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Transcript

Lesson 5: Parallel DC Circuits and Kirchhoffs Current Law (KCL) (Chapter 6) 1 Learning Objectives • • • • • • • • Restate the definition of a node and demonstrate how to measure voltage and current in parallel circuits. Solve for total circuit resistance of a parallel circuit. State and apply KCL in the analysis of simple parallel circuits. Demonstrate how to calculate the total parallel resistance given various resistors connected in parallel. Evaluate why homes, businesses and ships are commonly wired in parallel rather than series. Demonstrate how to calculate the total current and branch currents in a parallel circuit using the current divider equation. Determine the net effect of parallel combining voltage sources. Compute the power dissipated by each element in a parallel circuit, and calculate the total circuit power. 2 Parallel Circuits • Two or more elements are in parallel if they are connected to the same two nodes and consequently have the same voltage across them. 2-, 3-, and 2A are in parallel 3 Parallel Circuits • Remember: nodes are connection points between components. • Notice each component has two terminals and each is connected to one of the nodes above. 4 Parallel Circuits • House circuits contain parallel circuits. • The parallel circuit will continue to operate even though one component may fail open – an advantage over the series circuit. 5 Parallel Circuits vs. Series Circuits • In a series circuit, failure of a single components can disable all components in the circuit. Series Circuit • In a parallel circuit, failure of one component will still allow other components to operate. Parallel Circuit 6 Circuit Breaker Panel • Homes and ships are usually wired in parallel instead of series. • All components can operate at rated voltage independent of other loads when wired in parallel. AC 7 Series - Parallel Circuits • Circuits may contain a combination of series and parallel components. 8 Parallel Circuits • To analyze a particular circuit − First identify the node. − Next, label the nodes with a letter or number. − Then, identify types of connections. 9 Example Problem 1 Determine which elements are connected in parallel and which are connected in series. (A||B) || (C+(D||F)+E) (A+B)+ C||F+(D+E) A+ (B||C||D) A||B||C||D 10 Kirchhoff’s Current Law (KCL) • Kirchhoff’s Current Law states that the algebraic sum of the currents entering and leaving a node is equal to zero. I I 0 0 • Currents entering a node are positive and those leaving a node are negative. N I n 1 n I1 ( I 2 ) ( I 3 ) ( I 4 ) I 5 0 11 Kirchhoff’s Current Law (KCL) • KCL can also be stated as “the sum of currents entering a node is equal to the sum of currents leaving the node.” Iin I out I1 I 5 I 2 I 3 I 4 12 Parallel vs. Series Current Flow • Understanding how fluid flows may help with your understanding of KCL. • When water flows in a pipe, the amount of water entering a point is equal to the amount that leaves that point. • However, it should be noted that more water flows down the pipe with the lowest resistance. 13 Kirchhoff’s Current Law (KCL) • In technology, the term node is commonly used to refer to a junction of two or more branches. Two-node configuration Four-node configuration 14 Kirchhoff’s Current Law (KCL) • Determine I3 and I5 using KCL. KCL states: Iin I out Therefore, at node a: I1 I 2 I 3 4 A 3 A I3 7 A I3 And at node b: I3 I 4 I5 I5 I3 I 4 I 5 7 A 1A I5 6 A 15 Kirchhoff’s Current Law (KCL) • Determine Is using KCL. Parallel Circuit I s I1 I 2 I 3 I s 8mA 10mA 2mA 20mA 16 Direction of Current • Assume a current direction and draw current arrows. • If this assumption is incorrect, calculations will show that the current has a negative sign. • Negative sign simply indicates that the current flows in the opposite direction to the arrow you drew. 17 Example Problem 2 Determine the magnitude and direction of each current: Ohm's Law: I1 I1 v1 R1 5V 2mA 2.5k I 2 I1 I 3 I 3 I 2 I1 I 3 4mA 2mA I 3 2mA 18 Battery Cells Series vs Parallel Cells connected in series increases available voltage. Cells connected in parallel increases available current. Recall that voltages are additive (ET = E1+E2+…+EN) in series. However, in parallel circuits, voltages will be equal across the parallel elements. 