How to Calculate the Voltage Across a Resistor

The voltage across a resistor is the difference in electric potential between its two terminals. To calculate it, use Ohm’s law, but first check how the resistor is connected: it may have the full battery voltage across it, or only part of the voltage supplied by the battery.

Because no circuit diagram or numerical values are specified here, there is no single numerical answer for a particular textbook example. The method below shows how to decide which voltage to calculate.

Use Ohm’s law for that resistor

For an ohmic resistor, Ohm’s law is

$$V_R=I_RR,$$

where $V_R$ is the voltage magnitude across the resistor, $I_R$ is the current through it, and $R$ is its resistance. Use the current through that particular resistor—not necessarily the total current entering the circuit.

For example, if a $2\,\text{k}\Omega$ resistor carries $3\,\text{mA}$, then

$$V_R=(3\,\text{mA})(2\,\text{k}\Omega)=6\,\text{V}.$$

The units work conveniently because $1\,\text{mA}\times1\,\text{k}\Omega=1\,\text{V}$. If the current is not given, you must find it from the circuit or use another circuit rule.

When does the resistor get the whole battery voltage?

If a resistor is the only load connected across an ideal battery, the voltage across its terminals equals the battery’s voltage. For instance, connect a $1\,\text{k}\Omega$ resistor directly across an ideal $9\,\text{V}$ battery. The resistor has $9\,\text{V}$ across it, and the current is

$$I=\frac{V}{R}=\frac{9\,\text{V}}{1\,\text{k}\Omega}=9\,\text{mA}.$$

A resistor does not need to “consume” the battery voltage in order to have current. The battery establishes a potential difference across the circuit; current through the resistor and its resistance are related by Ohm’s law. The resistor transfers electrical energy into other forms, usually heat.

Voltage in a series circuit

In a series circuit, the same current flows through every resistor. The source voltage is divided among them. For a simple circuit with an ideal source and series resistors, first calculate the total resistance and current:

$$R_{\text{total}}=R_1+R_2+\cdots, \qquad I=\frac{V_{\text{source}}}{R_{\text{total}}}.$$

Then calculate the voltage across the resistor you care about using $V_R=IR$. Equivalently, for two series resistors,

$$V_1=V_{\text{source}}\frac{R_1}{R_1+R_2}.$$

Worked example: A $12\,\text{V}$ ideal battery is connected to $2\,\text{k}\Omega$ and $4\,\text{k}\Omega$ resistors in series. Their total resistance is $6\,\text{k}\Omega$, so the current is

$$I=\frac{12\,\text{V}}{6\,\text{k}\Omega}=2\,\text{mA}.$$

The voltage across the $2\,\text{k}\Omega$ resistor is $(2\,\text{mA})(2\,\text{k}\Omega)=4\,\text{V}$. The voltage across the $4\,\text{k}\Omega$ resistor is $(2\,\text{mA})(4\,\text{k}\Omega)=8\,\text{V}$. The drops add to the battery voltage: $4\,\text{V}+8\,\text{V}=12\,\text{V}$.

Parallel connections and more complex circuits

In a parallel circuit, components connected between the same two nodes have the same voltage across them. Thus, if a resistor is connected in parallel with an ideal battery, its voltage equals the battery voltage. The current can differ between parallel branches, depending on each branch’s resistance.

For a more complicated circuit, identify the two nodes at the resistor’s terminals, then solve for their potential difference. You might need to combine resistors, apply Kirchhoff’s rules, or find the circuit current first. Kirchhoff’s voltage law says that the signed voltage changes around a closed loop sum to zero; in a simple battery-and-resistors loop, this means the resistor drops add up to the source voltage.

A real battery can have internal resistance. In that case, the voltage at its terminals while supplying current can be lower than its emf. The resistor connected to the battery’s terminals has the terminal voltage across it, not necessarily the battery’s labeled emf. Textbook problems usually say whether to treat the battery as ideal or provide its internal resistance.

Common mistakes

• Using the battery voltage across every resistor in series. The source voltage is shared among series components; calculate the current first.
• Using the wrong current in $V=IR$. Use the current through the resistor in question. In a series path it is the same through each component; in parallel branches, currents can differ.
• Confusing voltage with current. Voltage is measured between two points. Current is the rate of charge flow through a component.
• Ignoring the circuit diagram or battery model. A stated emf, terminal voltage, and resistor voltage are not automatically interchangeable when internal resistance is included.

If a worked example seems inconsistent, check whether the resistors are actually in series or parallel, whether the battery is ideal, and whether the stated current is the current through the resistor. Those details determine which calculation applies.