EECE 210 Problems Exam-Solution, Spring 2013

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 1. An electric field ξ is applied in a region containing both positive and negative charges. A current IN flows due to the negative charges, and a current IP flows due to the positive charges. Which of the following statements is, or are, true? (Note: if a false statement is marked true, the answer to this Question 1 is considered incorrect). A. IN and IP are in the same direction as ξ. B. IN and IP are in the opposite direction to ξ. C. IP is in the direction of ξ and IN is in the opposite direction. D. IN is in the direction of ξ and IP is in the opposite direction. E. The total current is zero. Solution: The positive charges will flow in the direction of ξ and the negative charges will flow in the opposite direction. The currents of both types of charge will be in the direction of ξ. 2. Which of the following statements is, or are, true? (Note: if a false statement is marked true, the answer to this Question 2 is considered incorrect). A. An ideal capacitor always absorbs power. B. An ideal inductor always delivers power. C. An ideal passive resistor always absorbs power. D. An ideal, dependent voltage source always delivers power. E. An ideal, dependent current source always absorbs power. Solution: Ideal capacitors, inductors and sources can absorb or deliver power. An ideal passive resistor always absorbs and dissipates power. 3. Which of the following statements is, or are, true? (Note: if a false statement is marked true, the answer to this Question 3 is considered incorrect). A. The connection of current sources is valid, but the connection of voltage sources is invalid. B. The connection of current sources is invalid, but the connection of voltage sources is valid. C. The connections of current sources and of voltage sources are valid. D. The connections of current sources and of voltage sources are invalid. E. the circuit cannot be made valid by changing the values of the sources. Solution: The connection of current sources is invalid because it violates conservation of charge at the node in the middle. The connection of voltage of sources is invalid because it violates conservation of energy around the outer loop. + – 2 A 1 A 5 A 2 V 5 V 10 V 4. In the circuit shown, I0 = 1 A and the 6 V source does not absorb or deliver any power. Determine the power absorbed or delivered by the source I0. Note that the value of R need not be known. A. 9 W absorbed B. 9 W delivered C. 24 W absorbed D. 24 W delivered E. No power absorbed or delivered. Solution: Since the 6 V source does not absorb or deliver any power, it follows that the current through the 6 V source and the 10 Ω resistor is zero, which means that the voltage across R is 6 V, and the current through R and the 3 Ω resistor is I0. It follows that the power dissipated in the resistors is 6I0 + 3 2 0I = 9 W, which is also the power delivered by the source. 5. For the assigned positive directions shown, the voltage-current relationship for an ideal capacitor is: A. i = Cdv/dt B. i = -Cdv/dt C. v = Cdi/dt D. v = -Cdi/dt E. v = -(1/C)di/dt. Solution: The assigned positive direction of current is that of a voltage rise across the capacitor. According to the passive sign convention, the i-v relation is written with a negative sign. 6. An ideal, parallel-plate capacitor of 10 μF is charged to VC = 1 V. if the distance between the parallel plates is doubled, what will be the voltage across the capacitor? A. 3 V B. 6 V C. 2 V D. 4 V E. 