EECE 210 Problems, Fall 2010

54.81 kW 120 V, 20 A t2/3200 W, t is in s approx Hint: redraw as a bridge circuit Hint: transform current sources to voltage sources VA = 12 V, VB = 10 V, VC = 22 V Refer to figures below A 50 V 8% 2. Determine Ix in the circuit shown. + 2 Ω – + – Is = 5Ix A 2Is V Ix 4 V 2 Ω A. 2 A B. 4 A C. -2 A D. -4 A E. None of the above Solution: KCL at the upper node gives a current of 4Ix in the 2 Ω resistor; 2Is = 10Ix; from KVL around the right mesh: 10Ix = 8Ix + 4, so that Ix = 2 A. + 2 Ω – + – Is = 5Ix A 2Is V Ix 4 V 2 Ω 4 Ix Problem 1 Find Rab. A) 12 ohms B) 7.5 ohms C) 10 ohms D) 15 ohms E) None of the above Problem 2 Find the power consumed by the 50 ohm resistor in the circuit shown below A) P= 0.246 W B) P= 0.692 W C) P= 2.358 W D) P= 5.100 W E) None of the above Problem 3 Find V0 in the 30 Ohm resistor in the circuit shown below A) V0 = 6 V B) V0 = 66 V C) V0 = 72 V D) V0 = 78 V E) None of the above 1 Problem 4 In the circuit shown, find the voltage denoted by Vs1 A) 300 V B) 150 V C) -150 V D) 75 V E) None of the above Problem 5 In the circuit shown above, find the value of the load resistance RL in terms of R such that Vo is 50 V. A) R/3 B) 3R  C) R D) 2R E) None of the above 2 1. The current in a 1 μF capacitor is shown in the figure as a function of time. The total energy stored in μJ is: i 10 mA t 1 ms 4 ms A. 40 B. 100 C. 200 D. 50 E. 25 Solution: q at 4 ms is 20 2 4 10 = × μC. The energy in μJ is W = C C 200 2 (20) 2 = , where C is in μF. 2. If VSRC = 10 V, determine Rx so that Ix = 0. – + VSRC 5 Ω – + 5 V Rx Ix A. 5 Ω B. 1.25 Ω C. 2.5 Ω D. 1 Ω E. 1.67 Ω Solution: When Ix = 0, 5 5 = + SRC x x V R R , or SRC − 5 x V 25 = R Ω. 3. If R etermine the ratio ρ /α so that I1 = I2. Ω E. 8 Ω Solution: = 10 Ω, d A. 4 Ω B. 10 Ω C. 6 Ω D. 5 R I I 2 1 6 − ρ = , R RI I 1 2 6 −α = , or R RI R I 1 1 6 6 α ρ − = − , which gives ρ /α = R. – + – + ρ α – + V1 + 6 V R R I2 V1 – I2 I1 1/5 4. In the figure shown, the 24 V source having a source resistance of 1 Ω is replaced by the equivalent current source, the load resistance RL being the same. If RL = 5 Ω, the ratio of the power delivered by the ideal current source to the power delivered by the ideal 24 V source is: 24 V – + 1 Ω RL A. 5 B. 11 C. 7 D. 14 E. 9 Solution: The power delivered by the ideal voltage source is 1 24 24 + × RL . The equivalent current source is an ideal current source of 24 A in parallel with 1 Ω. The power delivered by the current source is 1 1 24 24 + × × L L R R . The ratio of the powers is numerically equal to RL. – + 3VO V _ + VO 2 A 4 A 2 Ω 2 Ω 4 Ω 4 Ω R 5. Determine VO in the circuit shown if R = 1 Ω A. 18 V B. 12 V C. 30 V D. 6 V E. 24 V Solution: The current through R is 6 A, so that VO = 6R. – + – + – + – + – + a b IX 0.5IX V 0.8Vab A 20 V 10 A 6. Given the source connections shown. Determine the actual power delivered or absorbed by each source. Solution: IX = 0.8×20 = 16 A. Current in 20 V source is 6 A in