Solving Electrical Engineering Problems: Voltage, Current, and Power Calculations, Exams of Electrical and Electronics Engineering

Download Solving Electrical Engineering Problems: Voltage, Current, and Power Calculations and more Exams Electrical and Electronics Engineering in PDF only on Docsity! 332:221 Principles of Electrical Engineering I – Fall 2007 Hourly Exam 1 – October 3, 2007 This is a closed-book closed-notes exam. Do all your work on these sheets. If more space is required, do your work on the back side of the sheets and indicate accordingly so that the grader does not miss it. Problem Page Maximum Points Description # Points earned 1 1 20 KCL, KVL & Power 2 2 15 KCL, KVL & Power 3 3 20 Voltage divider 4 4 20 Solving a simple circuit 5 5 25 Delta - Y transformations Total maximum points = 100 Total points earned by the student = Note: Any equation that contains wrong signs for any of its terms is obviously a wrong equation. Not much credit can be given to such an equation. Think of this way. If a deposit in your bank account comes out in the column of your withdrawals, will you complain to the bank or not? 1 Problem 1: (20 points) − + − 12V + is = 2vx + vg − − vdc + 1.5Ω − vx + 0.4Ω − vc + 10Ω G e a bcd icd Some one partially and correctly solved the circuit shown above and determined vc as 10 V. With this knowledge, determine the following: (a) Determine vx and then the current is = 2vx of the controlled source. To determine vx you may make use of the KVL around the loop abcGa. The KVL around the loop abcGa is given by 12 − vx − vc = 0 ⇒ vx = 12 − vc = 12 − 10 = 2 V. Thus the the current is = 2vx = 4 A. (b) What is the current icd that flows through the 1.5Ω resistance? To determine this you may make use of the KCL at the node d. We note that icd = is = 4 A. (c) Once the current icd is known, determine the voltage vdc. We note that vdc = icd1.5 = 6 V. (d) Determine vg, the voltage of the controlled source. To determine vg you may make use of the KVL around the loop GcdeG. The KVL around the loop GcdeG is given by vc − vdc − vg = 0 ⇒ vg = vc − vdc = 10 − 6 = 4 V. (e) Determine the power associated with the controlled source. Is this power consumed or generated by the controlled source? The power consumed by the controlled source = vgis = (4)(4) = 16 W. 4 Problem 4: (20 points) − + − va + − + − vb + Ra Rb RL + vout − ia ib iL Consider the circuit shown on the left where Ra = 20Ω, Rb = 20Ω, and RL = 30Ω. We would like to solve first for iL and then for vout = iLRL in terms of va and vb. To do so, write down the following laws and then solve for iL and then for vout = iLRL. 1. KCL to relate ia, ib, and iL. 2. KVL around the loop containing the source va, resistance Ra, and resistance RL. 3. KVL around the loop containing the source vb, resistance Rb, and resistance RL. Solution: It is easy to write the following three equations: ia + ib = iL KCL (1) va − iaRa − iLRL = va − ia20 − iL30 = 0 KVL (2) vb − ibRb − iLRL = vb − ib20 − iL30 = 0. KVL (3) There exists a lot of symmetry in the last two equations. By adding the last two equations and recognizing ia + ib = iL, we get va + vb − (ia + ib)20 − iL60 = 0 ⇒ va + vb − iL20 − iL60 = 0. This implies that iL = va + vb 80 ⇒ vout = iLRL = (va + vb)30 80 = (va + vb)3 8 . In general while keeping Ra, Rb, and RL as symbols, we can solve the equations as follows. We note that there exists a lot of symmetry between the equations (2) and (3). We can exploit such a symmetry. Multiplying equation (2) with Rb and similarly multiplying equation (3) with Ra, we get vaRb − iaRaRb − iLRLRb = 0 and vbRa − ibRbRa − iLRLRa = 0. By adding the above two equations and recognizing ia + ib = iL, we get vaRb + vbRa − iL(RaRb + RaRL + RbRL) = 0. The above equation implies that iL = Rb RaRb + RaRL + RbRL va + Ra RaRb + RaRL + RbRL vb. Then, we immediately get vout = iLRL = RbRL RaRb + RaRL + RbRL va + RaRL RaRb + RaRL + RbRL vb. For the given values of Ra = 20Ω, Rb = 20Ω, and RL = 30Ω, we have vout = 3 8 va + 3 8 vb == (va + vb)3 8 . 5 The ∆ and Y circuit equivalent relationships are as shown: R1 = RbRc Ra+Rb+Rc Ra = R1R2+R2R3+R3R1 R1 R2 = RaRc Ra+Rb+Rc Rb = R1R2+R2R3+R3R1 R2 R3 = RaRb Ra+Rb+Rc Rc = R1R2+R2R3+R3R1 R3 A B C Rb Ra Rc A B C R1 R2 R3 Problem 5: (25 points) Transform the ∆ circuit between the terminals C, D, and E to an equivalent Wye circuit. Then, utiliz- ing series and parallel equivalents, determine the resistance seen to the right of terminals A and B. Draw clearly all the equivalent circuits. A B D E F C30Ω 32Ω 30Ω 12Ω 24Ω 10Ω60Ω We can simplify the above circuit by transforming the ∆ circuit between the terminals C, D, and E to an equivalent Wye as follows: A B D E F C 30Ω 32Ω 18Ω 12Ω 24Ω 3Ω 6Ω By series and parallel equivalents, we can simplify the above circuit sequentially as illus- trated below: A B 30Ω 50Ω 15Ω30Ω A B D 30Ω 50Ω 10Ω ⇒ A B 30Ω 60Ω It should relatively be easy now to compute the resistance seen to the right of terminals A and B as 20Ω.