5 V. Solution: The charge on the capacitor is 10×VC = 10VC C. If the distance between the plates is doubled, the charge remains the same, but the capacitance is halved to 5 μF. The new voltage is 10VC/5 = 2VC V. v C – + i – + I0 R 6 V 3 Ω 10 Ω 7. If all voltages and currents are dc, determine VX assuming VSRC = 2 V. A. 1.6 V B. 2.4 V C. 3.2 V D. -1.6 V E. -2.4 V. Solution: Under dc conditions, the inductor behaves as a short circuit and the capacitor as an open circuit. The circuit reduces to that shown. From voltage division, VX = 40 10 40 + VSRC = 0.8VSRC = 1.6 V. 8. The current through a 2 μF capacitor is shown as a function of time. Determine the charge on the capacitor at t = 3.4 s. A. 5.6 C B. 5.2 C C. 4.8 C D. 4.4 C E. 4 C. Solution: The charge at t > 3 s is the net area under the curve, which gives q = 2×3 – 1×(t – 3) = (9 – t) = 5.6 C. 9. If the current in Problem 8 is applied to a resistor R = 5 Ω, determine the average power dissipated in the resistor over the interval from t = 0 to t = 5 s. A. 50 W B. 14 W C. 70 W D. 24 W E. 12.5 W. Solution: The power dissipated from t = 0 to t = 3 s is (2)2×R = 4R; the power dissipated from t = 3 s to t = 5 s is (-1)2×R = R; the average power dissipated is P = (4R×3 + R×2)/5 = 2.8R = 14 W 10 Ω 40 Ω VSRC + – + – VX i, A 1 t, s 3 5 2 -1 0 1 H 10 Ω 40 Ω VSRC + – 1 μF + – VX 10. Determine the resistance between nodes a and b if each of the resistances connected in Y is R and each of the resistances connected in the outer Δ is 3R, with R = 1 Ω. A. 15 Ω B. 20 Ω C. 15 Ω D. 1 Ω E. 10 Ω. Solution: If the Δ is transformed to Y, each of the resistances in Y is R Ω, and the two Y connections are in parallel. For each Y, the resistance between nodes a and b is 2R Ω, and these in parallel give R = 1 Ω. Alternatively, if the Y is transformed to Δ, the resistances connected in Δ are 3R Ω. The two Δ connections in parallel will give a Δ connection of 1.5R Ω resistors. The resistance between nodes a and b will be 1.5R Ω in parallel with 3R Ω, which again gives R = 1 Ω. 11.  Determine ISRC if VSRC = 10 V. Solution: If the nodes are labeled, it is seen that the three resistances are effectively connected between two nodes, which means that they are in parallel; 30 Ω in parallel with 60 Ω is (30)(60)/90 = 20 Ω, and 20 Ω in parallel with 20 Ω is 10 Ω. Hence, ISRC = VSRC/10 = 0.1VSRC = 1 V. 12.  Determine IX using source transformation, and assuming ISRC = 1 A. Solution: The 20 V source in series with 20 Ω is transformed to a current source of (20 V)/(20 Ω) = 1A in parallel with 20 Ω. Applying KCL: ISRC + 1 = 12 30 20 X X X V V V = + . It follows that VX = 12(ISRC + 1), and )1 0.4( 30 + = = SRC X X I V I = 0.8 A. 30 Ω 60 Ω 20 Ω VSRC + – ISRC 20 Ω 30 Ω 20 V + – ISRC IX + – VX 30 Ω ISRC IX 20 Ω 1 A a b 30 Ω 60 Ω 20 Ω b a VSRC + – ISRC b a Alternatively, the two current sources can be combined into a single current source (ISRC + 1) A. From current division, )1 0.4( )1 30 ( 20 20 + = + + = SRC SRC X I I I = 0.8 A. 13. In the circuit shown, the current in the top connection is zero. Determine: 10% (a) R 10% (b) Vx 10% (c) Vy. Solution: (a) From KCL at node b, Ibe = 5 A. From KCL at node e, Ide = 5 A. From KCL at node c, Ica = 15 A. From KCL at node a, Iad = 15 A. As a check, KCL at node d is satisfied. From ohm’s law, Vbe = 30 V, Vad = 15 V, and the voltage across R is 5R. Applying KVL to the outer loop starting from node e and going CW: 5R + 15 – 30 = 0, which gives 5R = 15, and R = 15/5 = 3 Ω. (b) From KVL around the mesh abc, starting from node a and going CW: -V1 + 20 = 0, or V1 = 20 V. From KVL around mesh cbe, starting from node c and going CW: V1 – 30 + Vx = 0, or Vx = 10 V. (c) From KVL around the mesh acd, starting from node a and going CW: -20 + Vy + 15 = 0, or Vy = 5 V. As a check, KVL around the mesh ced, starting from node c and doing CW: -Vx + 15 – Vy = 0, or -10 + 15 -5 = 0. 