**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*1 + *v*2 + … *v**N*

and Ohm’s Law:

*v *= *R*1*i *+ *R*2*i *+ … *R**N** **i*

= (*R*1 + *R*2 + … *R**N *)*i*

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

*v *= *R**eq** . **i*

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

*R**eq *= *R*1 + *R*2 + …+ *R**N*

**Parallel Resistors**

We apply KCL:

*i *= *i*1 + *i*2 + … *i**N*

and Ohm’s Law:

*i *= *G*1*v *+ *G*2*v *+ …*G**N **v*

= (*G*1 + *G*2+ …*G**N *)*v*

whereas the equivalent circuit shown in above Figure

*i *= *G**eq** **v*

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

*G**eq *= *G*1 + *G*2 + … + *G**N *(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 *= *G*1 + *G*2 then we may deduce that:

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

Hence:

or:

*R**eq *< *R*1 and *R**eq *< *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 by their equivalent resistances. We begin by noting that the 1 W are in series. Combining them we obtain: are in series. We begin by noting that the 1 ohm 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. Since the two 4 ohms resistors are connected in parallel, we can further simplify the circuit as shown below:

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

Then, from Ohm’s Law, we have:

**i = 28/7 = 4A**