Combinations of Resistors Series, Parallel and Kirchhoff PHY 202 (Blum)

Simplifying circuits using series and parallel equivalent resistances PHY 202 (Blum)

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. PHY 202 (Blum)

Look for series combinations Req=3k Req=3.6 k PHY 202 (Blum)

Look for parallel combinations Req = 1.8947 k Req = 1.1244 k PHY 202 (Blum)

Look for series combinations Req = 6.0191 k PHY 202 (Blum)

Look for parallel combinations Req = 2.1314 k PHY 202 (Blum)

Look for series combinations Req = 5.1314 k PHY 202 (Blum)

Equivalent Resistance I = V/R = (5 V)/(5.1314 k) = 0.9744 mA PHY 202 (Blum)

Backwards 1 V= (3)(.9744) = 2.9232 V= (2.1314)(.9744) = 2.0768 PHY 202 (Blum)

Backwards 2 V = 2.0768=I (3.3) I=0.629mA V = 2.0768=I (6.0191) PHY 202 (Blum)

Backwards 3 V=(.345)(3)=1.035 V=(.345)(1.8947)=0.654 PHY 202 (Blum)

When series and parallel combinations aren’t enough Kirchhoff’s Rules When series and parallel combinations aren’t enough PHY 202 (Blum)

Some circuits have resistors which are neither in series nor parallel They can still be analyzed, but one uses Kirchhoff’s rules. PHY 202 (Blum)

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. PHY 202 (Blum)

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. PHY 202 (Blum)

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 I1   I2  I3 The node rule is about currents! PHY 202 (Blum)

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 PHY 202 (Blum)

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 PHY 202 (Blum)

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 PHY 202 (Blum)

Branch version PHY 202 (Blum)

Neither Series Nor Parallel Assign current variables to each branch. Draw loops such that each current element is included in at least one loop. PHY 202 (Blum)

Apply Current (Node) Rule I1.5  I1  * * I3  I1-I3  I1.5+I3  *Node rule applied. PHY 202 (Blum)

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 PHY 202 (Blum)

5 = 1 • I1 + 2.2 • (I1 – I3) I1.5  I1  I3  I1-I3  I1.5+I3  PHY 202 (Blum)

1 • I1 + 3 • I3 = 1.5 • I1.5 I1.5  I1  I3  I1-I3  I1.5+I3  PHY 202 (Blum)

2.2 • (I1 – I3) = 3 • I3 + 1.7 • (I1.5 + I3) I1.5  I1  I3  I1-I3  PHY 202 (Blum)

Simplified Equations 5 = 3.2 • I1 - 2.2 • I3 I1 = 1.5 • I1.5 - 3 • 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. PHY 202 (Blum)

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 I3  0.174 mA PHY 202 (Blum)

Comparison with Simulation PHY 202 (Blum)

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 PHY 202 (Blum)

Loop version PHY 202 (Blum)

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. PHY 202 (Blum)

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 PHY 202 (Blum)

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 PHY 202 (Blum)

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 PHY 202 (Blum)

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 PHY 202 (Blum)

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. PHY 202 (Blum)

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. PHY 202 (Blum)

Branch Currents from Loop currents I1 = JA – JB = 3.1522 – 1.4699 = 1.6823 mA I1.5 = JB = 1.4699 mA PHY 202 (Blum)

Matrix equation PHY 202 (Blum)

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 PHY 202 (Blum)

Enter matrix in Excel, highlight a region the same size as the matrix. PHY 202 (Blum)

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 PHY 202 (Blum)

Result of matrix inversion PHY 202 (Blum)

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 PHY 202 (Blum)

Results of Matrix Multiplication PHY 202 (Blum)