MECH 231 MECH 231 MECH 231 Circuit Fundamentals Circuit Fundamentals Circuit Fundamentals Dr. Maria Moussa Dr. Maria Moussa Dr. Maria Moussa MECH 231 MECH 231 MECH 231 Circuit Fundamentals Circuit Fundamentals Circuit Fundamentals MECH 231 – Circuits Fundamentals Dr. Maria Moussa Maria Moussa DC Circuits - Basic Dr.Laws Dr. Maria Moussa MECH 231 MECH 231 Fall 2023-2024 MECH 231 Circuit Fundamentals Circuit Fundamentals Circuit Fundamentals Dr. Maria Moussa Dr. Maria Moussa Dr. Maria Moussa MECH 231 MECH 231 Chapter 2 – BasicCircuit Laws Circuit Fundamentals Fundamentals MECH 231 Circuit Fundamentals Dr. Maria Moussa Dr. Maria Moussa Dr. Maria Moussa MECH 231 MECH 231 MECH 231 Circuit Fundamentals Circuit Fundamentals Circuit Fundamentals Dr. Maria Moussa Dr. Maria Moussa Dr. Maria Moussa 1. Ohm’s Law 2. Nodes, Branches and Loops 3. Kirchhoff’s Laws 4. Series Resistors and Voltage Division MECH 231 MECH 231 MECH 231 Circuit Fundamentals Circuit Fundamentals Circuit Fundamentals 5. Parallel Resistors and Current Division 6. Wye-Delta Transformations Dr. Maria Moussa 9/6/2023 Dr. Maria Moussa Chapter 2 – Basic Laws Dr. Maria Moussa 2 MECH 231 Ohm’s Law Circuit Fundamentals MECH 231 MECH 231 Circuit Fundamentals Circuit Fundamentals • Materials have resistive behavior to electric charge. Dr. Maria Moussa Dr. Maria Moussa π • The resistance of any material is calculated as π = π where Dr. Maria Moussa π΄ • π is the resistivity of the material in Ωπ, MECH 231 • π is its length in π, Circuit Fundamentals • π΄ is its cross-sectional area in π2 . MECH 231 MECH 231 Circuit Fundamentals Circuit Fundamentals • The circuit element used to model the current-resisting behavior of a material is Dr. Maria Moussa Dr. Maria Moussa Dr. Maria Moussa the resistor. • The constant of proportionality is defined to be the resistance π . MECH 231 MECH 231 9/6/2023 MECH 231 Circuit Fundamentals Circuit Fundamentals Circuit Fundamentals Dr. Maria Moussa Dr. Maria Moussa Dr. Maria Moussa Chapter 2 – Basic Laws 3 9/6/2023 MECH 231 MECH 231 MECH 231 Circuit Fundamentals Circuit Fundamentals Circuit Fundamentals Dr. Maria Moussa Dr. Maria Moussa Dr. Maria Moussa MECH 231 MECH 231 MECH 231 Circuit Fundamentals Circuit Fundamentals Circuit Fundamentals Dr. Maria Moussa Dr. Maria Moussa Dr. Maria Moussa MECH 231 MECH 231 MECH 231 Circuit Fundamentals Circuit Fundamentals Circuit Fundamentals Dr. Maria Moussa Dr. Maria Moussa Dr. Maria Moussa Chapter 2 – Basic Laws 4 MECH 231 MECH 231 MECH 231 Circuit Fundamentals Circuit Fundamentals Circuit Fundamentals Dr. Maria Moussa Dr. Maria Moussa Dr. Maria Moussa Example 1: An electric iron draws its resistance. MECH 231 2π΄ at 120 π. Find MECH 231 Solution: Form Ohm’s law, Circuit Fundamentals π£ Fundamentals 120 Circuit π = = = 60Ω π 2 MECH 231 Circuit Fundamentals Dr. Maria Moussa of a toasterDr.isMaria Moussaelement (a resistor) Dr. Maria Example 2: The essential component an electrical thatMoussa converts electrical energy to heat energy. How much current is drawn by a toaster with resistance 15Ω at 110 π? Solution: i = 7.333 π΄.MECH 231 9/6/2023 MECH 231 MECH 231 Circuit Fundamentals Circuit Fundamentals Circuit Fundamentals Dr. Maria Moussa Dr. Maria Moussa Dr. Maria Moussa Chapter 2 – Basic Laws 5 MECH 231 MECH 