Series and Parallel
08/18/16
“How can different elements in a circuit be hooked up and what are the effects on current?”
When studying electronics, one might wonder, “What are the different ways that we hook up different resistors in a circuit, and how do they affect the circuit current itself?”. Well, let’s think about it.
One way we could hook up everything is to directly connect each element in series. This way, the voltage from the power source will pass through each individual part, giving an associated drop at each one. Due to the fact that they are all directly connected, each resistive element will have the same current pass through it. This makes calculating the final current easily, because we can solve symbolically as follows. Let’s say we have a circuit with 3 resistors, all of different values R1, R2, and R3. Each one of them will have the same current I. Because the voltage drop through all of them combined must be equal to the total voltage V, we can construct the algebraic equation I*R1+ I*R2 + I*R3=V. Due to a common factor of I, we can simplify this equation to be I*(R1+R2+R3)=V. We can then divide the voltage by the total resistance to find the current I = V/(R1+R2+R3). This pattern holds for any number of elements in series. Let’s do a numerical example to cement our knowledge. Let’s take R1=1 ohm, R2= 2 ohm, R3 = 3 ohm, and V = 12 volts. If we do our math right, then we should end up with I=12/(1+2+3) → I=12/6 → I = 2 amps.
Another example that we could do is to to elements, hook them up directly to the voltage source, but do not directly connect them, only have them in parallel. Let’s work out the framework for these paradigm. Since each element is directly hooked up the voltage source, not only must it provide a current to go through each element, but the voltage drop must be the same as the voltage source. So how can we find out the current? Well, it’s actually surprisingly simple. First we must notice that each of the elements obtain an individual current, corresponding to the voltage divided by the resistance, or v/r. We must then notice that the total current I will be all of the individual currents added up, I = V/R1+V/R2+V/R3+….Then, since there is a common factor on each of these elements V (as I = V/R), we can divide everything by the Voltage V, to obtain I/V=1/R1+1/R2+1/R3+… , and if we simply notice that I/V is equal to the inverse of the total resistance Req, we can then represent this equation as 1/Req=1/R1+1/R2+1/R3.. We can obtain an equivalent resistance for all of the elements in parallel, and find the total current by setting it equal to the total voltage or I=V/Req, and find our answer! As one can observe, and a parallel setup, the more elements one adds, the higher the current will be, because all of those elements will need to be supplied with the same voltage drop
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