Electric Circuits I Simple Resistive Circuit Dr. Firas Obeidat 1
Resistors in Series The equivalent resistance of any number of resistors connected in series is the sum of the individual resistances. It is often possible to replace relatively complicated resistor combinations with a single equivalent resistor without changing all the current, voltage, and power relationships in the remainder of the circuit. series combination of N resistors v s v 1 +v 2 +v 3 +v 4 + +v N v s R 1 i+r 2 i+r 3 i+r 4 i+ +R N i v s (R 1 +R 2 +R 3 +R 4 + +R N )i v s R eq i R eq R 1 +R 2 +R 3 +R 4 + +R N R eq : equivalent resistor 2
Resistors in Parallel The equivalent resistance of two parallel resistors is equal to the product of their resistances divided by their sum. It is often possible to replace relatively complicated resistor combinations with a single equivalent resistor without changing all the current, voltage, and power relationships in the remainder of the circuit. Parallel combination of N resistors i s i 1 +i 2 +i 3 +i 4 + +i N + + + + + i s Req + + + + + 3
Resistors in Parallel A parallel combination is routinely indicated by the following shorthand notation The special case of only two parallel resistors is encountered fairly often, and is given by + + + 4
Examples Example: find R ab? the 2Ω and 3Ω resistors are in parallel This 1.2 Ω resistor is in series with the 10 Ω resistor. Example: find R ab? Answer: 19Ω 5
Examples Example: find R eq? 6
Examples Example: find R eq? Answer: 10Ω Example: find R ab? The 3Ω and 6Ω resistors are in parallel because they are connected to the same two nodes c and b. The 12Ω and 4Ω resistors are in parallel since they are connected to the same two nodes d and b. Also the 1Ω and 5Ω resistors are in series 7
Examples Example: find R bc,r bc,and R bc? + + + +. ( + + + ) [ + ] + Example: find R eq and i?. +... A 8
Examples Example: determine the current i in the figure and the power delivered by the 80 V source. We first interchange the element positions in the circuit. combine the three voltage sources into an equivalent 90 V source, and the four resistors into an equivalent 30 Ω resistance + + + + -90+30i0 3A The desired power in the voltage source 80V is 80V 3A 240W 9
Examples Example: Calculate the power and voltage of the dependent source in the figure. The two 6Ω resistors are in parallel and can be replaced with a single 3Ω resistor in series with the 15Ω resistor. Thus, the two 6Ω resistors and the 15Ω resistor are replaced by an 18Ω resistor. The controlling variable i 3 depends on the 3Ω resistor and so that resistor must remain untouched. Also 9Ω 18Ω6 Ω Applying KCL at the top node. + + ( ) Employing Om s law. (2) Solve equation (1) and (2), then i3(10/3)a Thus, the voltage across the dependent source, v3i 3 10V p D v 0.9i 3 10 0.9 10/330W 10
Voltage Division Voltage division is used to express the voltage across one of several series resistors. So v s v 1 +v 2 R 1 i+r 2 i (R 1 +R 2 )i + Thus + Or + And the voltage across R1 is similarly + General result for voltage division across a string of N series resistors + + + + 11
Current Division Current division is used to express the current across one of several parallel resistors. The current flowing through R 2 is + Or And similarly + For a parallel combination of N resistors, the current through resistor R k is + + + + + 12
Examples Example: Determine v x in the circuit of shown figure. Combine the 6Ω and 3Ω resistors, replacing them with: 6 3/(6+3)2Ω Since v x appears across the parallel combination, the simplification has not lost this quantity. + Example: Determine v x in the circuit of shown figure. Answer: 2V. 13
Examples Example: write an expression for the current through the 3 resistor in the circuit of shown figure. The total current flowing into the combination is ( ) + + The desired current is given by current division: ( ) + Example: In the circuit of Fig. 3.39, use resistance combination methods and current division to find i1, i2, and v3. Answer: 100mA, 50mA, 0.8V. 14
