**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 *= *v*1Â + *v*2Â + â€¦Â *v**N*

and Ohmâ€™s Law:

*v *=Â *R*1*i *+Â *R*2*i *+ â€¦Â *R**N**Â **i*

=Â (*R*1 +Â *R*2Â + â€¦Â *R**N *)*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 *R**eq**Â .Â **i*

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

*R**eqÂ *= *R*1 + *R*2Â + â€¦+ *R**N*

**Parallel Resistors**

We apply KCL:

*i *= *i*1 + *i*2Â + â€¦Â *i**N*

and Ohmâ€™s Law:

*i *=Â *G*1*v *+Â *G*2*v *+ â€¦*G**N **v*

=Â (*G*1Â Â +Â *G*2+ â€¦*G**N *)*v*

whereas the equivalent circuit shown in above Figure

*i *=Â *G**eq**Â **v*

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

*G**eqÂ *= *G*1 + *G*2Â + â€¦ + *G**N 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 *G**eq Â *=Â *G*1 +Â *G*2Â then we may deduce that:

*G**eq Â *> *G*1 Â and Â *G**eq Â *> *G*2

Hence:

or:

*R**eq Â *< *R*1 Â and Â Â *R**eq Â *< *R*2

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**