EECE 210 Final, Fall 2018

EECE 210 Electric Circuits Final Exam – Dec 19, 2018 20% 1. Derive NEC looking into terminals ‘ab’, where the short-circuit current is directed from ‘a’ to ‘b’, assuming VSRC = 1 V. Solution: The dependent voltage source partitions the circuit into independent subcircuits. In the subcircuit on the RHS, no current passes though the upper 1  resistor, so that the voltage across this resistor is zero. The circuit reduces to that shown. On open circuit, KVL gives: -IX + VSRC – IX = 0, so that IX = VSRC/2 A, and Vab = VSRC/2 V. On short circuit, IX = 0 and the dependent sources are set to zero. It follows that ISC = VSRC/1 A. Hence, RN = 0.5  and IN = VSRC A. Version 1:VSRC = 1 V, IN = 1 A, RN = 0.5  Version 2:VSRC = 2 V, IN = 2 A, RN = 0.5  Version 3:VSRC = 3 V, IN = 3 A, RN = 0.5  Version 4:VSRC = 4 V, IN = 4 A, RN = 0.5  Version 5:VSRC = 5 V, IN = 5 A, RN = 0.5 . 20% 2. Determine vO(t), assuming VSRC = 1 V dc and iSRC(t) = 1.5sin1000t A. Solution:  = 1000 rad/s, jL = j . Under dc conditions, the inductor acts as a short circuit, the capacitor and independent current source as open circuits. VO = VSRC, irrespective of IL. Under ac conditions, the independent voltage 1/5 + –  + – vO 2iL 100  1 mH 0.5 mF iSRC VSRC iL – + 10 A 2  6  IX + – 1  VSRC 1  IX a b IX – + IX + – 1  VSRC 1  IX a b IX source acts as a short circuit. The downward current in the 100  resistor is (ISRC + 2IL) A in the frequency domain. This current divides between the inductor and capacitor. IL = , or, , or, . VO = jLIL = . Substituting numerical values, VO = V. The ac component of vO(t) = = . It follows that: vO(t) = V. Version 1:VSRC = 1 V, Isrc = 1.5 A; vO(t) = V Version 2: VSRC = 2 V, Isrc = 3 A; vO(t) = V Version 3:VSRC = 3 V, Isrc = 4.5 A; vO(t) = V Version 4:VSRC = 4 V, Isrc = 6 A; vO(t) = V Version 5:VSRC = 5 V, Isrc = 7.5 A; vO(t) = V. 20% 3. Determine VO, assuming VSRC = V. 2/5 – + 2:1 VO _ + j2.5  10  -j25  40  j5  j5  k = 0.5 VSRC Solution: The impedance seen at the primary of the ideal transformer will be derived. The impedance on the secondary side of the ideal transformer is reflected to the primary side as a reactance of j10  in parallel with a resistance of 40 , with -2VO appearing across the parallel combination, as shown. The two series-coupled coils in series with the capacitor will have a reactance of: j10 – j5 – j25 = -j20 . The two reactances in parallel are . TEC seen by this reactance is in series with 20 , the circuit becoming as shown. It follows that: = and . Version 1:VSRC = = (-1 + j) V; = 0.25 V Version 2: VSRC = 2 = 2(-1 + j) V; = 0.5 V Version 3:VSRC = 3 = 3(-1 + j) V; = 0.75 V Version 4:VSRC = 4 = 4(-1 + j) V; = 1 V Version 5:VSRC = 5 = 5(-1 + j) V; = 1.25 V 20% 4. The switch is opened at t = 0 after being closed for a long time. Determine RL and C so that the voltage vS across the open switch is a constant for t  0+, assuming RC = 1  and L = 1 H. Solution: In the steady state, with the switch closed, the inductor acts as short circuit and the capacitor as an open circuit. A current 5/RC flows, establishing a voltage of 5 V across C. At t = 0+, the initial current 3/5 – + 40  VSRC 40  -j20  j10  -2VO _ + + –5 V + – vS C L RC RL t = 0 + –5 V + – vS C L RC RL + – + – vC vL iL through L is 5/RC and the initial voltage across C is 5 V. For t > 0+, L and C will each discharge independently through the parallel resistance, as first-order circuits, so that and . vS = 5 – vC – vL = 5 . For vS to be a constant, the two exponential terms must cancel out. It follows that RL = RC and , or . Version 1: RC = 1 , L = 1 H; RL = 1 , and C = 1 F Version 2: RC = 2 , L = 2 H; RL = 2 , and C = 2/4 = 0.5 F Version 3: RC = 3 , L = 3 H; RL = 3 , and C = 3/9 = 0.33 F Version 4: RC = 4 , L = 4 H; RL = 4 , and C = 3/16 = 0.25 F Version 5: RC = 5 , L = 5 H; RL = 5 , and C = 5/25 = 0.2 F. 20% 5. Switch S1 is opened at t = 0 with no initial energy stored in the circuit. Switch S2 is moved from position ‘a’ to position ‘b’ at t = t0, at the first zero of iL for t > 0+. Determine (a) t0, (b) iC(t) and vC(t) for t ≥ 0+, where t is in ms. Assume that ISRC = 1 mA and VSRC = 5 V. Solution: (a) 0  t  t 0: vC(t) = Acos0t + Bsin0t, since there can be no steady value of vC. At t = 0+, vC(0+) = 0, so that A = 0, and vC(t) = Bsin0t V. iC(t) = 0CBcos0t. At t = 0+, iC(0+) = ISRC, so that B = ISRC/0C, and iC(t) = ISRCcos0t. iL(t) = ISRC – iC(t) = ISRC(1 – cos0t) A. The first zero of iL(t) for t > 0+ occurs at t0 = 2/0, where rad/s, so that ms. (b) 0  t  t 0: vC(t) = (ISRC/0C)sin0t = (ISRC×10-3/(104×10-7)sin10t = ISRCsin10t V, where ISRC is in mA and t is in ms. iC(t) = ISRCcos10t mA. t  t 0: At t = t0, vC(t0) = 0 and iC(t) =ISRC mA. At this instant, switch S2 is moved to position ‘b’, which makes iC(t0) =ISRC + = (ISRC + 5) mA, and the final value is zero, when the capacitor is fully charged.  = 103×10-7  0.1 ms. Hence, mA, where t is in ms. The final value of vC(t) 4/5 t = 0 0.1 H ISRC + – VSRC 1 k iC + – vC t = t0 S1 S2 a b iL 0.1 F is: ISRC×1 + 5 = (ISRC + 5) V. Hence, V. Version 1:ISRC = 1 mA: iC(t) = cos10t mA, vC(t) = sin10t V, 0  t  t0 mA, V, t  t0. Version 2:ISRC = 2 mA: iC(t) = 2cos10t mA, vC(t) = 2sin10t V, 0  t  t0 mA, V, t  t0. Version 3:ISRC = 3 mA: iC(t) = 3cos10t mA, vC(t) = 3sin10t V, 0  t  t0 mA, V, t  t0. Version 4:ISRC = 4 mA: iC(t) = 4cos10t mA, vC(t) = 4sin10t V, 0  t  t0 mA, V, t  t0. Version 5:ISRC = 5 mA: iC(t) = 5cos10t mA, vC(t) = 5sin10t V, 0  t  t0 mA, V, t  t0. 5/5