Voltage Divider in an Unloaded Series Circuit

Welcome back. In the previous lesson, you used Kirchhoff’s Voltage Law to show that a series circuit’s resistor drops add to the supply voltage. A voltage divider is that same series-circuit idea turned into a quick calculation method.

For your exam, the key is to recognise the standard two-resistor layout, identify which resistor the requested voltage is across, and put that resistor in the numerator of the divider ratio. By the end of this lesson, you should be able to calculate an output voltage in an unloaded series voltage divider and check that the answer makes physical sense.

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The standard unloaded voltage divider

A voltage divider has two resistors in series across a supply. The output is usually taken from the junction between the resistors, measured relative to the lower supply terminal.

In this standard arrangement:

• $R_1$ is the top resistor.
• $R_2$ is the bottom resistor.
• $V_{\mathrm{out}}$ is measured across $R_2$.
• Both resistors carry the same circuit current, because they are in series.
• Unloaded means no device is connected across the output terminals to draw extra current.

The word unloaded matters. The usual voltage-divider formula is valid only when the output is open circuit, or when any measuring device draws such a tiny current that it can be ignored. For this lesson and for basic exam questions, “unloaded” means use the standard formula directly.

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Where the formula comes from

You do not need to re-derive the formula every time, but understanding it helps prevent the most common exam error: putting the wrong resistor in the numerator.

In the circuit shown, the total series resistance is:

$$R_{\mathrm{T}}=R_1+R_2$$

The current is therefore:

$$I=\frac{V_{\mathrm{S}}}{R_1+R_2}$$

The output voltage is the voltage drop across the lower resistor $R_2$. Using Ohm’s law:

$$V_{\mathrm{out}}=I{R}_2$$

Substitute the current expression:

$$V_{\mathrm{out}}=\left(\frac{V_{\mathrm{S}}}{R_1+R_2}\right)R_2$$

This gives the standard voltage-divider relationship:

$$\boxed{V_{\mathrm{out}}=V_{\mathrm{S}}\left(\frac{R_2}{R_1+R_2}\right)}$$

Read that formula in words:

Output voltage equals supply voltage multiplied by the resistance of the resistor across the output, divided by the total series resistance.

The fraction is a resistance ratio. Because $R_2$ is only part of the total resistance, the ratio is less than $1$. Therefore, for this standard circuit:

$$V_{\mathrm{out}}<V_{\mathrm{S}}$$

This also gives two quick checks:

• If $R_1=R_2$, the output is half the supply voltage.
• If $R_2$ is much larger than $R_1$, the output is close to the supply voltage.
• If $R_2$ is much smaller than $R_1$, the output is close to $0\mathrm{V}$.

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A worked exam-style calculation

Question: An unloaded voltage divider has:

$$V_{\mathrm{S}}=24\mathrm{V}$$ $$R_1=2\mathrm{k\Omega }$$ $$R_2=6\mathrm{k\Omega }$$

Calculate the output voltage across $R_2$.

1. Select the formula

The output is across the lower resistor $R_2$, so use:

$$V_{\mathrm{out}}=V_{\mathrm{S}}\left(\frac{R_2}{R_1+R_2}\right)$$

2. Substitute all values, including units

$$V_{\mathrm{out}}=24\mathrm{V}\left(\frac{6\mathrm{k\Omega }}{2\mathrm{k\Omega }+6\mathrm{k\Omega }}\right)$$

3. Add the resistances in the denominator

$$V_{\mathrm{out}}=24\mathrm{V}\left(\frac{6\mathrm{k\Omega }}{8\mathrm{k\Omega }}\right)$$

Because both resistances are in $\mathrm{k\Omega }$, the units cancel in the ratio:

$$V_{\mathrm{out}}=24\mathrm{V}\left(\frac{6}{8}\right)$$ $$V_{\mathrm{out}}=24\mathrm{V}\left(0.75\right)$$ $$\boxed{V_{\mathrm{out}}=18\mathrm{V}}$$

A calculator entry could be written as:

$$24\times \left(\frac{6}{2+6}\right)$$

Use brackets around the whole denominator. Without brackets, a calculator may perform the operations in an order you did not intend.

4. Check using the series method

The total resistance is:

$$R_{\mathrm{T}}=2\mathrm{k\Omega }+6\mathrm{k\Omega }=8\mathrm{k\Omega }$$

The circuit current is:

$$I=\frac{24\mathrm{V}}{8\mathrm{k\Omega }}$$ $$I=3.0\mathrm{mA}$$

Voltage across $R_2$ is:

$$V_{R2}=I{R}_2$$ $$V_{R2}=(3.0\mathrm{mA})(6\mathrm{k\Omega })$$ $$V_{R2}=18\mathrm{V}$$

The answer agrees.

