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2.7 Voltage Division and Current Division

Introduction 🙂

In circuits with several resistors, it is often useful to know how the total voltage or total current splits among them. Voltage division and current division are simple rules that tell you how a source voltage is shared in a series connection and how a source current is shared in a parallel connection. These tools let you calculate specific voltages and currents without having to reapply the full set of circuit laws every time.

This chapter assumes you already understand what series and parallel circuits are, and that you are familiar with Ohm’s law and basic resistance calculations. Here we focus on the special patterns that appear when resistors are combined in clean series or clean parallel connections.

Voltage Division in Series Circuits 🔋

When resistors are connected in series and a voltage is applied across the whole chain, the same current flows through each resistor. Because of that shared current, the voltage across each resistor is different and depends on its resistance. This is the idea behind voltage division.

Consider a simple series circuit with a voltage source $V_{\text{in}}$ and two resistors $R_1$ and $R_2$ in series. The total resistance is
$$R_{\text{total}} = R_1 + R_2$$
The current through the series string is
$$I = \frac{V_{\text{in}}}{R_{\text{total}}} = \frac{V_{\text{in}}}{R_1 + R_2}$$

The voltage across each resistor is then
$$V_1 = I R_1,\quad V_2 = I R_2$$

Substituting $I$ from the total current expression gives the voltage division form.

Voltage division rule for series resistors
For resistors in series with total applied voltage $V_{\text{in}}$:

  • Voltage across resistor $R_k$ is
    $$V_k = V_{\text{in}} \cdot \frac{R_k}{R_1 + R_2 + \dots + R_n}$$
    Each resistor gets a share of the source voltage proportional to its resistance.

So a larger resistor in series gets a larger share of the total voltage. A smaller resistor gets a smaller share.

Example of Voltage Division

Suppose you have a 12 V source and two series resistors, $R_1 = 2\ \text{k}\Omega$ and $R_2 = 4\ \text{k}\Omega$. The total resistance is $6\ \text{k}\Omega$, so using the voltage division rule:

$$V_1 = 12\ \text{V} \cdot \frac{2\ \text{k}\Omega}{2\ \text{k}\Omega + 4\ \text{k}\Omega} = 12\ \text{V} \cdot \frac{2}{6} = 4\ \text{V}$$

$$V_2 = 12\ \text{V} \cdot \frac{4\ \text{k}\Omega}{6\ \text{k}\Omega} = 12\ \text{V} \cdot \frac{4}{6} = 8\ \text{V}$$

Notice that the larger resistor $R_2$ ends up with the larger voltage.

The same method works with any number of series resistors. As long as they all share the same current and only a single path exists, voltage division applies directly.

Viewing a Series Chain as a Divider

A series chain of resistors with a voltage source is often called a voltage divider, because it divides the source voltage into smaller voltages that can be taken from points between the resistors.

For three resistors in series, with nodes labeled from the negative terminal to the positive terminal of the source, the intermediate node voltages can be found by applying the same proportional rule to the part of the chain between the node and the reference point (usually the negative terminal). Every tap point along the series string has a voltage that is proportional to the resistance between that point and the reference.

The idea is always the same. If the total voltage across a chain is known, the voltage across any portion of the chain is that total multiplied by the fraction of the total resistance that lies in that portion.

Loading Effect on a Divider

A pure voltage divider assumes that nothing else is connected to the node where you measure the divided voltage. In practice, if you connect another circuit to the divider output, you are effectively adding another resistance in parallel with one of the divider resistors. That changes the effective resistance at that part of the chain, and the divided voltages shift.

This is known as loading. The more current the attached circuit draws, the more it disturbs the original division.

The usual way to reduce loading is to make the divider resistances small enough that the current drawn by the attached circuit is small compared with the current already flowing in the divider. In this case the attached circuit has less influence on the division ratio.