19 Voltage Sources in Parallel • When two equal sources are connected in parallel − Each source supplies half the required current 20 Voltage Sources in Parallel Demonstrating the effect of placing two ideal supplies of the same voltage in parallel. 21 Voltage Sources in Parallel • Because the voltage is the same across parallel elements, voltage sources can be placed in parallel only if they have the same voltage. • The primary reason for placing two or more batteries or supplies in parallel is to increase the current rating above that of a single supply. 22 Voltage Sources in Parallel • Voltage sources with different potentials should never be connected in parallel. • Large currents can occur and cause damage. 23 Voltage Sources in Parallel • If, for some reason, two batteries of different voltages are placed in parallel, both will become ineffective or damaged because the battery with the larger voltage will rapidly discharge through the battery with the smaller terminal voltage. Examining the impact of placing two leadacid batteries of different terminal voltages in parallel. 24 Example Problem 3 Determine the currents I2 and I3. Knowing that voltage is equal across parallel elements we can first find v1… v1 I1 R1 v1 2 A *101 20V v2 20V v3 20V Now solve for each parallel current: I2 v2 20V 500mA R2 40 25 I3 v3 20V 200mA R3 100 Resistors in Parallel • For only two resistors connected in parallel, the equivalent resistance may be found by the product of the two values divided by the sum: R1 R2 RT R1 R2 • Often referred to as “product over the sum” formula. (98)(2) RT 1.96 98 2 26 Resistors in Parallel • For a circuit with 3 or more resistors: IT I1 I 2 I 3 E V1 V2 V3 RT R1 R2 R3 27 Resistors in Parallel • Since voltage across all parallel elements in a circuit are the same (E = V1 = V2=V3): E E E E RT R1 R2 R3 28 1 1 1 1 RT R1 R2 R3 Resistors in Parallel • Total resistance of any number of resistors in parallel: 1 RT 1 1 1 ... R1 R2 Rn 1 1 1 1 1 RT 20 180 90 60 60 29 Resistors in Parallel • Sometimes you can solve resistive problems involving parallel circuits through inspection. • Below are a few examples of how this can be done: 30 Current Divider Rule • Similar to the Voltage Divider Rule, we can use the Current Divider Rule to calculate current across a resistive element without knowing the voltage drop across that element: E = I T RT E Ix R X IT RT Ix RX 31 RT IT I x Rx Example Problem 4 Use the current divider rule to determine all unknown First, find an equivalent resistance (RT): currents: 1 RT 1 1 1 1 R1 R2 R3 R4 1 RT 1 1 1 1 4.7 k 3.3k 1.0k 2.2k RT 507 507 I1 60mA 6.5mA 4.7k IT 507 I2 60mA 9.2mA 3.3 k 507 I3 60mA 30.5mA 1.0 k 507 I4 60mA 13.8mA 2.2 k Now use CDR: RT Ix Rx Verify (KCL): IT I1 I 2 I 3 +I 4 6.5mA 9.2mA 30.5mA 13.8mA 60mA 32 Analysis of Parallel Circuits • Voltage across all parallel branches is the same as the source voltage. • Determine current through each branch using Ohm’s Law. • Find the total current using Kirchhoff’s Current Law. 33 Example Problem 5 a. Determine all unknown currents and total resistance. b. Verify KCL for node a. I I 6 5 First: understand your circuit: I1 270V I2 5.6kΩ I4 I3 3.9kΩ 4.3kΩ 2.7kΩ IS Now, find an equivalent resistance (RT): RT RT Next, find total current (IT): 1 IT = 1 1 1 1 R1 R2 R3 R4 1 1 1 1 1 4.7 k 3.3k 1.0k 2.2k E 270V = 280mA RT 963.5 Finally, use CDR and find the individual currents (Ix): E Ix RX RT 963.5 270V I1 48mA 5.6k 270V I3 62.8mA 4.3k 270V 270V I2 69.2mA I 4 2.7 k 100mA 3.9k Verify (KCL): IT I1 I 2 I 3 +I 4 48mA 69.2mA 62.8mA 100mA 280mA 34 Power Calculations • As before, to calculate the power dissipated by each resistor, use: VI, I2R, or V2/R. • Total power consumed is the sum of the individual powers: PT = P1 + P2 +…+PN 35 Example Problem 6 a) b) c) Solve for indicated currents. Determine power dissipated by each resistor. Verify total power = sum of all power dissipated. Find an equivalent resistance (RT): RT 1 1 1 1 1 1 1 1 30V R1 R2 R3 2k 3k 1.8k I1 I2 I3 IS RT 720 Next, find total current (IT): IT = Now, use CDR and find Ix: E Ix RX Finally, find Px: E 30V = 41.67 mA RT 720 30V I1 15mA 2k 30V I2 10mA 3k 30V I3 16.67 mA 1.8k 2 Px I x 2 P1 (15mA) * 2k 450mW 2 * RX P 2 (10mA) * 3k 300mW 2 P 3 (16.67 mA) *1.8k 500mW 2 2 P T I T * RT (41.7 mA) * (720) 1.25W P T P1 P2 P3 450mW 300mW 500mW 1.25W Verify (KCL): IT I1 I 2 I 3 15mA 10mA 16.67mA36 41.67mA QUESTIONS? 37