Examples Example: For the circuit shown in the figure, determine: (a) the voltage v o (b) the power supplied by the current source, (c) the power absorbed by each resistor. (a) The 6kΩ and 12k Ω resistors are in series so that their combined value is 6k+12k18k Ω. Applying the current division technique to find i 1 and i 2 + ( ) ( ) + The voltage across the 9-k and 18-k resistors is the same, and v o 9000i 1 18000i 2 180V. 15
Examples Example: For the circuit shown in the figure, determine: (a) the voltage v o (b) the power supplied by the current source, (c) the power absorbed by each resistor. (b) Power supplied by the source is.. (c) Power absorbed by the 12kΩ resistor is. p. Power absorbed by the 6kΩ resistor is.. Power absorbed by the 9kΩ resistor is. ( ). or. 16
Examples Example: Design the voltage divider of the figure such that V R1 4V R2. The total resistance is defined by.. Since V R1 4V R2 R 1 4R 2 + + 5kΩ 1kΩ R 1 4R 2 4kΩ 17
Examples Example: calculate the indicated currents and voltage of figure shown below. Redrawing the network after combining series elements yields,, (, ) (,,....... + +.. 18
Examples Example: Determine the voltages V 1, V 2, and V 3 network of the figure shown below. for the For path (1) 0 20V-8V12V For path (2) 0 V-12V-7V Indicating that V 2 has a magnitude of 7 V but a polarity opposite to that appearing in the figure. For path (3) + 0 V-(-7V)15V 19
Delta-Wye Conversion Parallel and series combinations of resistors can often lead to a significant reduction in the complexity of a circuit. There is another useful technique, called delta-wye (Δ-Υ ) conversion that can be used to simplify the circuit. These networks (Δ,Υ) occur by themselves or as part of a larger network. They are used in threephase networks, electrical filters, and matching networks. Two forms of the same network: (a) Y, (b) T. Two forms of the same network: (a) Δ,(b) Π. 20
Delta to Wye Conversion To obtain the equivalent resistances in the wye network, we compare the two networks and make sure that the resistance between each pair of nodes in the Δ (or Π) network is the same as the resistance between the same pair of nodes in the Y (or T) network. For terminals 1 and 2. (Y) + (Δ) ( + ) Setting (Y) (Δ) + ( ) (1) Similarly + ( ) (2) + ( ) (3) 21
Delta to Wye Conversion Subtracting eq. (3) from eq. (1) ( ) + + Adding eq. (2) from eq. (4) (4) (5) + + Subtracting eq. (4) from eq. (2) (6) + + Subtracting eq. (5) from eq. (1) Each resistor in the Y network is the product of the resistors in the two adjacent Δ branches, divided by the sum of the three Δ resistors. + + (7) 22
Wye to Delta Conversion To obtain the conversion formulas for transforming a wye network to an equivalent delta network. Multiply eq.(5)&eq.(6), eq.(6)&eq.(7), eq.(7)&eq.(1). Then add all the three equation together. + + ( + + ) ( + + ) Dividing eq. (8) by each of equations (5), (6) and (7) leads to the following equations + + (8) (9) (10) (11) Each resistor in the Δ network is the sum of all possible products of Y resistors taken two at a time, divided by the opposite Y resistor. 23
Delta Wye Conversion The Y and Δ networks are said to be balanced when, Under these conditions, conversion formulas become or 3 Example: Obtain the equivalent resistance for the circuit in the figure below and use it to find current i. In this circuit, there are two Y networks and three Δ networks. Transforming just one of these will simplify the circuit. If we convert the Y network comprising the 5Ω, 10Ω, and 20Ω resistors, we may select 24
Delta Wye Conversion Example: Obtain the equivalent resistance for the circuit in the figure below and use it to find current i. 10Ω, 20Ω, 5Ω, 35Ω 17.5Ω 70Ω 25
Delta Wye Conversion Example: Obtain the equivalent resistance for the circuit in the figure below and use it to find current i. Combining the three pairs of resistors in parallel, we obtain 70 30 70 30 70 + 30 21Ω 12.5 17.5 12.5 17.5 12.5 + 17.5 7.292Ω 15 35 15 35 15 + 35 10.5Ω (7.292 + 10.5) 21 17.792 21 17.792 + 21 9.632Ω 120 9.632 12.458Ω 26
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