231 MECH 231 Example 3: In the circuit of Fig. 2.8, calculate the current I, the conductance G, and the power p. Circuit Fundamentals Circuit Fundamentals Circuit Fundamentals Solution: The voltage across the resistor is the same as the source voltage (30V) because the resistor and the voltage source are connected to the same pair of terminals. Dr. Maria Moussa Dr. Maria Moussa Hence, the currentDr. is Maria Moussa π£ 30 π= = = 6 ππ΄. π 5 × 103 MECH 231 MECH 231 The conductance is MECH 231 1 1 πΊ= = 3 = 0.2 ππ. π 5 × 10 Circuit Fundamentals Circuit Fundamentals Circuit Fundamentals We calculate the power in various ways π = π£π = 30 6 × 10−3 = 180 ππ Dr. Maria Moussa Dr. Maria Moussa Dr. Maria Moussa Or π = π 2 π = (6 × 10−3)25 × 103 = 180 ππ Or MECH 231 π = π£ 2πΊ = (30) MECH 231 2 0.2 × 10−3 = 180 ππ MECH 231 9/6/2023 Circuit Fundamentals Circuit Fundamentals Circuit Fundamentals Dr. Maria Moussa Dr. Maria Moussa Dr. Maria Moussa Chapter 2 – Basic Laws 6 MECH 231 MECH 231 Nodes Branches and Loops MECH 231 • Since the elements of Fundamentals an electric circuits canCircuit be connected in different ways, we need to understand some basic Circuit Fundamentals Circuit Fundamentals concepts of network topology. Dr. Maria Moussa Dr. Maria Dr. Maria Moussa • In network topology, we study the properties relating to Moussa the placement of elements in the network and the geometric configuration of the network. MECH 231 9/6/2023 MECH 231 MECH 231 Circuit Fundamentals Circuit Fundamentals Circuit Fundamentals Dr. Maria Moussa Dr. Maria Moussa Dr. Maria Moussa MECH 231 MECH 231 MECH 231 Circuit Fundamentals Circuit Fundamentals Circuit Fundamentals Dr. Maria Moussa Dr. Maria Moussa Dr. Maria Moussa Chapter 2 – Basic Laws 7 MECH 231 MECH 231 MECH 231 Circuit Fundamentals Circuit Fundamentals Circuit Fundamentals Dr. Maria Moussa Dr. Maria Moussa Dr. Maria Moussa Example 4: Determine the number of branches and nodes in the circuit shown in Fig. 2.12. Identify which elements are in series and which are in parallel. 231elements in the circuit, MECH MECH 231 Solution: Since thereMECH are four the231 circuit has four branches: 10π, 5Ω, 6Ω, πππ 2π΄. The circuit has three nodes as identified in Fig. 2.13. The 5Ω resistorCircuit is in series with the 10π voltage source because the same current would flow in both. Fundamentals Circuit Fundamentals Circuit Fundamentals The 6Ω resistor is in parallel with the 2π΄ current source because both are connected to the same nodes 2 and 3. 9/6/2023 Dr. Maria Moussa Dr. Maria Moussa Dr. Maria Moussa MECH 231 MECH 231 MECH 231 Circuit Fundamentals Circuit Fundamentals Circuit Fundamentals Dr. Maria Moussa Dr. Maria Moussa Dr. Maria Moussa Chapter 2 – Basic Laws 8 MECHLaws 231 Kirchhoff’s MECH 231 MECH 231 Fundamentals Circuit Fundamentals Circuit Fundamentals • These laws areCircuit formally known as Kirchhoff’s current law (KCL) and Kirchhoff’s voltage law (KVL). 