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quiz2soln

quiz2soln - E245B Exam II November 6 2001 Name SSN Pledge...

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Unformatted text preview: E245B Exam II November 6, 2001 Name: SSN: Pledge: Questions are worth 20 points each. !"#\$%&'((1 Find Io in the circuit shown using linearity and the assumption that Io = 1mA. I2 12K 4K I3 4K 4mA 4K 4K I1 2K + - V1 - )*++&,-&.()#%*-/#0( + Io = 1 A k I2 12K 4K V2 I3 IT 4K 4K 4K I1 2K If Io = 1 k A 1 Then V 2 = k (4 K + 2 K ) = 6V , I 1 = Then I 2 = I 1 + Io = 5 mA and V 2 = V 1 + 4 KI 2 = 16V 2 Then I 3 = So 7/2mA 1mA V2 4K + 12K 4mA x = 1mA IT = I 2 + I 3 = 7 mA 2 + - V1 - + 4/K A Io = 1 A k 6 4k = 3 mA 2 = 8 x = Io = 7 mA !"#\$%&'(2( Find RL for maximum power transfer and the maximum power that can be transferred in the network shown. )*++&,-&.()#%*-/#0( A Find RAB A 3K 2mA 2K RL 3K RAB = 2K + 3K + 5K = 10K 6K 5K 5K B B Find maximum load tranfer 3K I1 2mA I3 RL I3 - I1 = 2mA I2 = 1mA (2K) I1 + (10K) I3 + 5K (I3 - I2) + 3K (I1 - I2) = 0 yields: I3 = 0.9mA PL = I23 RAB 6K I2 5K PL = 8.1 mW !"#\$%&'(3( The voltage across a 50-F capacitor is shown in the figure. Determine the current waveform. v(t)(V) 10 0 2 4 6 8 10 12 t(ms) Figure P5.12 )*++&,-&.()#%*-/#0( C = 50 F Time(ms) i (t ) = C dv dt dv(V/ms) dt i(t)(mA) 0t 2 2t 4 4t 8 8 t 10 10 t 12 t > 12 v(t)(V) 10 5 0 -5 0 5 0 250 0 - 250 0 250 0 0 2 4 6 8 10 12 t(ms) !"#\$%&'(4 For the underdamped circuit shown in the figure determine the voltage v(t) if the initial conditions on the storage elements are iL(0)=1 A and vC (0) = 10 V. )*++&,-&.()#%*-/#0( iL (0) + - - Find V(t) if i L (0)=1A and vc (0)=10v d 2 v(t ) 1 dv(t ) 1 + + v(t ) = 0 dt 2 RC dt LC The characteristic equation is s 2 + 8s + 20 = 0 s = -4 2 j so v(t ) = k1e -4 t cos 2t + k2 e -4t sin 2t at t = 0, vc = 10 v(t ) = k1 = 10 then dv(t ) = -2k1e-4t sin 2t - 4k1e-4t cos 2t + 2k2 e -4t cos 2t - 4k2 e -4t sin 2t dt at t=0 dv(t ) = -4k1 + 2k2 = -40 + 2k2 dt also cdv(t ) v(t ) + + iL (t ) = 0 dt R at t=0 dv(t ) 1 -v(t ) 1 -10 = ( - iL (t )) = ( - 1) dt c R c R if -40 + 2k2 = -120 finally v(t ) = 10e -4t cos 2t - 40e -4t sin 2t !"#\$%&'(5( Find Z(j) at a frequency of 60 Hz for the network shown. 2 4 z 10mH 500F )*++&,-&.()#%*-/#0( Z L = j L = j (377)(10 10-3 ) = j 3.77 1 106 = = - j 5.3 j C j 377(5) j 3.77(4 - j 5.3) j15.08 + 20 Z = 2+ = 2+ = 5.1 + j 4.96 j 3.77 + 4 - j 5.3 4 - j1.535 Zc = ...
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