the direction of a voltage rise. Voltage across dependent voltage source is 0.5×16 = 8 V. Voltage across 2/5 dependent current source is 20 – 8 = 12 V. It follows that: Power delivered by 20 V source is 20×6 = 120 W Power delivered by 10 A source is 20×10 = 200 W Power absorbed by dependent current source is 12×16 = 192 W Power absorbed by dependent voltage source is 8×16 = 128 W VSRC – + ISRC + _ 4 V 1.5 A 12 Ω 6 Ω 22 Ω 8 Ω 7. Determine VSRC, ISRC, and the voltages across the four resistors in the circuit shown. (Four grade points for each answer plus 1 bonus grade). Solution: Going CCW around the upper mesh: VSRC – + ISRC + _ 4 V 1.5 A 12 Ω 8 Ω 6 Ω 22 Ω I1 I2 I1+1.5 I2 -1.5 x 4 +12I1 – 8I2 = 0 Going CW around the lower mesh: 4 – 6(I1 + 1.5) + 22(I2 – 1.5) = 0 The two equations reduce to: 3I1 – 2I2 = -1 -3I1 + 11I2 = 19 This gives: I1 = 1 A; I2 = 2 A. Hence, Voltage across 12 Ω resistor: 12 V Voltage across 6 Ω resistor: 15 V Voltage across 8 Ω resistor: 16 V Voltage across 22 Ω resistor: 11 V VSRC = 27 V ISRC = 3 A. 3/5 1. Determine the power dissipated in the circuit, assuming I = 1 A. I 1 Ω 1 Ω 1 Ω 1 Ω 1 Ω 3 Ω 3 Ω 2 Ω 2 Ω 0.5 Ω Solution: The 1 Ω Y is paralleled with a 3 Ω Δ, so that it effectively becomes a 0.5 Ω Y, and the circuit reduces to that shown. The resistance seen by the current source is 1||1 + 2.5 = 3 Ω, so that the power dissipated in the circuit is P = 3I2 W. I 1 Ω 0.5 Ω 0.5 Ω 0.5 Ω 2 Ω 0.5 Ω 2. Determine the power delivered by the 3 V source, assuming ρ = 2 V/A. Solution: The upper node is at 5 V with respect to the lowest node, the middle node is at 3 V. hence, Ix = 0.5 A and the current in the 6 Ω resistor is also 0.5 A. The current supplied by the 3 V source is (3 – 0.5ρ)/2 and the power delivered by the source is P = 1.5(3 – 0.5ρ) = 4.5 – 0.75ρ W. – + – + – + 10 Ω 4 Ω 6 Ω 2 Ω 5 V 3 V 10 A Ix ρIx – + 10 V R 10 Ω 2 Ω – + 2 V – + Ix 5 V Ix 6. Determine the power absorbed or delivered by the dependent source assuming R = 1 Ω. Solution: The current in the 2 Ω resistor is 2Ix flowing downwards. From KVL in the mesh on the left, 10 = 4Ix + 2, or Ix = 2 A. The voltage rise Vx across the dependent source is given by: Vx – RIx = 5, or Vx = 2R 4/5 + 5; The power P delivered by the source is P = 2(2×R + 5). 1 5/5 6. Determine VO. ce is replaced by a 10 x Solution: The 2Vx sour A source. The current in the 2 Ω resistor is Ix. The current in the dependent source is 5 – 2Ix, so that the current in the 1 Ω resistor is 15 – Ix. From KVL around the mesh abcd, 2I + 15 – Ix = 4Ix, which gives Ix = 5 A. It follows that VO = 15 – Ix = 10 V. – + 5 A 10 A 2 Ω 4 Ω 1 Ω – + VO Ix 5 - 2Ix 1 Ω a Ix 2Ix 2Ix 15 - Ix b d c – + 5 A 2Ix 2Vx 2 Ω 4 Ω 1 Ω Ix – + VO 1 Ω – + Vx Problem 1 Find the equivalent resistance between B and E. 4Ω 5Ω 5Ω 4Ω C B A 10Ω D 12Ω E A) 15.11Ω B) 16Ω C) 8.33Ω D) 13.61Ω E) None of the above Problem 2 Find V0. 2KΩ 1KΩ 2KΩ − + 12V 6mA 2KΩ + V0 − A) 12V B) 7.5V C) -12V D) -7.5V E) None of the above