3.1 Equivalent Resistance

Consider the circuit of figure 3.1  which has 12 resistors and a battery, and contemplate the effort involved in analyzing this circuit. (Recall that  “analyze a circuit” usually means determining the currents and voltages associated with all circuit elements.) The circuit has 13 elements so there are 26 voltage and current variables involved. The battery voltage is given, so there are 25 unknowns.  Suppose our analysis goal was simply to determine the current, I, drawn from the battery. Repeated application of KVL, KCL, and Ohm’s law, applied with a degree of trial and error, and a good deal of patience, would ultimately lead to I. In this section, we consider  techniques for replacing resistive networks with simpler equivalent resistances.  We will show that the 12 resistors shown below can be replaced with a single resistor having value \frac{R_1}{4} + \frac{R_2}{4} + \frac{R_3}{4}\Omega, from which the current, I, can be readily obtained via Ohm’s law.

Figure 3.1 Example circuit having 25 unknown values

Resistors in series. Consider the series combination of N resistors shown in fig. 3.2. We want to determine the value of the single resistor R_{eq} that is equivalent to this series combination, in the sense that the series combination and  R_{eq} each have the same i-v relationship , v=iR, as indicated in fig. 3.3.

Figure 3.2 Set of N series-connected resistors (left). We want to determine the value of the single resistor R_{eq} (right) such that both have the same resistance


Figure 3.3 Series-connected resistances (left) and equivalent resistance (right). The two circuits are equivalent if each have the same i-v relationship

The i-v relationship of the series combination is found by applying KVL to the loop in the left-most circuit of fig. 3.3, and noting that the same current i flows through all N series resistors, R_{1}, R_{2}, R_{3}, ..., R_{N}.  Thus,

(1)   \begin{equation*}  v=iR_{1}+iR_{2}+...+iR_{N} = i(R_{1}+R_{2}+...+R_{N}) \end{equation*}

from which we find the i-v relationship,

(2)   \begin{equation*}   \frac{v}{i}=R_{1}+R_{2}+...+R_{N} \end{equation*}


For the equivalent circuit on the right side of fig. 3.3, we can write the i-v expression by immediate application of Ohm’s law, namely,

(3)   \begin{equation*}  \frac{v}{i}=R_{eq} \end{equation*}

The two i-v relationships of equations (2) and (3) are equivalent if

(4)   \begin{equation*}   R_{eq}=R_{1}+R_{2}+...+R_{N} \end{equation*}

and we see that the series combination of resistors is equivalent to a single resistor having resistance equal to the sum of the individual resistances.

The take-away point is that in any circuit, we are free to  replace any series combination of resistors with an equivalent resistor equal to the sum of the series resistances, without impacting the circuit.



Determine the equivalent resistance for a series connection of five 100\Omega resistors shown in fig. 3.4.

Solution: The single 500\Omega resistor on the right side of fig. 3.4 has a resistance equivalent to the five series-connected 100\Omega resistors.

Figure 3.4 Five 100\Omega resistors connected in series are equivalent to a single 500\Omega resistors

Resistors absorb energy and consequently they are limited in the amount of power they can handle without potentially heating up causing thermal problems in a circuit. Resistors are rated to handle a certain amount of power, and \frac{1}{8}W, \frac{1}{4}W, and  \frac{1}{2} W rated resistors are common. The resistors in the ECE361 kits are all rated at \frac{1}{4}W. Although the two circuits shown in fig. 3.4 are electrically equivalent, the circuit on the left can handle more power, and hence higher voltage and current, than the circuit on the right, as we explore in the following example: 


Determine maximum voltage for the two circuits shown fig. 3.4 if all the resistors are rated at a maximum power of \frac{1}{4}W.  

Solution: Consider the 500\Omega resistor. The maximum power and voltage are obtain via: P=\frac{V^{2}}{R} = \frac{V^{2}}{500}=\frac{1}{4}  from which we obtain V=\sqrt{\frac{500}{4}}=11.2 volts. Now consider the circuit on the left of fig. 3.4. Each of the 100\Omega resistors can drop a maximum voltage of \sqrt{\frac{100}{4}}=5V corresponding to V=5\cdot5=25V.  So the circuit comprised of 5 100\Omega resistors can operate with up to 25 volts without overheating, whereas the circuit comprised of the single 500\Omega resistor can only operate up to 11.2 volts without overheating. This should make sense when we consider that 5 resistors in series have 5 times the mass and surface area, and hence 5x the heat carrying capacity as a single resistor.  If overheating is a concern, then use of a larger power-handling capacity 500\Omega resistor would be appropriate. 




Resistors in parallel.  We now consider the resistance equivalent to a parallel connection of N resistors. Proceeding analogous to the case for series resistors, we seek to obtain the i-v relationships for the two circuits depicted in fig. 3.5 below. We begin by applying KCL at the top node of the circuit on the left, obtaining

Figure 3.5 Set of N resistors connected in parallel (left) and a single resistor having equivalent resistance (right).

