hw_4s08_soln - . 08 cos 20 tA (b) v c ( t ) =-60 sin(80 t +...

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EECE 519 Homework #4 (Due Thursday, February 28) 1. Find the equivalent capacitance. 20 μF 60 μF 120 μF 50 μF 70 μF 50 μF and 70 μF in parallel = 50+70 = 120 μF , 60 μF and 120 μF in series = (60)(120) 60+120 = 40 μF 40 μF and 20 μF in series = 40+20 = 60 μF 120 μF and 60 μF in series = 120(60) 120+60 = 40 μF , C eq = 40 μF 2. Express the following voltages in polar form. (a) v 1 ( t ) = - 10 cos( ωt + 50 o ) v 1 ( t ) = 10 cos( ωt +50 o - 180 o ) = 10 cos( ωt - 130 o ) , V 1 = 10 n - 130 o or V 1 = 10 n 230 o V (b) v 2 ( t ) = 12 sin( ωt - 10 o ) v 2 ( t ) = 12 cos( ωt - 10 o - 90 o ) = 12 cos( ωt - 100 o ) , V 2 = 12 n - 100 o or V 2 = 10 n 260 o V 3. The voltage across a 100 μF capacitor is as shown below. Express the current through the capacitor in each case in standard form. (a) v c ( t ) = 40 cos(20 t - 90 o ) V i ( t ) = C dv dt = 100 × 10 6 d dt [40 cos(20 t - 90 o )] = - 100 × 10 6 × 40 × 20 sin(20 t - 90 o ) = - 0 . 08 cos(20 t - 90 o - 90 o ) = 0 . 08 cos(20 t - 90 o - 90 o + 180 o ) = 0
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Unformatted text preview: . 08 cos 20 tA (b) v c ( t ) =-60 sin(80 t + 30 o ) V i ( t ) = C dv dt = 100 10 6 d dt [-60 sin(80 t + 30 o )] =-100 10 6 60 80 cos(80 t + 30 o ) =-. 48 cos(80 t + 30 o ) = 0 . 48 cos(80 t + 30 o-180 o ) = 0 . 48 cos(80 t-150 o ) A 4. Solve for the current i(t) using phasor technique if v s ( t ) = 12 cos 3 t V.-+ v s ( t ) 3 3 H 1 3 F i ( t ) V s = 12 n o , Z L = jL = j (3)3 = j 9 , Z C = 1 jC =-j ( 1 1 3 3 ) =-j 1 Z eq = 3+ j 9-j 1 = 3+ j 8 = 8 . 544 n 69 . 44 o , I = V Z eq = 12 8 . 544 n 69 . 44 o = 1 . 404 n-69 . 44 o A i ( t ) = 1 . 404 cos(3 t-69 . 44 o ) A...
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