How to Find Voltage Across One Resistor in a Series Pair

A \(9.0\ \text{V}\) power supply is connected to two resistors in series: \(1.5\ \text{k}\Omega\) followed by \(3.0\ \text{k}\Omega\). You need to find the voltage across the second resistor. Would it be \(4.5\ \text{V}\), simply because there are two resistors?

No. The \(3.0\ \text{k}\Omega\) resistor has twice the resistance, so it gets twice the voltage drop. Its voltage is \(6.0\ \text{V}\), leaving \(3.0\ \text{V}\) across the first resistor.

The method is to find how much of the total resistance belongs to the resistor you care about. For two fixed resistors in series, with negligible current drawn by anything connected across them:

\[ V_B=V_s\frac{R_B}{R_A+R_B} \]

Here, \(V_B\) is the voltage across resistor B, \(V_s\) is the supply voltage across the pair, and \(R_A\) and \(R_B\) are their resistances.

This is called the voltage-divider rule, but it isn’t a separate law to memorise. It follows from conservation of charge and energy. Understanding that reasoning also helps you recognise when the shortcut gives the wrong answer.

01Start with conservation, not resistance addition

Imagine following the conventional current from the positive terminal of the supply.

It passes through resistor A, reaches the connection between the resistors, passes through resistor B, and returns to the negative terminal.

There is only one path.

1. Current at the junction: charge must be conserved

Suppose \(2.0\ \text{mA}\) enters the connection between the resistors. How much current leaves towards resistor B?

The same \(2.0\ \text{mA}\).

Charge cannot continuously accumulate at that connection in a steady DC circuit. The rate of charge entering must equal the rate leaving.

This is Kirchhoff’s current law, or the junction rule.

For our series pair:

\[ I_A=I_B=I \]

Here, \(I_A\) and \(I_B\) are the currents through the two resistors, and \(I\) is their common current, measured in amperes (\(\text{A}\)).

This doesn’t mean the resistors have equal voltages. It means the same amount of charge passes through each every second.

2. Voltage around the loop: energy must be conserved

Voltage measures energy transferred per unit charge.

If the supply provides \(9.0\ \text{J}\) of energy to each coulomb of charge, the resistors must together transfer \(9.0\ \text{J}\) per coulomb to their surroundings.

So the two voltage drops must add to the supply voltage:

\[ V_s=V_A+V_B \]

This is Kirchhoff’s voltage law, or the loop rule.

Choose the direction of conventional current as positive around the loop. Travelling through the source from negative to positive is a voltage rise. Travelling through either resistor in the direction of current is a voltage drop.

The signed loop equation is therefore:

\[ +V_s-V_A-V_B=0 \]

For our opening circuit, the supply produces a \(9.0\ \text{V}\) rise, resistor A produces a \(3.0\ \text{V}\) drop, and resistor B produces a \(6.0\ \text{V}\) drop.

3. Use Ohm’s law to find the individual drop

For a resistor that obeys Ohm’s law:

\[ V=IR \]

Here, \(V\) is its potential difference in volts, \(I\) is its current in amperes, and \(R\) is its resistance in ohms (\(\Omega\)).

The current through both series resistors is identical, so:

\[ V_A=IR_A \]

\[ V_B=IR_B \]

Substitute these into the loop equation:

\[ \begin{aligned} V_s&=IR_A+IR_B\\ &=I(R_A+R_B) \end{aligned} \]

Therefore:

\[ I=\frac{V_s}{R_A+R_B} \]

Now multiply the current by the resistance of the component you want:

\[ V_B=V_s\frac{R_B}{R_A+R_B} \]

This also explains why the voltage divides in proportion to resistance. Both resistors carry the same current, so a larger resistance needs a larger potential difference to carry that current.

The relationship assumes steady DC conditions, fixed ohmic resistors, negligible connecting-wire resistance, and no significant extra current drawn from the junction. The supply voltage must be the actual voltage across the pair, not necessarily the labelled emf of a battery with internal resistance.

02Testing the prediction: can a voltmeter change the voltage?

Here’s a measurement problem that makes the voltage-divider rule more interesting.

Suppose you build a circuit with:

• A regulated \(12.0\ \text{V}\) supply.
• Resistor A: \(2.0\ \text{k}\Omega\).
• Resistor B: \(4.0\ \text{k}\Omega\).
• A voltmeter connected across resistor B.