(5)   \begin{equation*}   i = i_{1} + i_{2} + .. + i_{N} \end{equation*}


Since the resistors are all connected in parallel, they all have the same voltage drop across their terminals, v. Therefore, we can re-write the KCL equation using Ohm’s law for each resistor:

(6)   \begin{equation*}  i = \frac{v}{R_1} + \frac{v}{R_2} +...+ \frac{v}{R_N}\end{equation*}

from which we obtain

(7)   \begin{equation*}  v = i [ \frac{1} {\frac{1}{R_1}+\frac{1}{R_2}+...+\frac{1}{R_N} } ] \end{equation*}

For the equivalent circuit we have

(8)   \begin{equation*}  v = i R_{eq} \end{equation*}

and by equating equations (7) and (8) we have the equivalence relationship

(9)   \begin{equation*}  R_{eq} = \frac{1}{\frac{1}{R_1}+\frac{1}{R_2}+...+\frac{1}{R_N}} \end{equation*}

Two resistors in parallel. A simplified version of eq (9) is obtained for two resistors in parallel:

(10)   \begin{equation*}  R_{eq} = \frac{1}{\frac{1}{R_1}+\frac{1}{R_2}} \end{equation*}

multiplying numerator and denominator by R_{1}R_{2} we have

(11)   \begin{equation*}  R_{eq} = \frac{1}{\frac{1}{R_1}+\frac{1}{R_2}} \frac{R_{1}R_{2}}{R_{1}R_{2}} \end{equation*}

which simplifies to

(12)   \begin{equation*}  R_{eq} =\frac{R_{1}R_{2}}{R_{1}+R_{2}} \end{equation*}

for the special case where R_{1}=R_{2}=R we find that R_{eq}=\frac{R}{2}




Exercise: Determine the equivalent resistance of the circuit comprised of two parallel resistors when R_{1}=1k\Omega and (a) R_{2}=1k\Omega, (b) R_{2}=100\Omega, (c) R_{2}=10k\Omega

Figure 3.6 Set of N resistors connected in parallel (left) and a single resistor having equivalent resistance (right).


(a) since R_{1}=R_{2}=1k\Omega, we have R_{eq}=\frac{R}{2}=500\Omega.

(b) R_{eq}=\frac{R_{1}R_{2}}{R_{1}+R_{2}}=\frac{1000 \cdot 100}{1000 +  100}=90.1\Omega

(c) R_{eq}=\frac{R_{1}R_{2}}{R_{1}+R_{2}}=\frac{1000 \cdot 10,000}{1000 +10,000`} =909\Omega


Exercise: Which resistors are in series? Which are in parallel?  Determine the equivalent resistance of the network.

Figure 3.7


Solution: R_{2} and R_{3} are in parallel. The parallel combination of R_{2} and R_{3} is in series with R_{1}. The equivalent resistance of the network of resistors is R_{1}+\frac{R_{2}R_{3}}{R_{2}+R_{3}}.


Exercise:  Determine the equivalent resistance of the circuit shown if all resistances have a value of 100\Omega

Figure 3.8a


Solution: We begin by noting that R_{3} and R_{4} are in series and the series combination of these two resistors is in parallel with R_{5} and R_{2}. This can be seen more clearly when the circuit is re-drawn in fig. 3.8b.

Figure 3.8b


Figure 3.8c shows a simplified circuit when resistors R_{2}, R_{3}, R_{4}, and R_{5} are combined. A shorthand notation is used to depict R_{2} being in parallel with R_{5} and with the series combination R_{3}+R_{4}.


Figure 3.8c


Using the value of 100\Omega for each resistor, the series combination R_{3}+R_{4}=200\Omega. This resistance, in parallel with R_{5} gives \frac{100\cdot 200}{100+200}=66.7\Omega. This resistance, in parallel with R_{2} gives \frac{100\cdot 66.7}{100+66.7}=40\Omega. Finally, the series combination of 40\Omega with R_{1} and R_{6} gives an equivalent resistance R_{eq}=100+100+40 = 240\Omega

Exercise. Determine the equivalent resistance of the circuit shown if each resistor is 10\Omega.


Figure 3.9a


Solution: We first re-draw the circuit in figure 3.9b to illustrate that R_{1} and R_{2} are in parallel. Likewise, R_{3} and R_{4} are in parallel. The combined resistance is \frac{10}{2}\Omega+\frac{10}{2}\Omega = 10\Omega


Figure 3.9b



Icon for the Creative Commons Attribution 4.0 International License

Applied Electrical Engineering Fundamentals by David J. McLaughlin is licensed under a Creative Commons Attribution 4.0 International License, except where otherwise noted.

Share This Book