Electronics tutorial: equivalent parallel 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 parallel configuration and calculate the current and voltage drop using Ohm's law.

Plumber circuit

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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.

Next we add a second pipe in parallel with exactly the same characteristics. Although both are resistive, the water can now flow through any of them, which has the effect that now twice the amount of water can flow through the circuit if the pump continues working at the same pressure. Remember that the wide pipes offer no resistance and therefore are perfect conductors of water.

You might understand this in other words. With one resistive pipe the water has to flow through a section of area A during a distance equal to its length L. Resistivity to the flow of water increases as the section is reduced and the length increased. By adding a second pipe we double the area, and now the water has to be pushed through a section of area 2A and distance L, so it is only half as resistive.

Naturally we can add a third pipe so that three time more water can flow through the circuit. Note that through each individual pipe flows the exact amount of water as in the first scenario, because the pump works at the same pressure, but the total amount of water (that evidently flows through the pump) has been multiplied by 3.

Electronic circuit

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Now let us talk in electronic terms. The simple plumbing circuit represented is a 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.

If we add a second resistor in parallel it will also have a 10V voltage drop, and according to Ohm's law, also consume 0.1A. Because both resistors are draining 0.1A each, the voltage supply will have to deliver 0.2A.

We know that we have two parallel resistors, but the voltage source does not. It only knows that it is delivering 10V and 0.2A, and therefore, it thinks that it is powering a 50Ω resistor (50Ω = 10V/0.2mA). Because now electrons have two ways they can go, the circuit becomes less resistive. In this scenario the equivalent parallel resistance is equal to the half of both resistances, but that is only true because both resistors are equal and does not always hold as we will see later.

 

We can repeat the exercise now with three resistors. This time all parallel resistors behave as if they were a single 33.3Ω resistor, and the voltage supply has to deliver 0.3A

Notice that the value of the equivalent parallel resistor is not lineal with the value of each resistor, although it evidently depends on them. When explaining the plumbing example we saw that adding parallel pipes is equal to increasing the section of the first pipe and we can do something similar here. The conductance G is how permissive a resistor is to the pass of electrons, equivalent to the section of the pipe and equal to the invert of the resistance R, and is measured in Siemens [S] The equivalent parallel resistor is a resistor which has the same conductance as all conductance in parallel added together.

So, if we wanted to calculate the equivalent value of a 100Ω, a 500Ω and a 75Ω resistor in parallel, we first calculate their conductance.

R1 = 100 Ω
R2 = 500 Ω
R3 = 75 Ω

G1 = 1 / R1 = 1/100 =  0.01 S
G2 = 1 / R2 = 1/500 =  0.002 S
G3 = 1 / R2 = 1/75 =  0.0133 S

Gparallel = G1 + G2 + G3 = 0.0253 S
Rparallel = 1/Gparallel = 1/0.0253 S = 39.5Ω

Equations

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Adding resistors in parallel is equivalent to a single resistor with value equal to the invert of the sum of the inverted resistance for each resistor.Or in other words, a resistor with a conductance equal to the sum of each individual conductance. Remember that the conductance is the invert of the resistance.