The other resistor must drop:

$$V_{R1}=24\mathrm{V}-18\mathrm{V}=6\mathrm{V}$$

KVL checks the result:

$$24\mathrm{V}-6\mathrm{V}-18\mathrm{V}=0\mathrm{V}$$

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The “resistor of interest” rule

Do not memorise the formula only as “put $R_2$ on top.” That works only when the question uses the standard layout and asks for voltage across the bottom resistor.

The more reliable rule is:

The resistor across which you want the voltage goes in the numerator.

If a question asks for voltage across the top resistor $R_1$, use:

$$V_{R1}=V_{\mathrm{S}}\left(\frac{R_1}{R_1+R_2}\right)$$

For example, with a $12\mathrm{V}$ supply, $R_1=4.7\mathrm{k\Omega }$, and $R_2=3.3\mathrm{k\Omega }$:

$$V_{R1}=12\mathrm{V}\left(\frac{4.7}{4.7+3.3}\right)$$ $$V_{R1}=12\mathrm{V}\left(\frac{4.7}{8.0}\right)$$ $$V_{R1}=7.05\mathrm{V}$$

The voltage across $R_2$ is then:

$$V_{R2}=12\mathrm{V}-7.05\mathrm{V}$$ $$V_{R2}=4.95\mathrm{V}$$

Notice that $R_1$ is larger, so it has the larger voltage drop. In a series circuit, voltage divides in proportion to resistance.

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Units and calculation traps

Resistances must use the same prefix before adding

This is correct because both resistor values are in $\mathrm{k\Omega }$:

$$2\mathrm{k\Omega }+6\mathrm{k\Omega }=8\mathrm{k\Omega }$$

But this is not correct:

$$470\mathrm{\Omega }+1.0\mathrm{k\Omega }=1.47\mathrm{k\Omega }$$

You must first convert one value. For example:

$$470\mathrm{\Omega }=0.470\mathrm{k\Omega }$$

Now the values can be added safely.

Example: A $12\mathrm{V}$ supply has $R_1=470\mathrm{\Omega }$ and $R_2=1.0\mathrm{k\Omega }$. Find the output across $R_2$.

$$V_{\mathrm{out}}=12\mathrm{V}\left(\frac{1.00\mathrm{k\Omega }}{0.470\mathrm{k\Omega }+1.00\mathrm{k\Omega }}\right)$$ $$V_{\mathrm{out}}=12\mathrm{V}\left(\frac{1.00}{1.47}\right)$$ $$\boxed{V_{\mathrm{out}}=8.16\mathrm{V}}$$

The answer is reasonable: $R_2$ is the larger resistor, so it receives more than half of the $12\mathrm{V}$ supply.

Avoid these common mistakes

Mistake	Why it is wrong	Better habit
Using $R_1$ in the numerator when the output is across $R_2$	It calculates the voltage across the wrong resistor.	Circle the requested resistor before writing the formula.
Forgetting to add both resistors in the denominator	The current depends on the total series resistance.	Write the denominator brackets first: $(R_1+R_2)$.
Adding $\mathrm{\Omega }$ and $\mathrm{k\Omega }$ directly	The values represent different sized units.	Convert to matching prefixes before adding.
Giving an output larger than the supply	An unloaded two-resistor divider cannot increase voltage.	Check that $V_{\mathrm{out}}<V_{\mathrm{S}}$.
Assuming the junction is always half the supply voltage	Half supply occurs only when the two resistors are equal.	Compare the resistor ratio.

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An exam routine you can repeat

For a basic unloaded voltage-divider question, use this layout:

1. Identify the supply voltage and the two series resistance values.
2. Mark which resistor has the requested voltage across it.
3. Ensure resistance units match before adding them.
4. Write the divider relationship with the requested resistor in the numerator.
5. Substitute values with brackets around the total resistance.
6. Calculate and state the answer in volts.
7. Check reasonableness: the two voltage drops must add to the supply voltage, and the larger series resistor should have the larger drop.

For the usual lower-resistor output arrangement, your method-mark working should look like:

$$V_{\mathrm{out}}=V_{\mathrm{S}}\left(\frac{R_2}{R_1+R_2}\right)$$ $$V_{\mathrm{out}}=\text{values substituted}$$ $$\boxed{V_{\mathrm{out}}=\text{answer in volts}}$$

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Key takeaways

A voltage divider is an unloaded series circuit that produces a chosen fraction of the supply voltage across one resistor.

For output measured across the lower resistor $R_2$:

$$\boxed{V_{\mathrm{out}}=V_{\mathrm{S}}\left(\frac{R_2}{R_1+R_2}\right)}$$

Remember these three rules:

• The two resistors are in series, so they carry the same current.
• Put the resistor across the required voltage in the numerator.
• Use matching resistance units before adding values in the denominator.

This formula is a shortcut built from Ohm’s law and the KVL relationship you used previously. Next, you will move from one current path to multiple paths by calculating total resistance and branch currents in parallel circuits.