9/6/2023 Dr. Maria Moussa Dr. Maria Moussa Dr. Maria Moussa MECH 231 MECH 231 MECH 231 Circuit Fundamentals Circuit Fundamentals Circuit Fundamentals Dr. Maria Moussa Dr. Maria Moussa Dr. Maria Moussa MECH 231 MECH 231 MECH 231 Circuit Fundamentals Circuit Fundamentals Circuit Fundamentals Dr. Maria Moussa Dr. Maria Moussa Dr. Maria Moussa Chapter 2 – Basic Laws 9 9/6/2023 MECH 231 MECH 231 MECH 231 Circuit Fundamentals Circuit Fundamentals Circuit Fundamentals Dr. Maria Moussa Dr. Maria Moussa Dr. Maria Moussa MECH 231 MECH 231 MECH 231 Circuit Fundamentals Circuit Fundamentals Circuit Fundamentals Dr. Maria Moussa Dr. Maria Moussa Dr. Maria Moussa MECH 231 MECH 231 MECH 231 Circuit Fundamentals Circuit Fundamentals Circuit Fundamentals Dr. Maria Moussa Dr. Maria Moussa Dr. Maria Moussa Chapter 2 – Basic Laws 10 MECH 231 MECH 231 MECH 231 Circuit Fundamentals Circuit Fundamentals Circuit Fundamentals Dr. Maria Moussa Dr. Maria Moussa Example 5: For the circuit in Fig. 2.21(a), find voltages π£1 and π£2. MECH 231 MECH 231 Dr. Maria Moussa MECH 231 Solution: To find π£1 and π£2 we apply Ohm’s law and Kirchhoff’s voltage law. Assume that current π flows throughCircuit the loop as shown in Fig. 2.21(b). Circuit Fundamentals Fundamentals Circuit Fundamentals From Ohm’s law, π£1 = 2π, π£2 = −3π Maria Moussa Dr. Maria Moussa ApplyingDr. KVL around the loop gives Dr. Maria Moussa −20 + π£1 − π£2 = 0 Substituting the first 2 equations into the third one, we obtain MECH 231 MECH 231 MECH 231 −20 + 2π + 3π = 0 ππ 5π = 20 βΉ π = 4π΄ Substituting π in the first 2 equation finally gives Circuit Fundamentals Circuit π£1 Fundamentals = 8π, π£2 = Circuit −12π Fundamentals Dr. Maria Moussa 9/6/2023 Dr. Maria Moussa Chapter 2 – Basic Laws Dr. Maria Moussa 11 MECH 231 MECH 231 MECH 231 Circuit Fundamentals Circuit Fundamentals Circuit Fundamentals Dr. Maria Moussa Dr. Maria Moussa Example 6: Determine π£0 and π in the circuit shown in Fig. 2.23(a). MECH 231 MECH 231 Dr. Maria Moussa MECH 231 Solution: We apply KVL around the loop as shown in Fig. 2.23(b). The result is −12 + 4πFundamentals + 2π£0 − 4 + 6π = 0 Circuit Fundamentals Circuit Fundamentals Circuit Applying Ohm’s law to the 6Ω resistor gives π£0 = −6π Maria Dr. Maria Moussa Substituting π£0 Dr. in the firstMoussa equation yields Dr. Maria Moussa −16 + 10π − 12π = 0 βΉ π = −8π΄ πππ π£0 = 48π 9/6/2023 MECH 231 MECH 231 MECH 231 Circuit Fundamentals Circuit Fundamentals Circuit Fundamentals Dr. Maria Moussa Dr. Maria Moussa Dr. Maria Moussa Chapter 2 – Basic Laws 12 MECH 231 MECH 231 MECH 231 Series Resistors and Voltage Division 9/6/2023 Circuit Fundamentals Circuit Fundamentals Circuit Fundamentals Dr. Maria Moussa Dr. Maria Moussa Dr. Maria Moussa MECH 231 MECH 231 MECH 231 Circuit Fundamentals Circuit Fundamentals Circuit Fundamentals Dr. Maria Moussa Dr. Maria Moussa Dr. Maria Moussa MECH 231 MECH 231 MECH 231 Circuit Fundamentals Circuit Fundamentals Circuit Fundamentals Dr. Maria Moussa Dr. Maria Moussa Dr. Maria Moussa Chapter 2 – Basic Laws 13 Principle MECH 231of Voltage division MECH 231 – Voltage Divider MECH 231Circuit 9/6/2023 Circuit Fundamentals Circuit Fundamentals Circuit Fundamentals Dr. Maria Moussa Dr. Maria Moussa Dr. Maria Moussa MECH 231 MECH 231 MECH 231 Circuit Fundamentals Circuit Fundamentals Circuit Fundamentals Dr. Maria Moussa Dr. Maria Moussa Dr. Maria Moussa MECH 231 MECH 231 MECH 231 Circuit Fundamentals Circuit Fundamentals Circuit Fundamentals Dr. Maria Moussa Dr. Maria Moussa Dr. Maria Moussa Chapter 2 – Basic Laws 14 MECH 231 231 MECH 231 Parallel Resistors and MECH Current Division 