19-1 (SJP, Phys 2020) Electrical Circuits: Many real world electronic devices are just collections of wires, resistors, capacitors, and batteries, forming circuits that do something (flashlights, toasters, blowdryers, radios, amplifiers,...) It's important to understand (and predict) the currents and voltages in such circuits. E.g., consider first a simple “flashlight circuit”: The “R” here might represent the resistance of the flashlight bulb. Here, V=I*R1, or I = V*(1/R1) Now consider a slightly more complicated circuit: These resistors are in series. There will be a voltage drop V1= I*R1 across the first resistor, and V2 = I*R2 across the second. The TOTAL voltage drop from top to bottom is V = V1+V2 = I*(R1+R2) The resistances simply add up! In other words, this circuit is essentially equivalent to the following simpler circuit: Similarly, for many resistors in series: Requiv= R1+ R2+ R3+ … = Important: the current I is the SAME through each of these series resistors. (What goes in must come out: conservation of charge! This is not an approximation of any kind, it’s exactly true) However, that doesn’t mean the current I in circuit 1 is the same as the current in circuit 2. Those are different circuits.... I R1 V Circuit 1 R1 V R2 Circuit 2 Requiv = R1 + R2 V R1 R2 R3I Requiv = R1+R2+R3I19-2 (SJP, Phys 2020) Here’s a different circuit. We say R1 and R2 are “in parallel”: This time, the current I is NOT necessarily the same through R1 and R2. The current divides up: I1 goes left, I2 goes right. (Conservation of charge, however, does tell us that I = I1 + I2, can you see why that is? Charges can't be created or destroyed here - they must go SOMEWHERE, and current just "counts the charges that flow by/sec") It also says I at the bottom (going into the battery) is exactly the same as I at the top (leaving the battery) Note: the voltage across R1 is exactly the same as the voltage across R2! This is an important point, stare at the picture and try to understand why. Think of this as two different ski runs. Both have the same top and bottom (the same height, the same voltage), but they have different resistances, so different number of skiers/hour. (Different currents through each resistor) Or, you might think of water flowing through pipes: Here, the difference in pressure (like voltage difference) is exactly the same for both pipes (pressure at the top of either is identical, pressure at the bottom of either is identical, so the difference across either is identical) but the current through each will be different. The total current is just the sum of the two currents, I = I1+I2 fixed, lowpressure fixed, highpressure constricted pipe 1 pump pipe 2 I2 I1total flow, I19-3 (SJP, Phys 2020) The previous parallel circuit (#3) is essentially equivalent to the following simpler circuit: In this situation (resistors “in parallel”) I claim 1Requivalent=1R1+1R2+1R3+ … We can prove it mathematically (if you’re interested): It comes from the fact that I = V / Requiv, but conservation of charge says I = I1+ I2+ I3+ … =VR1+VR2+VR3+ …= V *1R1+1R2+1R3+ …⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = V / Requiv Examples of equivalent resistances: 1Requivalent=1100Ω+1100Ω= .02 Ω−1,i.e. Requivalent= 1 / (.02 Ω−1) = 50Ω 1Requivalent=12Ω+11Ω= 1.5 Ω−1,i.e. Requivalent= 1 / (1.5 Ω−1) = 0.67Ω 1Requivalent=12Ω+10Ω= (0.5 + ∞) Ω−1= ∞ Ω−1,i.e. Requivalent= 1 / (∞ Ω−1) = 0. Ω(The last is a short circuit, 0 resistance. All the current is happy to flow through the 0 Ω side!) Note that R_Equiv always comes out less than any of the individual parallel R's. That means, if there are two (or more) ways for the current to go, there is LESS overall resistance to flow. (More ways for current to flow makes it easier for the current to flow. More ski runs at a resort means you can get more people skiing: more current, less overall resistance.)100 Ω100 Ω = 50 Ω 2 Ω 1 Ω = 0.67 Ω 2 Ω 0 Ω => (short circuit!) = 0 Ω R1 R3R2 I Requiv V19-4 (SJP, Phys 2020) You can have both parallel and series together. To find the “single equivalent resistor”, just "break it apart" bit by bit: Drawing circuits requires some abstraction. The real situation may look different, but be equivalent. The fact that wires can have funny shapes makes things especially confusing. So, e.g. the following two circuits are exactly the same: Straight lines (even with bends) always represent ideal wires. So you need to think hard in lots of these pictures! E.g. here's another pair of circuits which are exactly the same. (I labeled some points, to help guide your eye). Study this and convince yourself that it's two different representations of the SAME CIRCUIT. If I asked for the "equivalent resistance" between a and b in the above two, the one on the left looks scary. But the one on the right makes it apparent that it's not so bad, just combine R3 and R4 (which are in series) to get this picture: Now you could combine "R3+R4" with "R2", which are manifestly in parallel (dotted circle), and add R1 to get the equivalent resistance. (It's not as hard as the 1st picture implied!) R V = V R R2 R3 R3 = R1 b R4 R4 a b R2 ad c d c R1 1 Ω 2 Ω 3 Ω 3 Ω = 0.67 Ω 3.67 Ω = R3+R4 R1 R219-5 (SJP, Phys 2020) Here's yet another example of identical circuits. ALL the pictures represent exactly the same situation! Again, study them all, and convince yourself that there's no difference! All those circuits are just 3 resistors in parallel. I personally prefer the first drawing, because it makes it visually obvious that the voltage V across all 3 resistors is exactly the same. (Like 3 parallel runs down the ski hill) By the way, this setup is pretty much how houses are wired up for appliances: here's one MORE drawing, which is pretty much equivalent to all the ones above: (V=120 V AC, the fuse shuts off current to all 3 appliances if I exceeds about 15-20 A) R2 R3 = R1 V V R1 R2 R3 V R3 R2 R1 = V R1 R219-6 (SJP, Phys 2020) Another example. Here's a more complicated circuit, let's find the currents I1, I2, and I3. And, let's find Vab, the voltage difference between points a and b. Look at R2 and R3. They're in parallel. So we can combine them into an effective single resistor. The circuit is equivalent to this one I know what R/ / is because 1R/ /=1R2+1R3=120Ω+120Ω= 0.1Ω−1,i.e. R/ /= 10Ω. But now you can
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