Why Two Identical Parallel Resistors Give Half the Resistance

A single \(100\ \Omega\) resistor connected across an ideal \(6.0\ \mathrm{V}\) supply draws \(0.060\ \mathrm{A}\). Now connect a second identical resistor in parallel with it. What happens to the total current?

It doubles to \(0.120\ \mathrm{A}\). The equivalent resistance of the two resistors is therefore \(50\ \Omega\), exactly half the resistance of either individual resistor.

But why? The resistors have not changed. Each is still \(100\ \Omega\). The explanation comes from two conservation laws: conservation of charge at junctions and conservation of energy around complete circuit loops.

Once you understand those two laws, the half-resistance result follows directly.

01The measurement problem: what does half the resistance actually mean?

First, be clear about what we are measuring.

The equivalent resistance of a network is the resistance of a single hypothetical resistor that would draw the same total current as the entire network at the same terminal voltage.

We calculate it using:

\[ R_{\mathrm{eq}}=\frac{V}{I_{\mathrm{total}}} \]

Here:

• \(R_{\mathrm{eq}}\) is the equivalent resistance, measured in ohms (\(\Omega\)).
• \(V\) is the potential difference across the entire resistor network, measured in volts (\(\mathrm{V}\)).
• \(I_{\mathrm{total}}\) is the total current supplied to the network, measured in amperes (\(\mathrm{A}\)).

Notice that we use the total current, not the current through just one resistor.

How could we test the prediction?

Imagine setting up an investigation using:

• A regulated \(6.00\ \mathrm{V}\) DC power supply.
• Two identical \(100\ \Omega\) resistors.
• A switch.
• An ammeter.
• A voltmeter.
• Connecting wires.

Connect the positive terminal of the power supply to the switch, then to an ammeter in the main circuit wire. The wire reaches junction A, where it divides into two branches.

Each branch contains one \(100\ \Omega\) resistor. The branches reconnect at junction B, and a return wire connects B to the negative supply terminal.

Connect the voltmeter across A and B. It measures the potential difference across the entire parallel network.

The ammeter in the main wire measures the total current entering the junction.

Start with just one resistor connected between A and B.

Using Ohm’s law:

\[ I=\frac{V}{R} \]

\[ I=\frac{6.00}{100} =0.0600\ \mathrm{A} \]

The resistance calculated from the supply voltage and measured current is:

\[ R=\frac{6.00}{0.0600} =100\ \Omega \]

Now switch off the supply and add the second resistor as another branch between A and B.

Before switching on again, predict what the main ammeter should read.

The ideal circuit model predicts that the total current will double.

Illustrative results

The following values are theoretical predictions for ideal components, not claimed experimental measurements.

The important observation is that the voltage stays the same, while the total current doubles.

The equivalent resistance consequently halves.

A limitation of the first experimental design

There is something the initial apparatus cannot establish.

The main ammeter measures the total current, but it cannot tell us how much current flows through each individual branch.

A reading of \(0.120\ \mathrm{A}\) could represent several possible combinations:

\[ 0.120=0.060+0.060 \]

or

\[ 0.120=0.040+0.080 \]

Both satisfy conservation of current.

To test whether the two identical resistors actually carry equal currents, revise the apparatus. Move the ammeter into the first branch and record its current. Then move it into the second branch, keeping the supply voltage and resistor arrangement unchanged.

Alternatively, use two ammeters, one in each branch.

The revised investigation tests two separate predictions:

1. The branch currents are equal because the resistors are identical and have the same voltage across them.
2. The total current equals the sum of the branch currents.

We can now explain why both predictions follow from the circuit laws.

02Use conservation before reducing the resistors

Before applying a parallel-resistance shortcut, establish what happens at the junctions and around the loops.

The two relevant laws are Kirchhoff’s current law and Kirchhoff’s voltage law.

1. Kirchhoff’s current law: what happens at a junction?

Imagine \(0.060\ \mathrm{A}\) entering junction A.

The current divides between the two resistor branches.