9/6/2023 Circuit Fundamentals Circuit Fundamentals Circuit Fundamentals Dr. Maria Moussa Dr. Maria Moussa Dr. Maria Moussa MECH 231 MECH 231 MECH 231 Circuit Fundamentals Circuit Fundamentals Circuit Fundamentals Dr. Maria Moussa Dr. Maria Moussa Dr. Maria Moussa MECH 231 MECH 231 MECH 231 Circuit Fundamentals Circuit Fundamentals Circuit Fundamentals Dr. Maria Moussa Dr. Maria Moussa Dr. Maria Moussa Chapter 2 – Basic Laws 15 Principle MECH 231of Current division MECH 231 – Current Divider MECH 231Circuit Circuit Fundamentals Circuit Fundamentals Circuit Fundamentals Dr. Maria Moussa Dr. Maria Moussa Dr. Maria Moussa MECH 231 MECH 231 MECH 231 Circuit Fundamentals Circuit Fundamentals Circuit Fundamentals equivalent Dr.The Maria Moussa 9/6/2023 conductance of resistors connected in series: Dr. Maria Moussa Dr. Maria Moussa MECH 231 MECH 231 MECH 231 Circuit Fundamentals Circuit Fundamentals Circuit Fundamentals Dr. Maria Moussa Dr. Maria Moussa Dr. Maria Moussa Chapter 2 – Basic Laws 16 Example 7: Find π ππ for the circuit MECH 231 shown in Fig. 2.34. MECH 231 MECH 231 Solution: To get π Circuit combine resistors in series and in parallel. The 6Ω and 3Ω Fundamentals Circuit Fundamentals Circuit ππ, we Fundamentals resistors are in parallel, so their equivalent resistance is 6×3 Dr. Maria Moussa Maria Moussa Dr. Maria Moussa 6Ω β 3Ω = =Dr.2Ω 6+3 Also the 1Ω and 5Ω resistors are in series, hence their equivalent resistance is MECH 231 1Ω + 5Ω = 6Ω MECH 231 MECH 231 The circuit in Fig. 2.34 is reduced to that in Fig. 2.35(a). We notice that the two 2Ω Circuit Circuit Fundamentals Circuit Fundamentals resistors are in series so, Fundamentals 2Ω + 2Ω = 4Ω This 4Ω resistor is now in parallel with the 6Ω resistor in Fig. 2.35(a) so, Dr. Maria Moussa Dr. Maria Moussa Dr. Maria Moussa 4×6 4Ω β 6Ω = = 2.4Ω 4+6 MECH 231 MECH 231 MECH 231 The circuit is now in Fig. 2.35(b) where the 3 resistors are in series, hence, π ππ = 4Ω + 2.4Ω + 8Ω = 14.4Ω Circuit Fundamentals Circuit Fundamentals Circuit Fundamentals Dr. Maria Moussa 9/6/2023 Dr. Maria Moussa Chapter 2 – Basic Laws Dr. Maria Moussa 17 Example 8: Calculate the equivalent the circuit MECH 231 resistance π ππ for MECH 231 shown in Fig. 2.37. Solution: The 3Ω Circuit and 6ΩFundamentals resistors are in parallelCircuit Fundamentals 6× 3 6Ω β 3Ω = = 2Ω 6+3 Dr. 4Ω Maria Moussa Similarly, the 12Ω and resistors are in parallel Dr. Maria Moussa 12 × 4 12Ω β 4Ω = = 3Ω 12 + 4 MECH 231 The 1Ω and 5Ω resistorsMECH are in231 series 1Ω + 5Ω = 6Ω Circuit Fundamentals Circuit Fundamentals In Fig. 2.38(a), a 3Ω in parallel with a 6Ω gives a 2Ω. This 2Ω is in series with the 1Ω resulting in 3Ω. Dr. Maria Moussa Dr. Maria Moussa In Fig. 38(b), a 2Ω in parallel with a 3Ω gives a 1.2Ω. The 1.2Ω in series with MECH the 10Ω gives an 11.2Ω. 231 MECH 231 βΉ π ππ = 10 + 1.2 = 11.2Ω Circuit Fundamentals Circuit Fundamentals Dr. Maria Moussa 9/6/2023 Dr. Maria Moussa Chapter 2 – Basic Laws MECH 231 Circuit Fundamentals Dr. Maria Moussa MECH 231 Circuit Fundamentals Dr. Maria Moussa MECH 231 Circuit Fundamentals Dr. Maria Moussa 18 MECH 231 MECH 231 MECH 231 Circuit Fundamentals Circuit Fundamentals Circuit Fundamentals Wye-Delta Transformations • Situations often arise in circuit analysis when the resistors