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Basic Electronics and Robotics Training Course: Part 2 (Page 2 of 3)

Resistances in Parallel

When resistors are placed in parallel, they behave differently than they do when they are in series.  So, if you have a resistor of a certain value and you place other resistors in parallel with it, the overall resistance will decrease instead of increasing like with series.  One way to look at resistors in parallel is to consider them as conductances instead. In parallel, conductances add, just as resistors add in series. If you change all the ohmic values to siemens, you can add these figures up and convert the final answer back to ohms.

The symbol for conductance is G. R is the symbol for resistance in ohms so you use these formulas:

G = 1/R, and
R = 1/G

Consider five resistors in parallel. Call them R1 through R5, and then call the total resistance R.

So, if R1 = 100Ω, R2 = 200Ω, R3 = 300Ω, R4 = 400Ω and R5 = 500Ω.
What is the total resistance, R, of this parallel combination?

Converting the resistances to conductance values, you get
 G1 = 1/100 = 0.01 siemens
 G2 = 1/200 = 0.005 siemens
 G3 = 1/300 = 0.00333 siemens
 G4 = 1/400 = 0.0025 siemens
 G5 = 1/500 = 0.002 siemens

So adding these gives:
G = G1 + G2 + G3 + G4 + G5
G = 0.01 + 0.005 + 0.00333 + 0.0025 + 0.002 = 0.0228 siemens.

The total resistance is therefore:

R = 1/G

R = 1/0.0228 = 43.8Ω

There is a bit more math involved with finding the resistance in parallel than in series, but now you understand how resistors in parallel work and how each is found. Go ahead and do some more on your own using the figure above and change their resistance so you really understand how this works.

The Purpose of a Resistor

 Resistors can play any of numerous different roles in electrical and electronic equipment.

Here are a few of the more common ways resistors are used.

Voltage Division

When designing voltage divider networks, the resistance values should be as small as possible, without causing too much current drain on the power supply. The optimum values depend on the nature of the circuit being designed.  The reason for choosing the smallest possible resistances is that, when the divider is used with a circuit, you do not want that circuit to upset the operation of the divider. The voltage divider “fixes” the intermediate voltages best when the resistance values are as small as the current-delivering capability of the power supply will allow.

This figure below illustrates voltage division.

The individual resistors are R1, R2, R3, … Rn.

The total resistance is R = R1 + R2 + R3 +... _ Rn.

The supply voltage is E

The current in the circuit is therefore I =  E/R { Remember Ohms Law: E=IR}.

At the various points P1, P2, P3, … Pn. So at each point the current Iη will differ at each point.

Voltages will be E1, E2, E3, ..., En. The last voltage, Eη, is the same as the supply voltage, E.

All the other voltages are less than E, so E1 < E2 < E3 < ... < En = E. (The symbol < means “is less than.”)

So, let’s do some math to help you out here:

Let’s say R1 = 100, R2 = 470, R3 = 1,000, R4 = 4700

R = R1 + R2 + R3 + R4
R = 100 + 470 + 1,000 + 4,700 =  6,270 ohms

Let’s say E = 9 volts

So what is our (current) I
E = IR
I = E / R
I = 9 volts / 6270 ohms
I = 0.00143 amps or 1.43mA

Now that we have our total current used in the circuit, we can find out what our voltage is on each of the rest of the resistor points.


P1 = R1*I
P1 = 100 * 0.00143
P1 = 0.143 Volts

P2 = (R1 + R2) * I
P2 = (100 + 470) * 0.00143
P2 = .8151 Volts

P3 = (R1 + R2 + R3) * 0.00143
P3 = (100 + 470 + 1000) * 0.0143
P3 = 2.2451 Volts

P4 = (R1 + R2 + R3 + R4) * I
P4 = (100 +470 + 1000 + 4700) * 0.00143
P4 = 8.9661 Volts

Note that P4 does not = 9 volts, this is because of rounding off the numbers but if you measured it on the meter it would equal the voltage you are supplying but if we round this number off then it will equal 9 volts.

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