Does charge disappear when it reaches the junction? No. In a steady-state circuit, charge cannot continuously accumulate there.

The total current entering a junction must equal the total current leaving it.

This is Kirchhoff’s current law (KCL):

\[ \sum I_{\mathrm{in}}=\sum I_{\mathrm{out}} \]

For our two-branch circuit:

\[ I_{\mathrm{total}}=I_1+I_2 \]

Here:

• \(I_{\mathrm{total}}\) is the current entering junction A.
• \(I_1\) is the current through the first resistor.
• \(I_2\) is the current through the second resistor.

The currents recombine at junction B.

This equation tells us how the currents relate, but it does not yet tell us whether they are equal.

2. Kirchhoff’s voltage law: why is the voltage equal?

Now follow a complete loop.

A charge travels from the negative terminal of the supply, through the battery, along one branch, and back to its starting point.

The battery provides an increase in electrical potential. The resistor produces a decrease.

Since the charge returns to its starting point, its net change in electric potential must be zero.

This is Kirchhoff’s voltage law (KVL):

\[ \sum \Delta V=0 \]

Use the sign convention that a potential rise is positive and a potential drop is negative.

For the loop containing the first resistor:

\[ +V_s-V_1=0 \]

Therefore:

\[ V_1=V_s \]

For the loop containing the second resistor:

\[ +V_s-V_2=0 \]

Therefore:

\[ V_2=V_s \]

Combining these results:

\[ V_1=V_2=V_s \]

Both resistors experience the same potential difference.

Why? They connect to the same two circuit nodes. The ideal connecting wires have no potential drop, so each resistor starts and finishes at the same electrical potentials.

If you are unsure whether two components really connect between the same two nodes, first trace the wires rather than judging their positions on the page.

This is an important skill in more complicated HSC circuit questions.

03Why two identical parallel resistors give half the resistance

We now have everything needed.

Two identical resistors in parallel have:

• The same voltage across them, from KVL.
• A total current equal to the sum of their branch currents, from KCL.
• The same resistance, because they are identical and operate at the same temperature.

Let’s combine these ideas.

Call the resistance of each individual resistor \(R\).

Since both resistors have the same voltage \(V\), Ohm’s law gives:

For the first resistor:

\[ I_1=\frac{V}{R} \]

For the second resistor:

\[ I_2=\frac{V}{R} \]

Apply Kirchhoff’s current law:

\[ I_{\mathrm{total}}=I_1+I_2 \]

Substitute the two branch currents:

\[ I_{\mathrm{total}} =\frac{V}{R}+\frac{V}{R} \]

Therefore:

\[ I_{\mathrm{total}}=\frac{2V}{R} \]

Now calculate the equivalent resistance of the entire network:

\[ R_{\mathrm{eq}}=\frac{V}{I_{\mathrm{total}}} \]

Substitute the expression for total current:

\[ R_{\mathrm{eq}}=\frac{V}{2V/R} \]

Simplifying:

\[ \boxed{R_{\mathrm{eq}}=\frac{R}{2}} \]

That’s the result.

Two identical resistors connected in parallel have an equivalent resistance equal to half the resistance of either individual resistor.

What actually causes the resistance to halve?

The crucial point is that the voltage across each resistor stays the same.

Each resistor therefore draws exactly the same current it would draw if it were connected by itself across that voltage.

Adding the second branch doubles the total current supplied by the source.

Since equivalent resistance is the ratio of voltage to total current, doubling the current at constant voltage halves the equivalent resistance.

The resistors themselves have not changed.

A useful way to picture it

Imagine two identical water hoses connected between the same high-pressure and low-pressure points.

Each hose allows a certain flow rate for that pressure difference. Adding the second hose allows the same flow through another path, so the total flow doubles.

In this analogy:

• Pressure difference represents voltage.
• Flow rate represents current.
• Restriction to flow represents resistance.

The analogy is not an exact description of electric charge. It simply helps explain why adding another identical path increases the total flow at the same driving difference.

Why the current does not get halved in each branch

This is a particularly common misconception.

