Electronics tutorial: equivalent series resistor


Resistors, as their name suggests, offer resistance to the pass of current through them. They are also the most basic electronic passive component.

You can connect several resistors one after another in series, or connect them in parallel. In this post we will study the effect of the series configuration and calculate the current and voltage drop using Ohm's law.

Plumber circuit

Let us go back to our basic plumber circuit. We have a pump that creates a constant and known pressure to move water through a very restrictive section of pipe, resulting in a constant flow rate.

If we now add a second restrictive pipe, we have doubled the resistance the pump has to work against. Because our pump works at constant pressure, we now get only half the flow rate compared with before. Note that the effect of adding a second restrictive pipe is the same as using only one restrictive pipe, but with double the length or alternatively, half the diameter, and thus two times more restrictive.

We can go one step further and add a third restrictive pipe. Again the pump works at the same pressure which can now push only a third of the amount of water compared with the first case. As you can see, the total restrictiveness of a single circuit will always be the sum of all the restrictive pipes that the water has to pass trough in a single run before it returns to the pump.

Electronic circuit

If we were to translate what we just saw into an electronic circuit, we would get the following. In the first case we had a simple circuit with a single resistance as depicted below. Let us assume that our voltage source is 10V (volts) and we have connected a 100Ω resistor. If you remember Ohm's law, if we apply 10V to a 100Ω resistor, it lets pass through 0.1A (amperes). Because the current that flows through the voltage source has to be the same than through the resistor (in this circuit electrons cannot go anywhere else), 0.1A flow through all the circuit.

Next, we added a second resistor. This time the electrons have to pass first through a 100Ω, and then through a second 100Ω resistor. But our voltage source is still 10V, thus intuitively the current will be reduced to the half, 0.05A instead of 0.1A, and this is because we could replace both resistors with an equivalent series resistor of a value exactly equal to the sum of both resistors. Applying Ohm's law to 200Ω and 10V, we get 0.05A.

Now we could also calculate the voltage drop at each resistor (how much voltage they consume, so to speak). Again applying Ohm's law but to get the voltage at a 100Ω resistor that lets 0.05A through yields 5V.

And finally, we added a third resistor, for which we can apply the same logic as above. We could replace all three resistors with a series resistor of a value equal to the sum of all three, that is 300Ω, and for 10V we get a total current of 0.033A. Again, we can calculate the voltage drop at each resistor to be 3.33V. Adding all voltage drops together we get again our 10V.


Adding resistors in series is equivalent to a single resistor with value equal to the sum of each individual resistor.

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