Resistors in Series and in Parallel

Resistors are probably the most commonly occurring components in electronic circuits. Practical circuits often contain very complicated combinations of resistors. It is, therefore, useful to have a set of rules for finding the equivalent resistance of some general arrangement of resistors. It turns out that we can always find the equivalent resistance by repeated application of two simple rules. These rules relate to resistors connected in series and in parallel.

Figure 18: Two resistors connected in series.

Consider two resistors connected in series, as shown in Fig. 18. It is clear that the same current $I$ flows through both resistors. For, if this were not the case, charge would build up in one or other of the resistors, which would not correspond to a steady-state situation (thus violating the fundamental assumption of this section). Suppose that the potential drop from point $B$ to point $A$ is $V$. This drop is the sum of the potential drops $V_1$ and $V_2$ across the two resistors $R_1$ and $R_2$, respectively. Thus,

$$V = V_1 + V_2.$$ (135)

According to Ohm's law, the equivalent resistance $R_{\rm eq}$ between $B$ and $A$ is the ratio of the potential drop $V$ across these points and the current $I$ which flows between them. Thus,

$$R_{\rm eq} = \frac{V}{I} = \frac{V_1+V_2}{I} = \frac{V_1}{I} + \frac{V_2}{I},$$ (136)

giving

$$R_{\rm eq} = R_1 + R_2.$$ (137)

Here, we have made use of the fact that the current $I$ is common to all three resistors. Hence, the rule is

The equivalent resistance of two resistors connected in series is the sum of the individual resistances.

For $N$ resistors connected in series, Eq. (137) generalizes to $R_{\rm eq} = \sum_{i=1}^N R_i$.

Figure 19: Two resistors connected in parallel.

Consider two resistors connected in parallel, as shown in Fig. 19. It is clear, from the figure, that the potential drop $V$ across the two resistors is the same. In general, however, the currents $I_1$ and $I_2$ which flow through resistors $R_1$ and $R_2$, respectively, are different. According to Ohm's law, the equivalent resistance $R_{\rm eq}$ between $B$ and $A$ is the ratio of the potential drop $V$ across these points and the current $I$ which flows between them. This current must equal the sum of the currents $I_1$ and $I_2$ flowing through the two resistors, otherwise charge would build up at one or both of the junctions in the circuit. Thus,

$$I = I_1 + I_2.$$ (138)

It follows that

$$\frac{1}{R_{\rm eq}} = \frac{I}{V} = \frac{I_1+I_2}{V} = \frac{I_1}{V} +\frac{I_2}{V},$$ (139)

giving

$$\frac{1}{R_{\rm eq}} = \frac{1}{R_1} + \frac{1}{R_2}.$$ (140)

Here, we have made use of the fact that the potential drop $V$ is common to all three resistors. Clearly, the rule is

The reciprocal of the equivalent resistance of two resistances connected in parallel is the sum of the reciprocals of the individual resistances.

For $N$ resistors connected in parallel, Eq. (140) generalizes to $1/R_{\rm eq} = \sum_{i=1}^N (1/R_i)$.