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.
(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.
(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).