ECE2040D_HW5_Soln

ECE2040D_HW5_Soln - ECE 2040D Solutions to HW#5 - Spring...

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Unformatted text preview: ECE 2040D Solutions to HW#5 - Spring 2009 P6.4-5 v1 va v1 v2 0 R1 R7 v2 vb v v 1 2 0 R2 R7 0 va 0 vb R1 R1 1 v1 v2 R7 R7 R2 R2 1 v2 v1 R7 R7 v v v 0 R6 0 vc vb b c c 0 R4 R6 R4 R6 v v v v R R a c c 0 0 0 v0 5 va (1 5 )vc R3 R3 R3 R5 v0 i0 R5 R1 R6 R3 R5 R R R R R R R (1 2 ) v2 5 (1 1 ) 6 3 5 2 v1 R7 R7 R3 R4 R6 R7 R3 R7 R3 R4 R6 R3 vc v0 " R5 Page 1 / 8 ECE 2040D Solutions to HW#5 - Spring 2009 P6.4-7 Apply KCL at the inverting input node of the op amp va 6 0 v 0 a 0 10000 30000 va 1.5 V 0 Apply KCL to the super node corresponding the voltage source: va 0 va 6 0 10000 30000 v 6 0 v v a b a 30000 10000 3va va 6 va vb 3 va 6 vb vb 2va 6 3V 0 0 Apply KCL at node b: vb v v v v v 6 vb b 0 a b a 10000 30000 30000 10000 3vb vb v0 va vb 3 va 6 vb v0 8vb 4va 18 12 V 0 0 Apply KCL at the output node of the op amp: i0 v0 v v 0 b 30000 30000 0 i0 0.7 mA Page 2 / 8 ECE 2040D Solutions to HW#5 - Spring 2009 P6.4-9 vb 12 vb 0 vb 4 V 40000 20000 The node voltages at the input nodes of an ideal op amp are equal, so vc KCL at node b: The node voltages at the input nodes of an ideal op amp are equal, so vd KCL at node g: v f vg vg 20e3 40e3 0 v 2 vf 3 vb 4 V . vc 0 10e3 4 V . KCL at node d: so vg 2 vf 3 15 V. 5 vd v f 20e3 vd 2 vf 3 20e3 0 vf 6 vd 5 24 V 5 The node voltages at the input nodes of an ideal op amp are equal, so vc vg 16 V. 5 Page 3 / 8 ECE 2040D Solutions to HW#5 - Spring 2009 P6.5-2 P6.5-3 Page 4 / 8 ECE 2040D Solutions to HW#5 - Spring 2009 P7.2-8 iR iC i v 200 dv C dt iR ic 1 12e2t u10-3 40 10 2 10 e 2t 25 1 2e2t P A 200 e 2 t P A 200 e2t 25 50 e 2t 25 150e 2t A P7.33 i (t ) p (t ) C dvc so read off slope of vc (t ) to get i (t ) dt vc (t ) i (t ) so multiply vc (t ) & i(t ) curves to get p (t ) Page 5 / 8 ECE 2040D Solutions to HW#5 - Spring 2009 P7.4-2 4P F in series with 4P F = 2 P F 2P F = 4P F 4P F in series with 4P F = 2P F i (t ) (2106 ) d (3 3 e 250t ) dt (2106 ) (0 3(250) e 250t ) 1.5 e 250t mA 4 F4 F 4P F+4P F 2P F P7.5-5 iL (t ) for 1 5 u103 1 5u103 v (W ) dW 0 s t 2 u106 0d t 1 P s vs (t ) t 4 mV 4u103 t 2u106 3 5u10 6 u106 5 iL (t ) iL (1s) 3 6 0 4 u10 dW 2u10 4u103 1u106 2u106 3 5u10 vs (t) 1 mV for 1P s dt 3 P s iL (t ) t 1u103 1 6 1u103 6 t 1 A 1u103 dW u106 (t 1u106 ) u106 3 5u103 1P s 5 5u103 5 5u10 iL ( 3P s ) 1u103 3u106 1u106 3 5u10 8 A 5 8 A 5 for 3P s d t vs (t ) 0 so iL (t ) remains Page 6 / 8 ECE 2040D Solutions to HW#5 - Spring 2009 P7.5-6 v(t ) d is (t ) (in general) dt d 1u103 t is (t ) = 1u103 for 0<t <1 P s is (t ) 1 6 dt 1u10 3 3 3 v(t ) (2u10 )1u10 t 4u10 1u103 2u106 t 4 V 2 u103 is (t ) 4 u 103 d is (t ) 0 dt v(t) = (2u103 )1u103 + 4u103 0 = 2 V is (t ) = 1 mA 3 for 1P s <t < 3P s 1u103 t for 3P s< t < 5P s is (t ) = 4u10 6 1u10 v(t ) = 4 2u106 t d 1u103 is (t ) = 1u106 dt when 5P s <t< 7P s v(t ) = 2 V when 7P s< t < 8P s is (t ) = 1u103 and d is (t ) = 0 dt is (t ) = v(t ) = 12 + 2 u 106 t when 8P s < t , then is (t ) v(t )= 0 1u103 d t 8u103 is (t) = 1u103 6 1u10 dt d is (t ) dt 0 0 P7.6-2 d 5 dt (4sin 2t ) (4sin 2t ) 5 (8cos 2t ) (4sin 2t ) 80 [2 cos 2t sin 2t ] 80 [sin(2t 2t ) sin(2t 2t )] 80 sin 4t W t t 80 t W (t)= p(W ) dW = 80 sin4W dW [cos 4W |0 ] 20 (1 cos4t) 0 0 4 p (t ) v(t ) i (t ) Page 7 / 8 ECE 2040D Solutions to HW#5 - Spring 2009 P7.7-3 L L 5 L and L L 2 2 2 d 5 5 L 25cos 250t (14103 sin 250 t ) L (14103 )(250) cos 250 t 2 dt 2 25 so L = 2.86 H 5 3 (1410 ) (250) 2 L L L L L L Page 8 / 8 ...
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