 (0) = 0 and 2 sI 2 - 2 i 2 (0) + 3 sI 1 - 3 i 1 (0) + I 2 = (2 s + 1) I 2 + 3 sI 1 - 3 i 1 (0) - 2 i 2 (0) = 0 Putting these equations in matrix form yields (10 s + 1) 3 s 3 s (2 s + 1) I 1 I 2 = 10 i 1 (0) + 3 i 2 (0) 3 i 1 (0) + 2 i 2 (0) = 1 - 3 Solving yields I 1 I 2 = (10 s + 1) 3 s 3 s (2 s + 1) - 1 1 - 3 = 1 11 s 2 + 12 s + 1 2 s + 1 - 3 s - 3 s 10 s + 1 1 - 3 = 1 (11 s + 1)( s + 1) 11 s + 1 - 33 s - 3 = 1 s + 1 1 - 3 Therefore, by inspection, i 1 ( t ) = e - t u ( t ) A and i 2 ( t ) = - 3 e - t u ( t ) A Remark: normally, i1(t) and i2(t) would have two exponential terms present. Because of the special choice of initial conditions, a pole cancelled out. (c) From equation 18.24 with the lower limit changed to zero and the upper limit changed to , we have

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5/31/01 Mag Crt Probs P18-2 © R. A. DeCarlo & P.M. Lin W (0, ) = v 1 i 1 + v 2 i 2 ( 29 0 dt = 0.5 L 1 i 1 2 ( ) + 0.5 L 2 i 2 2 ( ) + Mi 1 ( ) i 2 ( ) - 0.5 L 1 i 1 2 (0) + 0.5 L 2 i 2 2 (0) + Mi 1 (0) i 2 (0) From part (b) all currents at t = are zero, hence W (0, ) = - 0.5 L 1 i 1 2 (0) + 0.5 L 2 i 2 2 (0) + Mi 1 (0) i 2 (0) = - 5 J The result of part (a) indicates that the initial store energy is 5 J. The result of part (c) indicates that the energy returned to the circuit is also 5 J, i.e., the total energy accumulated in the inductors over [0, ) is –5 J. Hence 5 J is dissipated in the resistors. Remark: the interested student might computer the integral R 1 i 1 2 ( t ) + R 2 i 2 2 ( t ) ( 29 0 dt , the actual energy dissipated in the resistors over [0, ), and show that this is 5 J. S OLUTION P ROBLEM 18.35. (a) »L1 = 4; L2 = 9; M = 3; »I1 = 2; I2 = -3; »W = 0.5*L1*I1^2 + 0.5*L2*I2^2 - M*I1*I2 W = 6.6500e+01 (b) »K = 0.5*L1*I1^2 K = 8 »% Minimize (over I2) K + 0.5*9*I2^2 - 3*2*I2 »% Take Derivative and set to zero; then solve for I2. »% Derivative is: 9*I2 – 6 = 0 »% The result is I2 = 2/3 A. »I2 = 2/3; »Wmin = 0.5*L1*I1^2 + 0.5*L2*I2^2 - M*I1*I2 Wmin = 6 (c)
5/31/01 Mag Crt Probs P18-3 © R. A. DeCarlo & P.M. Lin (d) »L1 = 4; L2 = 9; M = 3; »k = M/sqrt(L1*L2) k = 5.0e-01 S OLUTION P ROBLEM 18.37. »k = 0.5; »L1 = 9; L2 = 4;L3 = 1; »M = 0.5*sqrt(L1*L2) M = 3 »Lcpld = L1 + L2 + 2*M Lcpld = 19 »Leq = Lcpld + L3 Leq = 20 »Imax = 2; »Wmax = 0.5*Leq*Imax^2 Wmax = 40 J S OLUTION P ROBLEM 18.41.

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