Designing Simple Voltage Dividers 🎛️

Voltage dividers are very common as simple voltage level generators. You might want to obtain a smaller voltage from a larger one, for example 3 V from a 9 V supply as a reference or signal for a high impedance input.

A basic two resistor divider has an input voltage $V_{\text{in}}$ across the series pair and an output voltage $V_{\text{out}}$ taken across the lower resistor. If $R_1$ is the resistor connected to the positive terminal and $R_2$ is the resistor connected to the negative terminal, then

$$V_{\text{out}} = V_{\text{in}} \cdot \frac{R_2}{R_1 + R_2}$$

Two resistor voltage divider formula
With $V_{\text{in}}$ across $R_1$ and $R_2$ in series, and $V_{\text{out}}$ taken across $R_2$:
$$V_{\text{out}} = V_{\text{in}} \cdot \frac{R_2}{R_1 + R_2}$$
The division ratio depends only on the resistor ratio, not on the absolute values.

This means that if you keep the ratio $R_2 / (R_1 + R_2)$ the same, you get the same output fraction of the input voltage.

The following table shows a few typical ratios for a two resistor divider.

Desired fraction $V_{\text{out}} / V_{\text{in}}$Example $R_1$Example $R_2$Check ratio
1/210 kΩ10 kΩ10 / (10+10) = 0.5
1/320 kΩ10 kΩ10 / (20+10) = 0.333
2/310 kΩ20 kΩ20 / (10+20) = 0.667
1/430 kΩ10 kΩ10 / (30+10) = 0.25

You can scale both resistor values up or down together to change how much current the divider uses without changing the division ratio, provided the loading effect from whatever you connect to $V_{\text{out}}$ is still small.

Current Division in Parallel Circuits 💧

While voltage division describes how voltage splits in a series chain, current division describes how current splits among parallel branches. In a parallel connection, the voltage across each branch is the same, so the currents through the branches depend on the branch resistances.

Consider a simple current source that forces a total current $I_{\text{in}}$ into a node, which then splits into two parallel resistors $R_1$ and $R_2$ that connect to a common return node. The same voltage appears across both resistors. Let that voltage be $V$.

The current through each resistor is given by Ohm’s law:

$$I_1 = \frac{V}{R_1}, \quad I_2 = \frac{V}{R_2}$$

The total current is the sum of the branch currents:

$$I_{\text{in}} = I_1 + I_2$$

From this, you can express the branch currents directly in terms of the total current and the resistances, which gives the current division rule.

Current division rule for two parallel resistors
For a total current $I_{\text{in}}$ entering two resistors in parallel:

  • Current through $R_1$ is
    $$I_1 = I_{\text{in}} \cdot \frac{R_2}{R_1 + R_2}$$
  • Current through $R_2$ is
    $$I_2 = I_{\text{in}} \cdot \frac{R_1}{R_1 + R_2}$$
    Each branch current is proportional to the other branch’s resistance. A lower resistance branch carries more current.

Notice how this is different from voltage division. In voltage division, the voltage on a resistor is proportional to its own resistance. In current division, the current in a branch is proportional to the resistance in the other branch because a smaller resistance draws more of the total current.

Example of Current Division

Imagine a 6 mA current source feeding two parallel resistors, $R_1 = 2\ \text{k}\Omega$ and $R_2 = 4\ \text{k}\Omega$. The total current is 6 mA. Using the current division rule:

$$I_1 = 6\ \text{mA} \cdot \frac{R_2}{R_1 + R_2}
= 6\ \text{mA} \cdot \frac{4\ \text{k}\Omega}{2\ \text{k}\Omega + 4\ \text{k}\Omega}
= 6\ \text{mA} \cdot \frac{4}{6}
= 4\ \text{mA}$$

$$I_2 = 6\ \text{mA} \cdot \frac{R_1}{R_1 + R_2}
= 6\ \text{mA} \cdot \frac{2\ \text{k}\Omega}{6\ \text{k}\Omega}
= 6\ \text{mA} \cdot \frac{2}{6}
= 2\ \text{mA}$$

The smaller resistor $R_1$ carries the larger current. This matches the basic idea that current prefers the path of least resistance.