5 A 10 A 10 A 20 V 6 Ω 1 Ω R I = 0 a b c d e 5 A 15 A 5 A 5 A 15 A 5 A 10 A 10 A 20 V 6 Ω 1 Ω R I = 0 a b c d e V1 + – 30 V + – 15 V Vx Vy + – 15 V 5 A 10 A 10 A 20 V 6 Ω 1 Ω R I = 0 Vx Vy 1. Determine IX, assuming ISRC = 1 A (Hint: write one KVL equation and one KCL equation). A. 4 A B. 2 A C. 3 A D. 1 A E. 5 A Solution: Let the current through the 10 Ω resistor be IY. From KVL around the mesh abcda: 5IX + 5IX – 10 – 10IY = 0, which gives, IY = IX – 1. From KCL at node a, ISRC = IY + IX, or ISRC = 2IX – 1. It follows that IX = (ISRC + 1)/2 = 1 A. 4. Determine the resistance between nodes a and b assuming that all resistances are 1 Ω. A. 4 Ω B. 2.4 Ω C. 3.2 Ω D. 1.6 Ω E. 0.8 Ω Solution: The resistances can be split into two halves in parallel, where the resistances in the middle become 2R. Applying ∆-Y transformation, the resistances become as shown. It follows that 2Rab = 0.5R + R||1.5R + 0.5R = R + 0.6R = 1.6R, so that Rab = 0.8R = 0.8 Ω. 3. In the circuit shown, the values of two resistors are expressed in terms of the resistance R, and the values of two resistors are expressed in terms of the conductance G. Determine the resistance between terminals a and b, assuming 1 = 1 R = G Ω. A. 4.5 Ω B. 3 Ω C. 3/2 Ω D. 7.5 Ω E. 6 Ω Solution: The resistance of a resistor whose conductance is 2G is 2 2 1 G = R , and the resistance of a resistor whose conductance is G is G = R 1 . It follows that the resistance of the two paralleled resistors is R/2, and the total resistance between terminals ab is Reqs = 3R/2. a b R 0.5R 0.5R 0.5R 0.5R 0.25R 0.25R a b 2R 2R R R R R R a b _ + ISRC 5IX 15 Ω 12 Ω 5 Ω IX 10 Ω _ + 10 Ω 10 V 2G R/2 G a b R 1. The figure shows the assigned voltage drop across a given circuit element, the current being carried by electrons flowing through the circuit element in the direction shown. Which of the following statements is, or are, true? (Note: if a false statement is marked true, the answer to this Question 1 is considered incorrect). A. The circuit element delivers power. B. The electric potential energy of electrons is higher at terminal a than at terminal b. C. If the flow of electrons is 3.2×1019/s, and the charge per electron is -1.6×10-19 C, the magnitude of the current is 2×1038 A. D. The difference between the electric potential energies of electrons at terminals a and b is independent of which of these terminals is grounded. E. If the element is a resistor and power is absorbed by the element, then if the rate of flow of electrons in particles per second is doubled, the power absorbed by the element will be doubled. Answers: A. The direction of conventional current flow is opposite that of electron flow, because electrons are negatively charged. The current is in the direction of a voltage drop and power is absorbed by the element. B. Because of their negative charge, electrons will have a higher electric potential energy at the more negatively charged terminal, b. C. The magnitude of the current is 3.2×1019×1.6×10-19 = 5.12 A. D. The difference in electric potential energy is independent of where the arbitrary zero of this potential energy is taken. E. When the rate of flow is doubled, the current is doubled, the voltage drop is doubled, and