Before connecting the meter, you predict:

\[ V_B=12.0\frac{4.0}{2.0+4.0} =8.0\ \text{V} \]

What should the voltmeter show?

You might expect exactly \(8.0\ \text{V}\). That prediction assumes the voltmeter measures the circuit without affecting it.

The original measurement design

Start with the positive supply terminal. Connect it to resistor A, then connect A to resistor B. Return B to the negative terminal.

To measure B’s potential difference, connect the voltmeter’s red lead to the junction between A and B. Connect its black lead to B’s other end, which returns to the negative terminal.

The voltmeter is connected in parallel with B, not in series.

An ideal voltmeter has infinite input resistance and draws no current. A real voltmeter has finite input resistance, although it is generally designed to be high.

For a deliberately exaggerated demonstration, suppose our first meter has an input resistance of only \(4.0\ \text{k}\Omega\). This is an illustrative instrument model, not a specification for a typical modern digital multimeter.

When the meter is connected, the current reaches the junction and splits.

Some current flows through B. Some flows through the voltmeter.

The original circuit had one path. The measurement has introduced another.

Kirchhoff’s current law now says:

\[ I_A=I_B+I_m \]

Here, \(I_m\) is the current through the meter.

The original voltage-divider calculation no longer describes the complete circuit.

Revising the measurement

The first design failed to control the current drawn by the measuring instrument.

To improve it, keep the supply and resistors unchanged, but replace the voltmeter with one having a much higher input resistance.

For example, suppose the revised meter has \(R_m=1.0\ \text{M}\Omega\), or \(1000\ \text{k}\Omega\).

It still forms a parallel branch, but this time its resistance is much larger than B’s resistance. The meter therefore draws very little current.

The expected results for the same circuit are:

These are calculated predictions from the idealised circuit model, not recorded experimental measurements.

The higher-input-resistance meter would be expected to give a reading much closer to the original \(8.00\ \text{V}\).

If a real test gives substantially different results, the next checks are the actual supply-terminal voltage, resistor values, and meter specifications. A battery with internal resistance, for example, may not maintain a constant terminal voltage as the current changes.

The experimental lesson is that a measuring instrument is part of the physical system. Agreement with the unloaded calculation improves when the measurement has less influence on the circuit.

03Which calculation should you use?

Before adding resistances, identify whether you’re dealing with the original series pair or a circuit that has been changed by a measurement or an extra connection.

For the broader treatment of series resistance, current, and power, see Series Circuits: Current, Voltage, Resistance and Power.

The key diagnostic is simple: if charge now has a second path, the original two-resistor divider equation needs revisiting.

04Questions and solutions

The questions become progressively less routine. The final question is an extension challenge: it asks what can be recovered about a circuit when measuring it changes the circuit itself.

Assume ideal connecting wires and steady DC conditions unless the question states otherwise. Treat the resistors as ohmic at each specified operating point.

Question 1

Two resistors, A and B, are connected in series across an ideal \(10.0\ \text{V}\) supply.

Their resistances are:

• \(R_A=1.0\ \text{k}\Omega\)
• \(R_B=4.0\ \text{k}\Omega\)

Calculate the current through the circuit and the voltage across each resistor. Explain why the voltage drops are unequal.

Solution 1

The circuit current is \(2.0\ \text{mA}\), with \(2.0\ \text{V}\) across A and \(8.0\ \text{V}\) across B.

Both resistors carry the same current because there is only one path.

First calculate the total resistance:

\[ R_{\text{total}}=1.0+4.0 =5.0\ \text{k}\Omega \]

Then use Ohm’s law:

\[ \begin{aligned} I&=\frac{10.0\ \text{V}} {5.0\ \text{k}\Omega}\\ &=2.0\ \text{mA} \end{aligned} \]

For resistor A:

\[ V_A=(2.0\ \text{mA})(1.0\ \text{k}\Omega) =2.0\ \text{V} \]

For resistor B:

\[ V_B=(2.0\ \text{mA})(4.0\ \text{k}\Omega) =8.0\ \text{V} \]

The sum is \(10.0\ \text{V}\), as conservation of energy requires.

The tempting mistake is to divide the voltage equally because there are two components. Equal voltage drops would require equal resistances when the components carry the same current.

