Consider the circuit in Fig. 2.31, where two resistors are connected in parallel and therefore have the same voltage across them. From Ohm’s law,

Figure 2.31 Two resistors in parallel

Applying KCL at node a gives the total current i as : 


Substituting Eq. (2.33) into Eq. (2.34), we get : 

where Req is the equivalent resistance of the resistors in parallel:



The equivalent resistance of two parallel resistors is equal to the product of their resistances divided by their sum.

It must be emphasized that this applies only to tw o resistors in parallel. From Eq. (2.37), if R1 = R2 then Req = R1 / R2. 
We can extend the result in Eq. (2.36) to the general case of a circuit with N resistors in parallel. The equivalent resistance is:


Note that  Req is always smaller than the resistance of the smallest resistor in the parallel combination. If R1 = R2 = ⋯=RN = R, then:


For example, if four 100- Ω resistors are connected in parallel, their equivalent resistance is 25 Ω.

It is often more convenient to use conductance rather than resistance when dealing with resistors in parallel. From Eq. (2.38), the equivalent conductance for N resistors in parallel is:


where Geq = 1/Req, G1  = 1/R1, G2  = 1/R2, G3  = 1/R3, … , GN  = 1/RN. Equation (2.40) states:

The equivalent conductance of resistors connected in parallel is the sum of their individual conductance.

This means that we may replace the circuit in Fig. 2.31 with that in Fig. 2.32. Notice the similarity between Eqs. (2.30) and (2.40). The equivalent conductance of parallel resistors is obtained the same way as the equivalent resistance of series resistors. In the same manner , the equivalent conductance of resistors in series is obtained just the same way as the resistance of resistors in parallel. Thus, the equivalent conductance Geq of N resistors in series (such as shown in Fig. 2.29) is:

Figure 2.32 Equivalent circuit to Fig. 2.31.

Given the total current i entering node a in Fig. 2.31, how do we obtain current i1 and i2 ? We know that the equivalent resistor has the same voltage, or 


Combining Eqs. (2.33) and (2.42) results in