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## Combinations of Resistors

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**Combinations of Resistors**Series, Parallel and Kirchhoff**Simplifying circuits using series and parallel equivalent**resistances**Analyzing a combination of resistors circuit**• Look for resistors which are in series (the current passing through one must pass through the other) and replace them with the equivalent resistance (Req = R1 + R2). • Look for resistors which are in parallel (both the tops and bottoms are connected by wire and only wire) and replace them with the equivalent resistance (1/Req = 1/R1 + 1/R2). • Repeat as much as possible.**Look for series combinations**Req=3k Req=3.6 k**Look for parallel combinations**Req = 1.8947 k Req = 1.1244 k**Look for series combinations**Req = 6.0191 k**Look for parallel combinations**Req = 2.1314 k**Look for series combinations**Req = 5.1314 k**Equivalent Resistance**I = V/R = (5 V)/(5.1314 k) = 0.9744 mA**Backwards 1**V= (3)(.9744) = 2.9232 V= (2.1314)(.9744) = 2.0768**Backwards 2**V = 2.0768=I (3.3) I=0.629mA V = 2.0768=I (6.0191) I=0.345mA**Backwards 3**V=(.345)(3)=1.035 V=(.345)(1.8947)=0.654 V=(.345)(1.1244)=0.388**Kirchhoff’s Rules**When series and parallel combinations aren’t enough**Some circuits have resistors which are neither in series nor**parallel They can still be analyzed, but one uses Kirchhoff’s rules.**Not in series**The 1-k resistor is not in series with the 2.2-k since the some of the current that went through the 1-k might go through the 3-k instead of the 2.2-k resistor.**Not in parallel**The 1-k resistor is not in parallel with the 1.5-k since their bottoms are not connected simply by wire, instead that 3-k lies in between.**Kirchhoff’s Node Rule**• A node is a point at which wires meet. • “What goes in, must come out.” • Recall currents have directions, some currents will point into the node, some away from it. • The sum of the current(s) coming into a node must equal the sum of the current(s) leaving that node. • I1 + I2 = I3 I2 I1 I3 The node rule is about currents!**Kirchhoff’s Loop Rule 1**• “If you go around in a circle, you get back to where you started.” • If you trace through a circuit keeping track of the voltage level, it must return to its original value when you complete the circuit • Sum of voltage gains = Sum of voltage losses**Batteries (Gain or Loss)**• Whether a battery is a gain or a loss depends on the direction in which you are tracing through the circuit Loop direction Loop direction Loss Gain**Resistors (Gain or Loss)**• Whether a resistor is a gain or a loss depends on whether the trace direction and the current direction coincide or not. I I Loop direction Loop direction Current direction Current direction Loss Gain**Neither Series Nor Parallel**I1.5 I1 I3 I1.7 I2.2 Assign current variables to each branch. Draw loops such that each current element is included in at least one loop.**Apply Current (Node) Rule**I1.5 I1 * * I3 I1-I3 I1.5+I3 *Node rule applied.**Three Loops**• Voltage Gains = Voltage Losses • 5 = 1 • I1 + 2.2 • (I1 – I3) • 1 • I1 + 3 • I3 = 1.5 • I1.5 • 2.2 • (I1 – I3) = 3 • I3 + 1.7 • (I1.5 + I3) • Units: Voltages are in V, currents in mA, resistances in k**5 = 1 • I1 + 2.2 • (I1 – I3)**I1.5 I1 I3 I1-I3 I1.5+I3 **1 • I1 + 3 • I3 = 1.5 • I1.5**I1.5 I1 I3 I1-I3 I1.5+I3 **2.2 • (I1 – I3) = 3 • I3 + 1.7 • (I1.5 + I3)**I1.5 I1 I3 I1-I3 I1.5+I3 **Simplified Equations**• 5 = 3.2 • I1 - 2.2 • I3 • I1 = 1.5 • I1.5 - 3 • I3 • 0 = -2.2 • I1 + 1.7 • I1.5 + 6.9 • I3 • Substitute middle equation into others • 5 = 3.2 • (1.5 • I1.5 - 3 • I3) - 2.2 • I3 • 0 = -2.2 • (1.5 • I1.5 - 3 • I3) + 1.7 • I1.5 + 6.9 • I3 • Multiply out parentheses and combine like terms.**Solving for I3**• 5 = 4.8 • I1.5 - 11.8 • I3 • 0 = - 1.6 I1.5 + 13.5 • I3 • Solve the second equation for I1.5 and substitute that result into the first • 5 = 4.8 • (8.4375 I3 ) - 11.8 • I3 • 5 = 28.7 • I3 • I30.174 mA**Other currents**• Return to substitution results to find other currents. • I1.5 = 8.4375 I3 = 1.468 mA • I1 = 1.5 • I1.5 - 3 • I3 • I1 = 1.5 • (1.468)- 3 • (0.174) • I1 = 1.68 mA**Neither Series Nor Parallel**JB JA JC Draw loops such that each current element is included in at least one loop. Assign current variables to each loop. Current direction and lop direction are the same.**Loop equations**• 5 = 1 (JA - JB) + 2.2 (JA - JC) • 0 = 1 (JB - JA) + 1.5 JB + 3 (JB - JC) • 0 = 2.2 (JC - JA) + 3 (JC - JB) + 1.7 JC • “Distribute” the parentheses • 5 = 3.2 JA – 1 JB - 2.2 JC • 0 = -1 JA + 5.5 JB – 3 JC • 0 = -2.2JA – 3 JB + 6.9 JC**Algebra**• JC = (2.2/6.9)JA + (3/6.9)JB • JC = 0.3188 JA + 0.4348 JB • 5 = 3.2 JA – 1 JB - 2.2 (0.3188 JA + 0.4348 JB) • 0 = -1 JA + 5.5 JB – 3 (0.3188 JA + 0.4348 JB) • 5 = 2.4986 JA – 1.9566 JB • 0 = -1.9564 JA + 4.1956 JB**More algebra**• JB = (1.9564/4.1956) JA • JB = 0.4663 JA • 5 = 2.4986 JA – 1.9566 (0.4663 JA) • 5 = 1.5862 JA • JA = 3.1522 mA**Other loop currents**• JB = 0.4663 JA = 0.4663 (3.1522 mA) • JB = 1.4699 mA • JC = 0.3188 JA + 0.4348 JB • JC = 0.3188 (3.1522) + 0.4348 (1.4699) • JC = 1.644 mA**Branch Variables**I1.5 I1 I3 I1.7 I2.2 Assign current variables to each branch. Draw loops such that each current element is included in at least one loop.**Loop Variables**JB JA JC Draw loops such that each current element is included in at least one loop. Assign current variables to each loop. Current direction and lop direction are the same.**Branch Currents from Loop currents**• I1 = JA – JB = 3.1522 – 1.4699 = 1.6823 mA • I1.5 = JB = 1.4699 mA**Loop equations as matrix equation**• 5 = 3.2 JA – 1 JB - 2.2 JC • 0 = -1 JA + 5.5 JB – 3 JC • 0 = -2.2JA – 3 JB + 6.9 JC**Enter matrix in Excel, highlight a region the same size as**the matrix.**In the formula bar, enter =MINVERSE(range) where range is**the set of cells corresponding to the matrix (e.g. B1:D3). Then hit Crtl+Shift+Enter**Prepare the “voltage vector”, then highlight a range the**same size as the vector and enter =MMULT(range1,range2) where range1 is the inverse matrix and range2 is the voltage vector. Then Ctrl-Shift-Enter. Voltage vector