Question 2

A battery has an emf of \(12.0\ \text{V}\) and an unknown internal resistance \(r\).

The internal resistance behaves as though it is in series with two external resistors:

• \(R_A=4.0\ \Omega\)
• \(R_B=6.0\ \Omega\)

A voltmeter drawing negligible current reads \(6.0\ \text{V}\) across resistor B.

1. Determine the circuit current, the battery’s internal resistance, and the voltage across the external pair.
2. Resistor A is replaced with a \(1.0\ \Omega\) resistor. The battery’s emf and internal resistance remain unchanged. Find the new voltage across B and the new battery-terminal voltage.
3. Explain how the voltage across B can increase even though the battery-terminal voltage decreases.

Solution 2

The original internal resistance is \(2.0\ \Omega\). Replacing A with \(1.0\ \Omega\) increases the voltage across B from \(6.0\ \text{V}\) to \(8.0\ \text{V}\), even though the external terminal voltage falls from \(10.0\ \text{V}\) to approximately \(9.33\ \text{V}\).

The critical distinction is between the battery’s emf and its terminal voltage.

1. Find the original current and internal resistance.

The voltmeter gives us the voltage across B directly:

\[ \begin{aligned} I&=\frac{V_B}{R_B}\\ &=\frac{6.0\ \text{V}}{6.0\ \Omega}\\ &=1.0\ \text{A} \end{aligned} \]

The current through A is also \(1.0\ \text{A}\).

Therefore:

\[ V_A=(1.0)(4.0)=4.0\ \text{V} \]

The terminal voltage across the external resistors is:

\[ V_{\text{terminal}}=4.0+6.0 =10.0\ \text{V} \]

The emf, written \(\mathcal{E}\), supplies \(12.0\ \text{V}\) around the complete circuit. The remaining \(2.0\ \text{V}\) is lost across the internal resistance.

From the loop rule:

\[ \mathcal{E}=I(r+R_A+R_B) \]

Therefore:

\[ \begin{aligned} r&=\frac{\mathcal{E}}{I}-R_A-R_B\\ &=\frac{12.0}{1.0}-4.0-6.0\\ &=2.0\ \Omega \end{aligned} \]

2. Recalculate after replacing resistor A.

The new complete-circuit resistance is:

\[ R_{\text{total}}=2.0+1.0+6.0 =9.0\ \Omega \]

The current becomes:

\[ \begin{aligned} I&=\frac{12.0\ \text{V}} {9.0\ \Omega}\\ &=1.33\ \text{A} \end{aligned} \]

Using the unrounded value \(I=4/3\ \text{A}\):

\[ V_B=IR_B =\frac{4}{3}(6.0) =8.0\ \text{V} \]

The new terminal voltage is:

\[ \begin{aligned} V_{\text{terminal}} &=I(R_A+R_B)\\ &=\frac{4}{3}(1.0+6.0)\\ &=9.33\ \text{V} \end{aligned} \]

3. Explain the apparently conflicting changes.

Reducing A’s resistance increases the current through B, so B’s voltage drop increases.

However, the higher current also increases the voltage lost across the battery’s internal resistance. That leaves less voltage available across the external pair as a whole.

The common mistake is to use the \(12.0\ \text{V}\) emf as though it were the voltage across the two external resistors. It isn’t, because the battery has internal resistance.

Question 3

An ideal \(12.0\ \text{V}\) supply is connected to two resistors in series.

Resistor A is adjustable and initially has resistance \(100\ \Omega\). Resistor B remains fixed at \(100\ \Omega\).

Initially, the voltage across A is \(6.0\ \text{V}\). After increasing A’s resistance, its measured voltage becomes \(9.6\ \text{V}\).

Determine:

1. The new resistance of A.
2. The circuit current before and after the adjustment.
3. The power dissipated by A before and after the adjustment.

A student argues that A must now dissipate more power because its voltage has increased. Evaluate that argument.

Solution 3

The new resistance of A is \(400\ \Omega\). Its voltage increases from \(6.0\) to \(9.6\ \text{V}\), but its current and power both decrease. The student’s claim is incorrect.

