Series and Parallel Resistors

Section 2.3 Series and Parallel Resistors

Let’s begin by applying Kirchhoff’s laws and Ohm’s law to the circuit shown in Figure 2.3.1.

At node A, we apply the KCL to find

\begin{equation} \sum\limits_i I_i=0 \Rightarrow I_1 - I_2 - I_3 = 0. \tag{2.3.1} \end{equation}

The KVL can then be applied to Loop I and Loop II. Examining Loop I first, we will start in the lower left corner and traverse the loop in the direction indicated by the arrow. Traversing this loop has us pass through battery \(V_b\) from the negative to positive terminal, resulting in \(\Delta V=+V_b\text{.}\) The next component that we encounter is \(R_1\text{,}\) which we travel across in the direction of current, so Ohm’s law tells us that the voltage falls by \(\Delta V=I_2 R_1\) in the direction of current. Likewise, the voltage falls by \(\Delta V=I_2 R_2\) as we travel across \(R_2\) in the direction of current on the way back to our starting point. The KVL for Loop 1 will thus give

\begin{equation} \sum\limits_{i\text{ in Loop I}} \left(\Delta V\right)_i = 0 \Rightarrow V_b - I_2 R_1 - I_2 R_2 = 0\tag{2.3.2} \end{equation}

where there are negative signs in front of \(I_2 R_1\) and \(I_2 R_2\) because voltage falls by these amount in the direction that we’ve traversed the loop. A similar application of the KVL to Loop II gives

\begin{equation} \sum\limits_{i\text{ in Loop II}} \left(\Delta V\right)_i = 0 \Rightarrow V_b - I_3 R_3 = 0\text{.}\tag{2.3.3} \end{equation}

Often, voltages provided by batteries (or a DC voltage source) and the resistances are known as these were the physical components assembled into the circuit. Thus, (2.3.1) - (2.3.3) represent a series of equations that must be solved to determine currents \(I_1,\ I_2, \text{ and } I_3\) in Figure 2.3.1.

Rather than solving this system of equations for the three currents, we will instead perform some intermediate analysis to investigate the behavior of resistor combinations. Starting with (2.3.2), we can see that

\begin{equation*} V_b - I_2\left(R_1 + R_2\right) = V_b - I_2 R_{1+2}\text{.} \end{equation*}

Resistors \(R_1\) and \(R_2\) are considered to be in series. Resistors are in series if the current in one resistor must be the same current in the other series resistors. In this case, \(R_1\) and \(R_2\) are in series because \(I_2\) must be the current through each of the resistors since there is no junction between them. We see that the voltage drop across the pair of series resistors is just \(I_2 R_{1+2}\) where \(R_{1+2}=R_1 + R_2\text{,}\) meaning that resistors in series act together as a single resistor that has a resistance equal to the sum of the individual series resistances. This relationship holds for any number of resistors that are in series such that

\begin{equation} R_\text{eq} = \sum\limits_\text{series} R_i \quad \text{(Resistors in series)}\text{.}\tag{2.3.4} \end{equation}

This means that currents \(I_1\text{,}\) \(I_2\text{,}\) and \(I_3\) will have identical values in all three circuits displayed in Figure 2.3.2

Figure 2.3.2. Original circuit (left) is simplified into an intermediate circuit (center) by combining series resistors \(R_1\) and \(R_2\) into effective resistance \(R_{1+2}\text{.}\) The circuit is further simplified (right) by combining parallel resistors \(R_{1+2}\) and \(R_3\) into effective resistance \(R_{(1+2)||3}\text{.}\)

If we now examine the simplified circuit in Figure 2.3.2b, we can see that both \(R_{1+2}\) and \(R_3\) both have the same voltage drop across them

\begin{align*} I_2 R_{1+2} \amp = V_b\\ I_3 R_3 \amp = V_b \end{align*}

so that

\begin{align*} I_2 \amp = \frac{V_b}{R_{1+2}} \\ I_3 \amp = \frac{V_b}{R_3} \text{.} \end{align*}

These expressions can be inserted into (2.3.1) to find

\begin{align*} I_1 \amp = I_2 + I_3 \Rightarrow \\ \amp = \frac{V_b}{R_{1+2}} + \frac{V_1}{R_3} \\ \amp = V_b \left(\frac{1}{R_{1+2}} + \frac{1}{R_3}\right) \end{align*}

so that

\begin{equation*} V_b = I_1 \left(\frac{1}{R_{1+2}} + \frac{1}{R_3}\right)^{-1}\text{.} \end{equation*}

This has the form of Ohm’s Law where \(R_{1+2}\) and \(R_3\) combine into an equivalent resistance

\begin{equation*} R_{(1+2)||3}^{-1}=R_{1+2}^{-1}+R_3^{-1}\text{.} \end{equation*}

We say that resistors \(R_{1+2}\) and \(R_3\) are in parallel because the potential difference across each resistor must be identical as each side of the resistors are connected together by wires. An equivalent resistance for a combination of any number of parallel resistors can be found using

\begin{equation} R_{eq}^{-1} = \sum\limits_\text{parallel} R_i^{-1} \quad \text{(Resistors in parallel)}\text{.}\tag{2.3.5} \end{equation}

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