Why is the total resistance calculated differently when resistors are in parallel, compared to when they are in a series? For example, $$ \dfrac{1}{R_t} = \dfrac{1}{R_1} + \dfrac{1}{R_2} + \ldots + \dfrac{1}{R_n} $$

Also, if there is a set value for the amount of resistance for example a 50 ohm resistor and a 20 ohm resistor in parallel, why would this not equate to 70 ohms if the resistance that the electrons are encountering are the exact same (50 ohms and 20 ohms) as if it were a series? I get that since there are two paths, the electrons will split making it less resistance overall, i just don't quite understand how this could happen if it is set that they encounter 50 ohms and 20 ohms of resistance.

  • 1
    \$\begingroup\$ How many people per minute can you get down a corridor? How many people per minute can you get down two parallel corridors? Answer: twice as many. The resistance is halved. Now repeat for different width corridors. You'll end up with the parallel resistor formula. \$\endgroup\$
    – Transistor
    Apr 30, 2016 at 9:52
  • \$\begingroup\$ @transistor if you put those two corridors in series you still get the same number of people down per minute. \$\endgroup\$
    – Andy aka
    Apr 30, 2016 at 9:55
  • \$\begingroup\$ I know, I know. I'm trying to avoid water analogies again. At least it will take them twice as long to get through. ;^) \$\endgroup\$
    – Transistor
    Apr 30, 2016 at 10:10

4 Answers 4


If you have a circuit with a 50 V battery and a single 50 ohm resistor there would be a current of 1A flowing through that resistance in accordance with Ohm's Law (V=IR). Now add in a 20 ohm resistor in parallel with this 50 ohm resistor. The current passing through the original 50 ohm resistor will be the same due to Ohm's Law (V=IR): The voltage drop across the resistor is the same (50V) and the resistance is the same (50 ohms) so you would have the same current. However, now the electrons can also flow through the other resistor so the overall current is larger. If you now zoom out and look at the overall circuit you have the same voltage as before but a bigger total current so the total (equivalent) resistance would have to be smaller (since from V=IR for V to be the same if I gets bigger R must get smaller). And this total resistance is given by that equation.

  • 1
    \$\begingroup\$ Thanks I think I get it. Is the overall larger current because there is a larger space for a larger amount of electrons to move through per second? \$\endgroup\$ Apr 30, 2016 at 8:46

The inverse of resistance is conductance. In series circuit you add all resistance and see the total resistance.... In parallel circuit you add all conductance and see the total conductance, and then inverse it again to convert it to the equivalent resistance.... Simple.


When they are in series, all current is encountering both resistances. When they are in parallel, some current is encountering each resistance. Exactly how much current flows through each depends on the voltage across them, and when you do the math for the system you end up with the same result as the equation in the question.


Suppose we have 4 resistors (R1, R2,R3 and R4) and a battery that can supply V volts.

Our wires are perfect conductors (no resistance so no voltage drops) and our battery is a perfect battery (no internal resistance, so it can supply as much current as we need without changing its terminal voltage)

We build two circuits, one with only series connections and another with only parallel connections. What we want to find is the value of a single (equivalent) resistor that would replace all the other resistors in those circuits and have exactly the same effect.

enter image description here

(1) Series connection:

The current through each resistor (Is) must be the same.

Why? Because what goes in at one end of the resistor must come out the other otherwise the resistor would have to store or supply charge which it physically can't do.

Energy is lost (as heat) in each resistor because 'it resists the flow of charge (i.e. 'a resistor'). This loss of (potential) energy from the electrical charge is measureable as a voltage drop across each individual resistor in the circuit.

Resistors don't produce voltage or generate current so the sum of all these individual voltage drops across each resistor must equal (exactly) the battery supply voltage, V.

Each voltage drop (V1,V2 etc.) can be (and usually is) different. It is proportional to the relative size of the resistor in that circuit, equal sized resistors will produce the same voltage drop.

This gives us

        V  = V1+V2+V3+V4  and is known as Kirchoff's voltage rule (or law)

We also know that each voltage drop can also be calculated using Ohm's law (V=IR)

So V = IsR1 + IsR2 + IsR3 + IsR4

          V  =  Is (R1 + R2 + R3 + R4) 

Suppose we now substitute ONE RESISTANCE (Rs) that produced exactly the same current as the series circuit (Is).

By Ohm's law V = Is (Rs )

but we also know V = Is (R1 + R2 + R3 + R4)

Comparing the two equations we can see that

            Rs    = R1 + R2 + R3 + R4

If we then repeat the experiment with N resistors (were N is a positive non-zero integer) we get:

            Rs     = R1 + R2 + R3 + R4 + ... RN  

i.e The equivalent single resistance to replace any number of resistors connected in series is simply the SUM of all the resistances.

enter image description here

(2) Parallel connection

Starting with the same battery (V) and resistors (R1,R2,R3,R4) we connect them as a parallel circuit.

In this case the current in each branch is different BUT the voltage across each resistor is the same (= V). What we know is total current supplied by the battery (Ip) must exactly equal the sum of all the currents. (What goes in must come out). This is known as kirchoff's current rule or law.

             Ip   =   I1 + I2 + I3 + I4

By Ohm's law (V=IR) we can easily calculate each current (I = V/R)

             Ip   =   V/R1  + V/R2  + V/R3 + V/R4

             Ip   =   V ( 1/R1  + 1/R2  + 1/R3  + 1/R4)

Once again we can substitute a single resistor (Rp) that would produce exactly the same current, Ip from the battery.

             Ip   =   V/Rp

Combing the two equations we get

             V/Rp  =   V ( 1/R1  + 1/R2  + 1/R3  + 1/R4)

The voltage cancels out and we get

              1/Rp  =    1/R1  + 1/R2  + 1/R3  + 1/R4  

If we then repeat the experiment with N resistors (were N is a positive non-zero integer) we get

              1/Rp  =    1/R1  + 1/R2  + 1/R3  + 1/R4  +  ... 1/RN

i.e. The reciprocal of a single equivalent resistor which replaces any number of restances connected in parallel is the sum of all the reciprocals of of each individual resistance.

Finally - does this all make sense? (reality check)

Series connections make the equivalant resistance value larger than any individual value.

Yes. e.g. if we double the length of a piece of wire we would expect it to double its resistance. ( because resistance is directly proportional to the length of a conductor)

Parallel connections make the equivalent resistance value smaller than any individual value.

Yes. e.g. Using thicker wire by paralleling two idendical strands of thin wire produces half the resistance of a single thin strand. (because resistance is inversely proportional to cross-sectional area of conductor).


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