When resistors are connected in series, a simplification of the circuit is possible. Consider the resistive circuit shown in Fig. 1(a). Since the resistors are in series, they both carry the same current. Ohm's law gives:

(1)

\[{v_1} = {R_1}i\] \[{v_2} = {R_2}i\]Applying KVL around the loop:

(2)

\[ - v + {v_1} + {v_2} = 0 \Rightarrow v = {v_1} + {v_2}\]Substituting equations (1) into equation (2) and solving for the current

*i*results in:(3)

\[i = \frac{v}{{{R_1} + {R_2}}}\]- Now consider the circuit of Fig. 1(b). Application of Ohm's law to this
circuit and solution for the current,
*i*, gives: - Comparing equation (3) with equation (4), we can see that the circuits of
Figs. 1(a) and 1(b)
*have the voltage-current characteristics at their terminals*if we select:

(4)

\[i = \frac{v}{{{R_{eq}}}}\](5)

\[{R_{eq}} = {R_1} + {R_2}\]
Figures 1(a) and 1(b) are called **equivalent circuits** if the equivalent
resistance of Fig. 1(b) is chosen as shown in equation (5). R_{eq} of
equation (5) is called the equivalent resistance of the series combination of
resistors R_{1} and R_{2}.

This result can be generalized to a series combination of N resistances as follows:

- A series combination of N resistors R
_{1}, R_{2}, ... , R_{N}can be replaced with a single equivalent resistance \({R_{eq}} = {R_1} + {R_2} + \cdots + {R_N}\). The equivalent circuit can be analyzed to determine the current through the series combination of resistors.

Replace any series resistors in the circuit below with their equivalent resistance.

All four resistors in the circuit are in series. We can add all four resistances to obtain the equivalent resistance, so the equivalent resistance is:

And the equivalent circuit is:

\[{R_{eq}} = 1k\Omega + 2.2k\Omega + 1.5k\Omega + 4.7k\Omega = 9.4k\Omega \]

Replace any series resistors in the circuit below with their equivalent resistance.

The two resistors to the far right—the 1.5 kΩ and 2.2 kΩ resistors—are not in series. However, recall that KCL can be applied over any enclosed surface. If we define a surface which encloses the 1.5 kΩ and 2.2 kΩ resistors, as shown below, we see that the 4.7k and 6.8k resistors

*are*in series and can be combined into a ingle 11.5 kΩ resistor.The 1k and 1.5k resistors to the left of the source are more obviously in series, and can be combined into a single 2.5k resistor. The final circuit is shown below.

- Where possible in the circuits below, replace series resistances with their equivalent resistance.
- Using only the fixed resistors from the Digilent
^{®}Analog Parts Kit, create resistors with resistances within 5% of the following values. (You may assume that the resistors in the Analog Parts Kit have resistances which are exactly their nominal values.)- 2.5 kΩ
- 3 kΩ
- 2 kΩ
- 12 kΩ

- Here are some possible solutions. (There are other combinations which will also meet the criteria.)
- 1 kΩ and 1.5 kΩ resistors in series
- Two, 1.5 kΩ resistors in series
- 1.5 kΩ and 470Ω resistors in series
- 4.7 kΩ, 6.8 kΩ, and 1.5 kΩ resistors in series. (Could also meet the criteria with a 10 kΩ resistor and a 2.2 kΩ resistor in series.)

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