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Chapter 2 Resistive Circuits 1 Overview of Chapter 2 2.1 Series Resistors and Parallel Resistors 2.2 Voltage Divider Circuit 2.3 Current Divider Circuit 2.4 Voltage and Current Measurement 2.5 Wheatstone Bridge 2.6 Wye-Delta Transformations. 2 Resistors 3 4 5 6 2.1 Series and Parallel Resistors (1) Series Resistors Two or more elements are in series if they are cascaded or connected sequentially and consequently carry the same current. The equivalent resistance of any number of resistors connected in a series is the sum of the individual resistances. N Req R1 R2 R N Rn n 1 7 Series Resistors in Circuit The potential difference (voltage) between the terminals of the battery (V) equals the sum of the potential differences across the resistors. KVL V VR1 VR 2 VR 3 I ( R1 R2 R3 ) IReq 8 2.1 Series and Parallel Resistors (2) Parallel Resistor Two or more elements are in parallel if they are connected to the same two nodes and consequently have the same voltage across them. The equivalent resistance of a circuit with N resistors in parallel is: 1 1 1 1 Req R1 R2 RN 9 Parallel Resistors in Circuit The supply current (I) equals the sum of the currents in the branches. KCL I I1 I 2 I 3 V V V R1 R2 R3 1 1 1 V R1 R2 R3 1 V R eq 10 2.1 Series and Parallel Resistors (3) Example Find Req in circuit below. 11 2.1 Series and Parallel Resistors (3) 4×6 4. 4||6 → 4+6 2.4Ω 5. 4 + 8 + 2.4 = 14.4Ω 1. 6||3 → 6×3 6+3 2Ω 2. 1 + 5 = 6Ω 3. 2 + 2 = 4Ω 12 2.1 Series and Parallel Resistors (4) Example Find Req in circuit below. Answer: 11 ohm 13 2.2 Voltage Divider Circuit (1) The voltage divider for N series resistors can be expressed as Rn vn v R1 R2 RN 14 2.2 Voltage Divider Circuit (2) To determine the voltage across each resistor. 15 2.2 Voltage Divider Circuit (3) Apply KVL to the circuit; From Ohm’s law, v v1 v2 0 @ v v1 v2 v1 iR1 and v2 iR2 From KVL, therefore, v iR1 iR2 Solving equation for current, i v R1 R2 R n Apply Ohm’s law to determine voltage across each resistor; vn v R1 R2 RN R1 R2 v2 v v1 v R1 R2 R1 R2 16 2.3 Current Divider Circuit (1) The total current, I is shared by the resistors in inverse proportion to their resistances. The current divider can be expressed as: v iReq in Rn Rn 17 2.3 Current Divider Circuit (2) To determine the current flow on each resistor. 18 2.3 Current Divider Circuit (3) 1 v v Apply KCL, i i1 i2 v R R1 R2 eq iR1 R2 Using ohm’s law on each branch, therefore: v iReq R1 R2 Current flow through each branch, i 1 iR2 R1 R2 iR1 i2 R1 R2 19 Example (voltage divider) Determine vo and io in the circuit below. Answer: Vo=4V, io=4/3A 20 2.4 Voltage and Current Measurement • An ammeter is an instrument designed to measure current and is placed in series with the circuit element. • A voltmeter is an instrument designed to measure voltage and is placed in parallel with the element. 