1. Use the loop rule to find the new voltage across B.

After the adjustment:

\[ V_B=12.0-9.6=2.4\ \text{V} \]

Since B remains at \(100\ \Omega\):

\[ \begin{aligned} I_{\text{after}} &=\frac{2.4\ \text{V}}{100\ \Omega}\\ &=0.024\ \text{A} \end{aligned} \]

So the new current is \(24\ \text{mA}\).

Both resistors have this same current.

2. Find A’s new resistance.

\[ \begin{aligned} R_A&=\frac{V_A}{I}\\ &=\frac{9.6\ \text{V}}{0.024\ \text{A}}\\ &=400\ \Omega \end{aligned} \]

3. Calculate the original current and power.

Originally, both resistances were \(100\ \Omega\).

\[ \begin{aligned} I_{\text{before}} &=\frac{12.0\ \text{V}}{200\ \Omega}\\ &=0.060\ \text{A} \end{aligned} \]

The power dissipated by A was:

\[ \begin{aligned} P_{\text{before}} &=V_AI_{\text{before}}\\ &=(6.0)(0.060)\\ &=0.360\ \text{W} \end{aligned} \]

Here, \(P\) is power, measured in watts (\(\text{W}\)).

After the adjustment:

\[ \begin{aligned} P_{\text{after}} &=V_AI_{\text{after}}\\ &=(9.6)(0.024)\\ &=0.2304\ \text{W} \end{aligned} \]

The resistor now dissipates approximately \(0.230\ \text{W}\), down from \(0.360\ \text{W}\).

4. Explain why power decreased.

The increased resistance gives A a larger share of the supply voltage, but it also increases the total series resistance.

That reduces the current through the whole circuit.

Since \(P=VI\), a higher voltage does not guarantee greater power if the current changes at the same time.

The tempting reasoning looks at A in isolation. The correct reasoning follows the entire loop and recalculates the current.

Question 4

A \(12.0\ \text{V}\) ideal supply feeds a \(2.0\ \text{k}\Omega\) resistor A and a \(4.0\ \text{k}\Omega\) resistor B in series.

Initially, the supply current is \(2.0\ \text{mA}\), and the voltage across B is \(8.0\ \text{V}\).

After a fault develops in the part of the circuit containing B, the supply current rises to \(3.0\ \text{mA}\), and the voltage across that part falls to \(6.0\ \text{V}\).

Two explanations are proposed:

• Model A: B’s resistance has fallen, but there is still only one current path.
• Model B: B remains \(4.0\ \text{k}\Omega\), but a conducting leakage path has appeared in parallel with it.

Assume the leakage path, if present, behaves as a fixed resistor.

1. Determine the resistance required by each model.
2. Can the supply-current and voltage measurements distinguish between these explanations?
3. Propose one additional electrical measurement that would distinguish them, and predict its result for each model.

Solution 4

Both models can reproduce the measurements exactly. The external readings establish an equivalent resistance of \(2.0\ \text{k}\Omega\) for the faulty region, but they do not reveal its internal arrangement.

An additional measurement of the current through B would distinguish the models.

1. Find the resistance seen by the source.

The voltage across A must be:

\[ V_A=12.0-6.0=6.0\ \text{V} \]

This agrees with the measured current:

\[ \begin{aligned} I_A&=\frac{6.0\ \text{V}} {2.0\ \text{k}\Omega}\\ &=3.0\ \text{mA} \end{aligned} \]

The faulty region has a voltage of \(6.0\ \text{V}\) and receives \(3.0\ \text{mA}\) in total.

Its effective resistance is:

\[ \begin{aligned} R_{\text{effective}} &=\frac{6.0\ \text{V}}{3.0\ \text{mA}}\\ &=2.0\ \text{k}\Omega \end{aligned} \]

2. Test Model A.

If there is still only one path, B itself must now have resistance:

\[ R_B=2.0\ \text{k}\Omega \]

All \(3.0\ \text{mA}\) passes through B.

The result matches the measurements.

3. Test Model B.

Here, B is still \(4.0\ \text{k}\Omega\), so the current through it is:

\[ \begin{aligned} I_B&=\frac{6.0\ \text{V}} {4.0\ \text{k}\Omega}\\ &=1.5\ \text{mA} \end{aligned} \]

The current entering the faulty region is \(3.0\ \text{mA}\).