are neither in parallel nor in series. Dr. Maria Moussa Dr. Maria Moussa Dr. Maria Moussa • Many circuits of this type can be simplified by using three-terminal equivalent networks. MECH 231 MECH 231 MECH 231 Circuit Fundamentals Circuit Fundamentals Circuit Fundamentals Dr. Maria Moussa Dr. Maria Moussa Dr. Maria Moussa MECH 231 MECH 231 MECH 231 • Our main interest is how to identify them when they occur as part of a network and how to apply wye-delta transformation in the analysis of that network Circuit Fundamentals Circuit Fundamentals Circuit Fundamentals Dr. Maria Moussa 9/6/2023 Dr. Maria Moussa Chapter 2 – Basic Laws Dr. Maria Moussa 19 9/6/2023 MECH 231 MECH 231 MECH 231 Circuit Fundamentals Circuit Fundamentals Circuit Fundamentals Dr. Maria Moussa Dr. Maria Moussa Dr. Maria Moussa MECH 231 MECH 231 MECH 231 Circuit Fundamentals Circuit Fundamentals Circuit Fundamentals Dr. Maria Moussa Dr. Maria Moussa Dr. Maria Moussa MECH 231 MECH 231 MECH 231 Circuit Fundamentals Circuit Fundamentals Circuit Fundamentals Dr. Maria Moussa Dr. Maria Moussa Dr. Maria Moussa Chapter 2 – Basic Laws 20 9/6/2023 MECH 231 MECH 231 MECH 231 Circuit Fundamentals Circuit Fundamentals Circuit Fundamentals Dr. Maria Moussa Dr. Maria Moussa Dr. Maria Moussa MECH 231 MECH 231 MECH 231 Circuit Fundamentals Circuit Fundamentals Circuit Fundamentals Dr. Maria Moussa Dr. Maria Moussa Dr. Maria Moussa MECH 231 MECH 231 MECH 231 Circuit Fundamentals Circuit Fundamentals Circuit Fundamentals Dr. Maria Moussa Dr. Maria Moussa Dr. Maria Moussa Chapter 2 – Basic Laws 21 9/6/2023 MECH 231 MECH 231 MECH 231 Circuit Fundamentals Circuit Fundamentals Circuit Fundamentals Dr. Maria Moussa Dr. Maria Moussa Dr. Maria Moussa MECH 231 MECH 231 MECH 231 Circuit Fundamentals Circuit Fundamentals Circuit Fundamentals Dr. Maria Moussa Dr. Maria Moussa Dr. Maria Moussa MECH 231 MECH 231 MECH 231 Circuit Fundamentals Circuit Fundamentals Circuit Fundamentals Dr. Maria Moussa Dr. Maria Moussa Dr. Maria Moussa Chapter 2 – Basic Laws 22 9/6/2023 MECH 231 MECH 231 MECH 231 Circuit Fundamentals Circuit Fundamentals Circuit Fundamentals Dr. Maria Moussa Dr. Maria Moussa Dr. Maria Moussa MECH 231 MECH 231 MECH 231 Circuit Fundamentals Circuit Fundamentals Circuit Fundamentals Dr. Maria Moussa Dr. Maria Moussa Dr. Maria Moussa MECH 231 MECH 231 MECH 231 Circuit Fundamentals Circuit Fundamentals Circuit Fundamentals Dr. Maria Moussa Dr. Maria Moussa Dr. Maria Moussa Chapter 2 – Basic Laws 23 Example 9: Convert theMECH β network equivalent 231 in Fig. 2.50(a) to an MECH 231 Y network MECH 231 Solution: Using the previous equations, we obtain Circuit Fundamentals Circuit Fundamentals π π π π 10 × 25 250 π 1 = = = = 5Ω π π + π π + π π 15 + 10 + 25 50 Dr. Maria Moussa Dr. Maria Moussa Circuit Fundamentals π π π π 25 × 15 π 2 = = = 7.5Ω π + π + π 50 π MECH π231 π MECH 231 π ππ π 15 × 10 π = = = 3Ω Circuit 3Fundamentals Circuit π π + π π + π π 50 Fundamentals The equivalent Y network is shown in the figure below. Dr. Maria Moussa Dr. Maria Moussa 9/6/2023 Dr. Maria Moussa MECH 231 Circuit Fundamentals Dr. Maria Moussa MECH 231 MECH 231 MECH 231 Circuit Fundamentals Circuit Fundamentals Circuit Fundamentals Dr. Maria Moussa Dr. Maria Moussa Dr. Maria Moussa Chapter 2 – Basic Laws 24