EECE 210 Exam 2 With Solution, Fall 2016

EECE 210 Electric Circuits Quiz 2 – Oct 22, 2016 20% 1. Determine: (a) VO (6 grades), (b) the short circuit current Iab (6 grades), (c) Thevenin’s resistance seen by a test current source applied between terminals ‘ab’ (8 grades). Assume R = 10 . Solution: (a) The current through R is 0.5 – 0.1VO. From KVL, (0.5 – 0.1VO)R + 20 = VO. It follows that . (b) When terminals ‘ab’ are short circuited, VO = 0, and the dependent source is zero, that is replaced by an open circuit. It follows that ISC = . (c) When a test current source is applied, with the independent sources set to zero, it follows that , so that . Alternatively, the dependent source is equivalent to a resistance of VT/(0.1VT) = 10 , so that RTh = 10||R = . Version 1: R = 10 , (a) V, (b) ISC = 2.5 A, (c) . Version 2: R = 15 , (a) V, (b) ISC = A, (c) . Version 3: R = 20 , (a) V, (b) ISC = A, (c) . Version 4: R = 30 , (a) V, (b) ISC = A, (c) . Version 5: R = 40 , (a) V, (b) ISC = A, (c) . 20% 2. Determine IO using NEC, assuming R = 1  (15 grades for NEC, 5 grades for IO). Solution: On open circuit, it follows from KVL around the 3  resistor and 3 V source that VTh – 3 – 3 = 0, so that VTh = 6 V. On short circuit, IN = 2 A, so that RTh = 3 . Alternatively, if the independent sources are set to zero, the 6  resistor is short circuited, so that the resistance seen by R is 3 . Version 1 - p1/4 + – a b 20 V + – VO R 0.1VO 0.5 A a b + – VT R 0.1VT IT R + – 20 V + – 3 V 3  6  1 A IO + – 20 V + – 3 V 3  6  1 A IN 1 A + – 20 V + – 3 V 3  6  1 A + – VTh + – 3 V NEC seen by R is as shown. It follows from current division that: . Version 1: R = 1 , A Version 2: R = 2 , A Version 3: R = 3 , A Version 4: R = 5 , A Version 5: R = 6 , A. 20% 3. Determine IS using superposition, assuming R = 5 . Solution: The dependent source is replaced by an independent source VY = 5VX. If the 1.5 A source is applied alone, with the voltage sources set to zero, VX1 = (80||20)×1.5 = 16×1.5 = 24 V. If the 10 V source is applied alone, with the other sources set to zero, no current flows in the 20  and 80  resistors, so that VX2 = 10 V. If the source VY is applied alone, it follows from voltage division that VX3 = . Substituting VY = 5VX, the superposition equation is: VX = 24 + 10 – VX, or VX = 17 V. From KVL in lower mesh on the LHS, , and . Version 1: R = 5 , = 16.5 A. Version 2: R = 10 , = 9 A. Version 3: R = 15 , = 6.5 A. Version 4: R = 20 , 5.25 A. Version 5: R = 25 , = 4.5 A. 2 A IO 3  R + – 10 V 80  20  1.5 A R – + IS 5VX + – VX + – 10 V 80  20  1.5 A R IS VY + – VX I1 + – 20% 4. Determine VO using the node-voltage method, assuming ISRC = 1 A. Solution: Node-voltage equation for node ‘a’, bearing in mind that the 5  resistor is redundant: (0.5 + 0.5)Va – 0.5Vb = Ib = 0.5Va. This gives Va = Vb. The node-voltage equation for node ‘b’ is: -0.5Va + (0.5 + 0.5)Vb – 0.5Vc = -ISRC. Substituting Vc = 18 V and Va = Vb, 0.5Va = 9 – ISRC, or, Va = VO = 18 – 2ISRC. Version 1: ISRC = 1 A, VO = 18 – 2ISRC = 16 V Version 2: ISRC = 2 A, VO = 18 – 2ISRC = 14 V Version 3: ISRC = 3 A, VO = 18 – 2ISRC = 12 V Version 4: ISRC = 4 A, VO = 18 – 2ISRC = 10 V Version 5: ISRC = 5 A, VO = 18 – 2ISRC = 8 V. 20% 5. Assuming a dc state and ISRC = 5 A, determine: (a) The charge on the capacitor; (b) the energy stored in the capacitor; (c) the flux linkage of the inductor; (d) the energy stored in the inductor; (e) if after reaching the dc state, the current source is replaced by a short circuit, determine the total power dissipated in the circuit as t   (4 grades for each part). Solution: (a) In the dc state, the capacitor acts as an open circuit and the inductor as a short circuit. ISRC flows through the 1  resistor and the inductor. The voltage across the capacitor is therefore V = 1×ISRC V. The charge on the capacitor is CV = 10-3×ISRC C  ISRC mC. (b) The energy stored in the capacitor is (1/2)qV = (1/2)(ISRC)2 mJ. (c) The flux linkage  = LISRC = 10-3×ISRC Wb-T  ISRC mWb-T. (d) The energy stored in the inductor is (1/2)ISRC = (1/2)(ISRC)2 mJ. (e) As t  , the energy dissipated in the circuit is, from conservation of energy, the energy stored in the capacitor and inductor, which totals (ISRC)2 mJ. The energy divided by infinite time is zero power. Version 1: ISRC = 1 A, (a) ISRC mC = 1 mC; (b) (1/2)(ISRC)2 mJ = 0.5 mJ; (c) ISRC mWb-T = 1 mWb-T; (d) (1/2)(ISRC)2 mJ = 0.5 mJ; (e) (ISRC)2 mJ = 1 mJ. + – Ib Ib + – VO 18 V ISRC 2  2  2  5  + – Ib Ib + – VO 18 V ISRC 2  2  2  5  Va Vb Vc ISRC 1 mF 1 mH 1  10  10  Version 2: ISRC = 2 A, (a) ISRC mC = 2 mC; (b) (1/2)(ISRC)2 mJ = 2 mJ; (c) ISRC mWb-T = 2 mWb-T; (d) (1/2)(ISRC)2 mJ = 2 mJ; (e) (ISRC)2 mJ = 4 mJ. Version 3: ISRC = 3 A, (a) ISRC mC = 3 mC; (b) (1/2)(ISRC)2 mJ = 4.5 mJ; (c) ISRC mWb-T = 3 mWb-T; (d) (1/2)(ISRC)2 mJ = 4.5 mJ; (e) (ISRC)2 mJ = 9 mJ. Version 4: ISRC = 4 A, (a) ISRC mC = 4 mC; (b) (1/2)(ISRC)2 mJ = 8 mJ; (c) ISRC mWb-T = 4 mWb-T; (d) (1/2)(ISRC)2 mJ = 8 mJ; (e) (ISRC)2 mJ = 16 mJ. Version 5: ISRC = 5 A, (a) ISRC mC = 5 mC; (b) (1/2)(ISRC)2 mJ = 12.5 mJ; (c) ISRC mWb-T = 5 mWb-T; (d) (1/2)(ISRC)2 mJ = 12.5 mJ; (e) (ISRC)2 mJ = 25 mJ.