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Series resistors 1 R1 2 R2 3 R3 I V1 0 Figure 1: Series circuit configuration. What is the relationship between V1 and I? π1 = πΌ(π 1 + π 2 + π 3 ) (1) This is the same as if I had: 1 V1 Req 0 Figure 2: Equivalent circuit configuration. Where: π ππ = π 1 + π 2 + π 3 . (2) This is true for all series configurations. 1 R1 2 R2 3 R3 V1 0 Figure 3: Series circuit configuration β four resistors. 4 R4 I π ππ = π 1 + π 2 + π 3 + π 4 , π1 = πΌ(π 1 + π 2 + π 3 + π 4 ) (3) How can I tell if I have a series configuration? The same current flows through all the resistors. Parallel Resistors 1 I R1 VS R2 I2 I1 R3 I3 0 Figure 4: Parallel circuit configuration. This time, the voltage across the resistors is the same. Note that for all of the resistors, one of their terminals is connected to Node 1 and the other is connected to Node 0 (or ground). In this case: 1 π 1 πΌ = ππ ( + 1 π 2 + 1 ) π 3 (4) Note that this makes Req: 1 VS Req 0 Figure 5: Equivalent circuit configuration. ππ πΌ = π , so: ππ 1 π ππ 1 1 1 =π +π +π 1 2 3 Note the difference between Equation (2) and Equation (5). (5) Analyzing Series Circuits 1 V1 R1 + VR1 - R2 2 R3 3 + VR2 - I + VR3 - 0 Figure 6: Calculating circuit variables in a series circuit. Calculate all the currents and voltages in the circuit. Step 1: Draw the equivalent circuit β calculate Req. 1 V1 I Req 0 Figure 7: Using the equivalent circuit in a series circuit. π ππ = π 1 + π 2 + π 3 These circuits are equivalent because for the same voltage, they provide the same current. Step 2: Solve for current I: πΌ= π1 π ππ Step 3: Go back to original circuit, and solve for VR1, VR2, VR3. ππ 1 = πΌ β π 1 , ππ 2 = πΌ β π 2 , ππ 3 = πΌ β π 3 Analyzing Parallel Circuits 1 I R1 VS R2 I2 I1 R3 I3 0 Figure 8: Calculating circuit variables in a parallel circuit. Step 1: For parallel circuits, the equivalent circuit looks the same: 1 VS Req 0 Figure 9: Using the equivalent circuit from a parallel configuration. 1 1 1 1 = + + π ππ π 1 π 2 π 3 Step 2: Calculate I: πΌ= ππ π ππ Step 3: Calculate I1, I2, I3: πΌ1 = ππ π 1 πΌ2 = ππ π 2 πΌ3 = ππ π 3 Note that you donβt need to go through Step 2 to get to Step 3. You could complete Step 3 without Step 2 and calculate I as πΌ = πΌ1 + πΌ2 + πΌ3 . More complex equivalent resistors: V2=VS 2 I R1 VS 1 R3 V1 R2 I1 I3 0 V0=0 Figure 10: A more complex parallel/series configuration β calculating the equivalent resistor. Note that the same current I1 goes through R1 and R2. R1 and R2 are in series: π ππ1 = π 1 + π 2 I 2 V2=VS Req1 VS 0 I1 R3 I30=0 V Figure 11: Incremental calculation of equivalent resistance. Youβve βlostβ the information on V1, but you can retrieve it later. Note that this does not change the value of I, I1 or I2. Note that now, the resistors Req1 and R2 are in parallel. So: 1 1 1 = + π ππ π ππ1 π 2 1 I VS Req 0 Figure 12: Final equivalent circuit. Note that this time, you have lost the information on I1 and I2, but you will retrieve that later as well. ππ πΌ=π . Solve for I: ππ Now, go backward: ππ From Figure 11, πΌ1 = π ππ1 ππ and πΌ2 = π . 2 From Figure 10, ππ 1 = πΌ1 π 1 and ππ 2 = πΌ1 π 2 . From Figure 10 again, πΌ = πΌ1 + πΌ3 What is the equivalent resistance of the following: R1 2 R4 R2 3 R5 1 R3 V1 4