21 2.5Wheatstone Bridge A Wheatstone bridge circuit is an accurate device for measuring resistance 22 Under balance condition where no current flow between BD, VAD VAB or I1 R1 I 2 R2 and VDC VBC or I 3 R3 I 4 R4 Current in each resistance arm, I1 I 3 and I 2 I 4 Therefore, R2 R4 R1 R3 R2 R3 R4 R1 23 2.6 Wye-Delta Transformations • Wye-delta transformation occurs when the resistors are neither in parallel or in series. • This circuit can be simplified to a three-terminal equivalent 24 Wye-Delta Transformations II Two topologies can be interchanged: Wye (Y) or tee (T) networks Delta (Δ) or pi (Π) networks Transforming between these two topologies often makes the solution of a circuit easier 25 Wye-Delta Transformations III The superimposed wye and delta circuits shown here will used for reference The delta consists of the outer resistors, labeled a,b, and c The wye network are the inside resistors, labeled 1,2, and 3 26 Y-Δ Transformations Rc Rb Ra Ra R1 R2 R2 R3 R3 R1 R1 Rb R1 R2 R2 R3 R3 R1 R2 Rc R1 R2 R2 R3 R3 R1 R3 27 Δ - Y Transformations R1 Rb Rc ( Ra Rb Rc ) R2 Rc Ra ( Ra Rb Rc ) Ra Rb R3 ( Ra Rb Rc ) 28 Example 1 Transform Y circuit to Δ circuit Ra = 15 Rb = 10 Rc = 25 29 Example #2 Find the current and power supplied by the 40 V sources in the circuit shown below. 30 Example #2 Solution: We can find this equivalent resistance easily after replacing either the upper Δ (100Ω, 125Ω, 25Ω) or the lower Δ (40Ω, 25Ω, 37.5Ω) with its equivalent Y. We choose to replace the upper Δ. Thus, 31 Example #2 100 125 R1 50 250 125 25 R2 12.5 250 100 25 R3 10 250 32 Substituting the Y-resistor into the circuit, Equivalent resistor, Req: 50 50 Req 55 80 100 33 Simplified circuit: Hence, current and power values are: V 40 i 0.5 A R 80 p V * I 40 0.5 20W 34 Another option: 100 Ra Rb Rb Rc 40 Rc Another option is transform from Why to Delta.. You can try this as well. 35 Example #3 (a) Find no load value of vo. (b) Find vo when RL = 150 kΩ (c) How much power is dissipated in the 25 kΩ resistor if the load terminals are short-circuited ? 36 Example #3 a) Find no load value of vo R v Apply ohm law: R R v 1 1 1 2 75k v0 200 150V 75k 25k b) Find vo when RL = 150 kΩ 75k 150k Req 50k 75k 150k 50k v0 200 133.33V 50k 25k 37 Example #3 c) How much power is dissipated in the 25 kΩ resistor if the load terminals are short-circuited ? V 200 3 I 8 10 A R 25k 3 P VI (200)(8 10 ) 1.6W 38 Example #4 • Find the power dissipated in the 6 Ω resistor. 39 Example #4 Solution: Req Equivalent resistance Req ( 4 6 ) 1.6 4 current io, iR2 i1 R1 R2 iR1 i2 R1 R2 16 i0 10 8A 16 4 40 Example #4 Note that io is the current in the 1.6Ω resistor. Use current divider to get current in the 6Ω resistor, i1 i0 iR2 R1 R2 i2 iR1 R1 R2 4 i6 8 3.2 A 46 Then the power dissipated by the resistor is P I R (3.2) (6) 61.44W 2 2 41 Example #5 • Find the voltage of vo and vg. 42 Example #5 - Solution step 1 simplified the circuit 1)60 30 20 2)20 30 50 3)50 25 75 Current in resistor 30Ω (at Vo) is: i1 i30 iR2 R1 R2 i2 iR1 R1 R2 (25)(75) 15 A 50 75 43 Voltage v0 where i0 is 15A and R0 is 20 v0 (15)(20) 300V Total voltage at the resistor, Vr: v0 30i30 300 450 750V 44 V12 Vr Voltage vg v g v12 vr v g 12(25) 750 v g 1050V 45 Example #6 Find the current of ig and io in the circuit 46 Example #6 • Find the current of ig and io in the circuit. Solution: • Equivalent resistance: 5 20 4 4 6 10 15 12 12 40 10 40 8 47 Example #6 The current values, 125 ig 12.5 A 82 i6 ( 40)(12.5) 10 A (15 12 13) (6 4) iR2 i1 R1 R2 Thus, iR1 i2 R1 R2 (5)(10) i0 2A 20 5 48 Exercise 1 Determine the value of io 49 Exercise 2 Find i and Vo 50 Exercise 3 Calculate the value of current; I. 51