the power absorbed is quadrupled. Only statement D is true. 2. Which of the following statements is, or are, true of the circuit shown? (Note: if a false statement is marked true, the answer to this Question 2 is considered incorrect). A. It is impossible to tell whether or not the source connections are valid. B. The connections of voltage sources and current sources are valid. C. The connection of voltage sources is not valid. D. The connection of current sources is not valid. E. The source connections become valid if the 5 Ω resistor is removed. Solution: From KCL, the currents in the voltage sources are the same as those of the current sources, and the current at node b is 1 A in the direction shown. The voltage across the 5 Ω resistor is 5 V in the polarity shown. From KVL, the voltage across the 2 A source is zero, and the voltage across the 1 A source is 5 V in the polarity shown. KVL and KCL are satisfied throughout the circuit, so the circuit is valid, + – v Direction of flow of electrons a b – + – + 5 V 10 V 2 A 1 A 5 Ω – + – + 5 V 10 V 2 A 1 A 5 Ω a b 1 A – + 5 V – + 0 – + 5 V 2 A 1 A 4. When the switch is closed, a current i flows that begins charging the capacitor. After a sufficiently long time, the capacitor is fully charged. Determine the energy stored in the capacitor when it is fully charged, assuming VB = 2 V. A. 4 µJ B. 16 µJ C. 36 µJ D. 25 µJ E. 9 µJ Solution: When the capacitor is fully charged, i = 0, and the voltage across the capacitor is VB, and the energy stored in the capacitor is 2 2 1 CVB W = . 5. Determine the total energy delivered by the battery in the preceding problem when the capacitor is fully charged. A. 32 µJ B. 72 µJ C. 18 µJ D. 50 µJ E. 8 µJ Solution: The instantaneous power delivered by the battery is p = VBi. The total energy delivered by the battery is ∫ ∫ ∞ ∞ = = 0 0 idt V V idt W B B , assuming that charging begins at t = 0. But ∫ ∞ 0idt is the total charge delivered to the capacitor, which is CVB. It follows that W = CVB2 . 6. Determine I assuming ρ = 0.9 Ω. A. 30 A B. 12 A C. 60 A D. 15 A E. 20 A Solution: From KCL at node a: I = I1 + I2; from KVL in the outer loop, 6 = 1×I1; from KVL in the RHS mesh, ρI = 1×I2; hence, I = 6 + ρI, or = 1− ρ 6 I A. 9. Determine VX, assuming VSRC = 2.5 V. A. 1 V B. 5 V C. 3 V D. 2 V E. 4 V Solution: From KVL, VSRC = 3I + Vx = 5I; I = VSRC/5, and VX = 2VSRC/5 = 0.4VSRC. + – VB 2 Ω 2 µF i – + – + I 6 V 1 Ω 1 Ω ρI V – + – + I 6 V 1 Ω 1 Ω ρI V I1 I2 a – + VSRC I 3 Ω – + 2I – + VX 7. A nonideal voltage source has an open-circuit voltage VSRC. When connected to a load resistor that draws a current of IL = 1 A, the power dissipated in the load is four times the power dissipated in the source resistance. Determine the short-circuit current of the equivalent nonideal current source. A. 12.5 A B. 5 A C. 7.5 A D. 15 A E. 10 A Solution: Since the power dissipated in the load is four times the power dissipated in the source resistance, the load resistance is 4Rsrc. The short-circuit current of the nonideal current source is src SRC SRC R V I = . From KVL, VSRC = 5RsrcIL. It follows that L src SRC SRC I R V I = 5 = . 