Conservation of charge therefore requires:

\[ I_{\text{leak}}=3.0-1.5 =1.5\ \text{mA} \]

The leakage path has the same \(6.0\ \text{V}\) across it, so:

\[ \begin{aligned} R_{\text{leak}} &=\frac{6.0\ \text{V}} {1.5\ \text{mA}}\\ &=4.0\ \text{k}\Omega \end{aligned} \]

Two \(4.0\ \text{k}\Omega\) paths in parallel have an effective resistance of \(2.0\ \text{k}\Omega\). This model also matches the measurements.

4. Identify a distinguishing measurement.

With the circuit safely opened for testing, insert an ammeter into B’s individual branch and restore the circuit.

Predictions:

The source-current and terminal-voltage measurements alone cannot distinguish the models because they only establish how the entire faulty region behaves from the outside.

The trap is assuming that equivalent resistance reveals the physical arrangement inside a circuit.

It doesn’t. Charge conservation at an internal junction provides the information needed to distinguish the possibilities.

Question 5

An ideal, regulated \(12.0\ \text{V}\) source is connected to two unknown, fixed ohmic resistors A and B in series.

A single voltmeter has a finite input resistance of \(6.0\ \text{k}\Omega\).

The following measurements are made:

• Measurement 1: The voltmeter is connected across A and reads \(4.0\ \text{V}\).
• Measurement 2: The voltmeter is disconnected from A and connected across B. It reads \(6.0\ \text{V}\).

The meter is never connected across both resistors simultaneously. Its input resistance is unchanged between measurements, and the source maintains exactly \(12.0\ \text{V}\) across the resistor pair.

Two students disagree.

Student 1 says the results violate conservation of energy because \(4.0+6.0=10.0\ \text{V}\), not \(12.0\ \text{V}\).

Student 2 says the readings cannot even reveal the original voltage ratio because the meter changes the circuit.

1. Determine whether either student is correct.
2. Find the resistance of A and B.
3. Calculate the voltage across each resistor when the meter is completely disconnected.
4. Suppose the voltmeter’s input resistance had not been supplied. Could the two readings still reveal the unloaded voltages? Explain why or why not.

Solution 5

Neither student is completely correct.

The readings do not violate conservation of energy, and surprisingly, their ratio does reveal the ratio of the original resistor voltages.

The resistances are:

\[ R_A=2.0\ \text{k}\Omega,\qquad R_B=3.0\ \text{k}\Omega \]

With the meter disconnected, the true voltage drops are \(4.8\ \text{V}\) across A and \(7.2\ \text{V}\) across B.

The important insight is that the two readings come from different circuits. Yet the same voltmeter introduces a particular relationship between them.

1. Apply the conservation laws to Measurement 1.

When the meter is across A, the measured voltage across A is \(4.0\ \text{V}\).

The loop rule requires the voltage across B, in that same measurement, to be:

\[ V_B=12.0-4.0=8.0\ \text{V} \]

Current passes through B before splitting into two branches across A and the voltmeter.

Current conservation therefore gives:

\[ \frac{8.0}{R_B} = \frac{4.0}{R_A} + \frac{4.0}{R_m} \]

The left side is the current through B. The two right-hand terms are the separate currents through A and the meter.

2. Apply the conservation laws to Measurement 2.

Now the meter has moved across B.

It reads \(6.0\ \text{V}\), so A also has \(6.0\ \text{V}\) across it in this measurement.

The current through A splits between B and the meter:

\[ \frac{6.0}{R_A} = \frac{6.0}{R_B} + \frac{6.0}{R_m} \]

We have two independent equations describing the two different circuit configurations.

3. Solve for the unknown resistances.

Use \(R_m=6.0\ \text{k}\Omega\).

For convenience, express every resistance numerically in kilohms, so the resulting currents are in milliamperes.

Dividing the first equation by \(4.0\), and the second by \(6.0\), gives:

\[ \begin{aligned} \frac{2}{R_B} &=\frac{1}{R_A}+\frac{1}{6.0}\\ \frac{1}{R_A} &=\frac{1}{R_B}+\frac{1}{6.0} \end{aligned} \]

Substitute the second relationship into the first:

\[ \begin{aligned} \frac{2}{R_B} &=\frac{1}{R_B}+\frac{2}{6.0}\\ \frac{1}{R_B} &=\frac{1}{3.0} \end{aligned} \]

Therefore:

\[ R_B=3.0\ \text{k}\Omega \]

And:

\[ \begin{aligned} \frac{1}{R_A} &=\frac{1}{3.0}+\frac{1}{6.0}\\ &=\frac{1}{2.0} \end{aligned} \]

So:

\[ R_A=2.0\ \text{k}\Omega \]

These resistances explain both meter readings.