You can check this result another way. The equivalent resistance of the parallel pair is

$$R_{\text{eq}} = \frac{R_1 R_2}{R_1 + R_2} = \frac{2\ \text{k}\Omega \cdot 4\ \text{k}\Omega}{2\ \text{k}\Omega + 4\ \text{k}\Omega} = \frac{8}{6}\ \text{k}\Omega \approx 1.333\ \text{k}\Omega$$

The voltage across both resistors is

$$V = I_{\text{in}} R_{\text{eq}} = 6\ \text{mA} \cdot 1.333\ \text{k}\Omega \approx 8\ \text{V}$$

Then $I_1 = V / R_1 = 8\ \text{V} / 2\ \text{k}\Omega = 4\ \text{mA}$ and $I_2 = 8\ \text{V} / 4\ \text{k}\Omega = 2\ \text{mA}$, which agrees with the division formulas.

Extension to More Than Two Branches

Current division can be extended to more than two parallel resistors. For three or more parallel resistors with a total current $I_{\text{in}}$, it is often easier to find the total equivalent resistance, then the voltage across the parallel network, and finally find each branch current from Ohm’s law individually.

There is also a direct general expression. If $R_1, R_2, \dots, R_n$ are in parallel with the same voltage $V$ across them and a total current $I_{\text{in}}$ entering the combination, then:

$$I_k = \frac{V}{R_k}$$

and

$$I_{\text{in}} = \sum_{k=1}^{n} I_k$$

If you want $I_k$ directly in terms of $I_{\text{in}}$ and the resistances without finding $V$ first, you can use the conductances $G_k = 1 / R_k$. The fraction of the total current through branch $k$ is $G_k$ divided by the sum of all conductances.

Comparing Voltage and Current Division 🧮

Voltage division and current division are related ideas, but they apply in different situations and follow opposite patterns when you compare resistance and the shared quantity.

The key contrasts are summarized in the table below.

FeatureVoltage Division (Series)Current Division (Parallel)
ConfigurationResistors in seriesResistors in parallel
Shared quantitySame current flows through all resistorsSame voltage across all resistors
Divided quantityVoltage across each resistorCurrent through each resistor
Relationship to resistanceHigher resistance gets higher voltageLower resistance gets higher current
Formula pattern (two elements)$V_1 = V_{\text{in}} \frac{R_1}{R_1+R_2}$$I_1 = I_{\text{in}} \frac{R_2}{R_1+R_2}$
Typical applicationVoltage level setting, reference generationCurrent sharing, analyzing parallel branches

Remember that both rules are shortcuts that come from Ohm’s law and Kirchhoff’s laws. They save time when a circuit section clearly matches a series or parallel pattern. If a circuit does not have a pure series or pure parallel structure, you may need to simplify it first or use more general analysis techniques.

Practical Use and Limitations 🧰

In practical circuits, voltage and current division are powerful tools, but some care is needed.

For voltage division, if you use very large resistor values, the divider current becomes very small. Any load you connect can then draw a current comparable to the divider current, which alters the effective division ratio. This causes the output voltage to drop from its intended value. For accurate division, the divider current should be noticeably larger than the load current.

For current division, real sources are often voltage sources rather than ideal current sources. In that case, current division is still useful, but you need to keep in mind that if the source voltage is fixed, the total current is set by the equivalent resistance of the parallel network. Changing one branch resistance changes not only how current splits but also the total current drawn from the source.

Despite these limitations, being able to recognize where voltage division and current division apply is an important skill. It allows quick estimates and simple calculations that reveal how different parts of a circuit behave relative to each other, using the proportional relationships that series and parallel connections naturally create.

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