8. Determine the energy stored in the inductor if L = 1 H assuming dc conditions, after all voltages and currents have assumed constant values. A. 0.75 J B. 1 J C. 1.5 J D. 2 J E. 0.5 J Solution: Under dc conditions, the current through the capacitors is zero, and the voltage across the inductor is zero. From KVL, a current equal to 5 Ω = 5 V 1A passes through the upper 5 Ω resistor and through L. The energy stored in the inductor is 2 12 W = 1 L = L/2 J. 13. Determine the power dissipated in the 5 Ω resistor, assuming ISRC = 1 A. Solution: The two current sources in parallel with their associated resistors can be transformed to voltage sources as shown. From KVL around the circuit: 3ISRC = 2VX + 5I, where VX = 5I. Substituting for VX, 3ISRC = 15I, and I = 0.2ISRC. The power dissipated in the 5 Ω resistor is P = 2 2 0.2 0.04 5 SRC SRC I I = × W – + VSRC IL Rsrc 4Rsrc – + 5 V 5 Ω 1 µF 5 Ω L 3 Ω 2 Ω ISRC 5 Ω 0.5VX A – + VX – + – + – + VX 3 Ω 2 Ω 5 Ω VX 3ISRC I 11. Determine R, so that no power is delivered or absorbed by the 1 A current source, assuming VSRC = -6 V. Solution: The solution proceeds in the following steps: 1. Since 1 A passes in branch ad, the voltage drop across the 2 Ω resistor is 2 V in the polarity shown. 2. Since no power is delivered or absorbed by the 1 A source, the voltage across this source is zero. From KVL around mesh abd, the voltage across R is also 2 V. 3. From KVL around mesh bcd, the voltage across branch cd is (2 + VSRC) V, since by going clockwise around this mesh, (2 + VSRC) – 2 – VSRC = 0. 4. From Ohm’s law, the current in the branch bc is (1 + VSRC /2) A directed towards node b. 5. From KCL at node b, the current flowing away from node b is: (2 + VSRC /2) A. 6. From Ohm’s law applied to R, R(2 + VSRC /2) = -2. It follows that R = 1 / 4 1 2 / 2 2 − = − − − SRC SRC V V . 13. Determine the power dissipated in the 5 Ω resistor, assuming ISRC = 1 A. Solution: The two current sources in parallel with their associated resistors can be transformed to voltage sources as shown. From KVL around the circuit: 3ISRC = 2VX + 5I, where VX = 5I. Substituting for VX, 3ISRC = 15I, and I = 0.2ISRC. The power dissipated in the 5 Ω resistor is P = 2 2 0.2 0.04 5 SRC SRC I I = × W 1 A VSRC 2 Ω 2 Ω R 3 Ω 2 Ω ISRC 5 Ω 0.5VX A – + VX – + – + – + VX 3 Ω 2 Ω 5 Ω VX 3ISRC I a b d c 1 A VSRC 2 Ω 2 Ω R 1 A 2 V + – 2 V (2 + VSRC) V (1+VSRC /2) A (2+VSRC /2) A 0 10. The figure shows the variation with time of the charge across a capacitor of 0.1 µF. Determine the average voltage across the capacitor. A. V B. 2.5 V C. 2 V D. 10 V E. 5 V Solution: The net area is ( ) 6 2 2 4 2 4 1 = × − × µC-min. The average charge is 6/6 = 1µC and the average voltage is V = (1µC)/(C µF) = 1/C V. 12. Determine the power delivered by ISRC = 1 A. Solution: The 1 Ω and the 2 Ω resistors on the extreme RHS can be combined into a 3 Ω resistor. This gives a set of 3 Ω resistors connected in ∆, these being connected in parallel with three 1 Ω resistors connected in Y. The three ∆-connected resistors can be transformed to three 1 Ω resistors connected in Y. When these are paralleled with the existing 1 Ω resistors connected in Y, the result is three 0.5 Ω resistors connected in Y, as shown. The two 0.5 Ω resistors in series, paralleled with the give a 0.5 Ω resistor. When this is added to the 0.5 Ω and 2.5 Ω resistors, the resistors on the RHS of the source reduce to a 3 Ω resistor. Alternatively, the three 1 Ω resistors connected in Y can be transformed to three 3 Ω resistors connected in ∆. These are paralleled with the three 3 Ω ∆- connected resistors to give three 1.5 Ω resistors connected in ∆. These resistors can then be combined with the 1 Ω resistor to give a 0.5 Ω as before. The two 3 Ω resistors in parallel with ISRC give a 1.5 Ω resistor. The power delivered by the source is 1.5 2 ISRC P = . 2 Ω 1 Ω 1 Ω 1 Ω 1 Ω 1 Ω 3 Ω 3 Ω 2 Ω 0.5 Ω 3 Ω ISRC 0.5 Ω 1 Ω 2 Ω 0.5 Ω 3 Ω ISRC 0.5 Ω 0.5 Ω min q µC 4 4 6 -2