4. Disconnect the voltmeter and reconstruct the original circuit.

Without the meter, there is no extra branch.

The current is:

\[ \begin{aligned} I_0&=\frac{12.0\ \text{V}} {5.0\ \text{k}\Omega}\\ &=2.4\ \text{mA} \end{aligned} \]

The unloaded voltage drops are:

\[ \begin{aligned} V_{A,0} &=(2.4)(2.0)\\ &=4.8\ \text{V} \end{aligned} \]

\[ \begin{aligned} V_{B,0} &=(2.4)(3.0)\\ &=7.2\ \text{V} \end{aligned} \]

They add to the required \(12.0\ \text{V}\).

Student 1’s mistake was adding readings from different circuit states.

When the meter is across A, A has \(4.0\ \text{V}\) and B has \(8.0\ \text{V}\).

When the meter is across B, A has \(6.0\ \text{V}\) and B has \(6.0\ \text{V}\).

Each complete loop still accounts for \(12.0\ \text{V}\).

5. The surprising part: the reading ratio is still useful.

Student 2 is right that the meter changes the circuit. But that doesn’t destroy every piece of information in the readings.

Let \(M_A\) be the meter reading when it is connected across A, and \(M_B\) the reading when it is connected across B.

Applying the junction and loop rules to each configuration gives:

\[ D=R_A+R_B+\frac{R_AR_B}{R_m} \]

where \(D\) is a common resistance expression.

The two meter readings can then be written as:

\[ M_A=\frac{V_sR_A}{D} \]

\[ M_B=\frac{V_sR_B}{D} \]

Notice something unexpected.

Both expressions have exactly the same denominator.

Dividing one by the other cancels the supply voltage and the entire common denominator:

\[ \frac{M_A}{M_B} =\frac{R_A}{R_B} \]

Therefore:

\[ \frac{4.0}{6.0} =\frac{R_A}{R_B} =\frac{2}{3} \]

Although both measured voltages are smaller than their unloaded values, their ratio is unchanged.

When the meter is disconnected, the two resistors again form an unloaded series pair. Their voltage drops must have that same \(2:3\) ratio and add to \(12.0\ \text{V}\).

So even without knowing \(R_m\), we can calculate:

\[ \begin{aligned} V_{A,0} &=12.0\frac{4.0}{4.0+6.0}\\ &=4.8\ \text{V} \end{aligned} \]

And:

\[ V_{B,0}=12.0-4.8 =7.2\ \text{V} \]

This method recovers the unloaded voltages without the meter’s resistance.

However, it cannot determine the absolute values of \(R_A\) and \(R_B\) without additional information. Their ratio alone does not tell us whether they are, for example, \(2\ \text{k}\Omega\) and \(3\ \text{k}\Omega\), or \(20\ \text{k}\Omega\) and \(30\ \text{k}\Omega\).

The result relies on the conditions in the question: a fixed ideal supply voltage, unchanged ohmic resistors, one meter with the same input resistance in both measurements, and no other loads.

If the source terminal voltage changed between measurements, or the two measurements used meters with different input resistances, that common-denominator cancellation would not generally hold.

That’s the real conceptual trap. A measurement can disturb a system while still preserving some relationships within it. The challenge is to establish which relationships survive, rather than assuming either that the readings are perfect or that they are completely useless.

05What this prepares you to measure next

The voltage-divider rule works beautifully when the junction between two resistors is effectively unloaded.

But that junction might eventually be connected to a sensor input, an amplifier, or another part of a circuit. If the connected device draws appreciable current, the voltage can change just as it did when we attached the low-resistance voltmeter.

Before trusting a calculated output voltage, ask two questions: Is the source maintaining the assumed terminal voltage, and does anything connected across the chosen resistor draw current?

Those questions lead naturally to analysing loaded voltage dividers and parallel networks, where conservation at junctions becomes just as important as conservation around loops.