# Series and Parallel Resistors

Combining Resistors

Relatively complicated resistor combinations can be replaced by a single equivalent resistor whenever we are not specifically interested in the current, voltage or power associated with any of the individual resistors.

Series Resistors

Consider the series combination of N resistors shown in (a) below:

We apply KVL:

v = v1Â + v2Â + â€¦Â vN

and Ohmâ€™s Law:

v =Â R1i +Â R2i + â€¦Â RNÂ i

=Â (R1 +Â R2Â + â€¦Â RN )i

and Â then Â compare Â this Â result Â with Â the Â simple Â equation Â applying Â to Â theÂ equivalent circuit shown in above Figure

v =Â ReqÂ .Â i

Thus, the value of the equivalent resistance for N series resistances is:

ReqÂ = R1 + R2Â + â€¦+ RN

Parallel Resistors

We apply KCL:

i = i1 + i2Â + â€¦Â iN

and Ohmâ€™s Law:

i =Â G1v +Â G2v + â€¦GN v

=Â (G1Â Â +Â G2+ â€¦GN )v

whereas the equivalent circuit shown in above Figure

i =Â GeqÂ v

and thus the value of the equivalent conductance for N parallel conductances is:

GeqÂ = G1 + G2Â + â€¦ + GN Â  Â  Â  Â  Â  Â Â (parallel)

In terms of resistance instead of conductance

The special case of only two parallel resistors is needed often:

Combining two resistors in parallel

Note that since Geq Â =Â G1 +Â G2Â then we may deduce that:

Geq Â > G1 Â and Â Geq Â > G2

Hence:

or:

Req Â < R1 Â  and Â Â Req Â < R2

Thus the equivalent resistance of two resistors in parallel is less than the value of either of the two resistors.

The special case of N resistors of equal value R in parallel is:

Example

We want to find the current i in the circuit below:

In order to find i, we can replace series and parallel connections of resistors byÂ their equivalent resistances. We begin by noting that the 1 WÂ are in series. Combining them we obtain:Â are in series.Â We begin by noting that the 1 ohmÂ are 3 ohm are in series. Combining them we obtain:

Note that it is not possible to display the original voltage v in this figure. SinceÂ the two 4 ohms resistors are connected in parallel, we can further simplify theÂ circuit as shown below:

Here, the 5 ohms and 2 ohms resistors are in series, so we may combine them intoÂ one 7 ohms resistor.

Then, from Ohmâ€™s Law, we have:

i = 28/7 = 4A