C26-act - ELECTRIC CIRCUITS ECSE-2010 Class 26 ACTIVITY...

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1 ELECTRIC CIRCUITS ECSE-2010 Class 26 ACTIVITY 26-1 i(t) 200 Ω 60 Ω 20 Ω 5 F μ 10 mH i(t) 1.6cos(4000t) Amps = 1 v + v + Draw and Label the Frequency Domain Diagram 2 v + AC Ladder Network ACTIVITY 26-1a 0 1.6/0 200 Ω 60 Ω 20 Ω j50 −Ω j40 Ω C 6 j Zj 5 0 (4000)(5x10 ) =− Ω 1 V +− V + 2 V + 3 L Z j(4000)(10x10 ) j40 == Ω 4000 ω = I Frequency Domain Diagram ACTIVITY 26-1a 0 1.6/0 200 Ω 60 Ω 20 Ω j50 Ω j40 Ω 1 V + V + 2 V + 4000 = I 2 Z 1 Z Z 21 Find Z , Z , Z, Y ACTIVITY 26-1a 0 1.6/0 200 Ω 60 Ω 20 Ω j50 j40 Ω 1 V V + 2 V + 4000 = I 2 Z 2 Z6 0 ( 2 0 j 4 0 ) =+ 1200 j2400 80 j40 + = + 80 j40 x 24 j18 =+ Ω 0 30 /36.9 ACTIVITY 26-1a I 0 1.6/0 4000 = 200 Ω j50 1 Z 0 12 Z j50 Z 24 j32 40 / 53.1 = −+=− Ω = − Ω 24 j18
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2 ACTIVITY 26-1b Z I 0 1.6/0 4000 ω = 200 Ω 1 Z2 4 j 3 2 =− Ω 1 Z 0 1 111 Y .02 j.02 S .028 /45 S Z 200 Z == + = + = ACTIVITY 26-1b Z I 0 1.6/0 4000 = 200 Ω 1 4 j 3 2 1 Z 0 Y .02 j.02 S =+ = 0 1 Z 25 j25 35.4 / 45 Y = = − Ω ACTIVITY 26-1b I Z 0 Z 25 j25 35.4 / = I 25 Ω j25 −Ω = ACTIVITY 26-1b I Z 0 Z 35.4 / =− Ω 0 1.6/0 4000 = V + 0 V Z x 1.6 /0 = 40 j40 V = 0 56.6 / 45 V 0 v(t) 56.6 cos (4000t 45 ) V => = ACTIVITY 26-1c I 0 1.6/0 4000 = V + 200 Ω j50 24 j18 1 V +− 0 1 2 j50 V x V 10 j70 V 70.7 / 81.9 V
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This note was uploaded on 10/21/2010 for the course ECSE 2010 taught by Professor Millard during the Spring '08 term at Rensselaer Polytechnic Institute.

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C26-act - ELECTRIC CIRCUITS ECSE-2010 Class 26 ACTIVITY...

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