You might reason:

There are now two branches, so the original current must split in half. Each resistor should receive half the original current.

That would be reasonable if the total current were fixed independently of the circuit.

But our supply is an ideal constant-voltage source.

Each branch still has the full supply voltage across it. Ohm’s law therefore gives the same current through each resistor as before.

The total current increases because the source now supplies both branch currents.

For our \(100\ \Omega\) example:

The branch current stays the same. The total current doubles.

04A useful contrast: the same resistors in series

Suppose we take the same two \(100\ \Omega\) resistors and connect them in series instead.

What changes?

In a series circuit, the same current passes through both resistors. The potential differences across the resistors add to the supply voltage.

For an ideal \(6.0\ \mathrm{V}\) supply:

\[ 6.0=V_1+V_2 \]

Because the resistors are identical, the voltage divides equally:

\[ V_1=V_2=3.0\ \mathrm{V} \]

The current is:

\[ I=\frac{3.0}{100} =0.030\ \mathrm{A} \]

The equivalent resistance is:

\[ R_{\mathrm{eq}}=\frac{6.0}{0.030} =200\ \Omega \]

The two resistors now have twice the resistance of either individual resistor.

The difference comes down to what is conserved in each arrangement.

This is why you should establish the circuit connections and apply KCL and KVL before deciding which resistance rule to use.

For a broader explanation of branch voltage, current, resistance, and power, see Parallel Circuits: Voltage, Current, Resistance and Power.

05Worked examples

06When does the half-resistance rule need qualification?

The result \(R_{\mathrm{eq}}=R/2\) depends on a few assumptions.

The resistors must have equal resistance.

If one resistor is \(100\ \Omega\) and the other is \(200\ \Omega\), the equivalent resistance is not half of either.

Instead:

\[ \frac{1}{R_{\mathrm{eq}}} =\frac{1}{100}+\frac{1}{200} \]

\[ R_{\mathrm{eq}}=66.7\ \Omega \]

The resistors must connect between the same two nodes.

Two resistors drawn beside each other are not automatically parallel. Their actual electrical connections determine the arrangement.

The resistors must retain their resistance.

The shortcut assumes ohmic resistors with constant resistance at their operating temperature. Significant heating or temperature changes can alter the resistance.

The ideal voltage supply assumption matters when discussing total current.

The equivalent resistance of two equal external resistors remains \(R/2\). However, a real battery has internal resistance.

When the total current increases, the voltage lost inside the battery can increase. Its terminal voltage may consequently fall.

This means the source current does not necessarily double, even though the external resistor network has half the resistance.

We will examine this distinction in the final question.

07Questions and solutions

Question 1

Two identical \(150\ \Omega\) resistors are connected in parallel across an ideal \(9.0\ \mathrm{V}\) supply.

Calculate:

a) The equivalent resistance.

b) The current through each resistor.

c) The total current supplied by the battery.

Assume constant resistor temperatures.

Solution 1

The equivalent resistance is \(75\ \Omega\), each resistor carries \(0.060\ \mathrm{A}\), and the total supply current is \(0.120\ \mathrm{A}\).

Step 1: Establish the voltage across each resistor.

Both resistors connect between the same two nodes.

Using KVL:

\[ V_1=V_2=9.0\ \mathrm{V} \]

Step 2: Calculate each branch current.

Using Ohm’s law:

\[ I_1=\frac{9.0}{150} =0.060\ \mathrm{A} \]

Since the resistors are identical:

\[ I_2=0.060\ \mathrm{A} \]

Step 3: Apply KCL.

\[ I_{\mathrm{total}}=I_1+I_2 \]

\[ I_{\mathrm{total}}=0.060+0.060 \]

\[ I_{\mathrm{total}}=0.120\ \mathrm{A} \]

Step 4: Calculate equivalent resistance.

\[ R_{\mathrm{eq}}= \frac{9.0}{0.120} \]

\[ R_{\mathrm{eq}}=75\ \Omega \]

The answer makes sense because \(75\ \Omega\) is half of \(150\ \Omega\).

A common mistake is to add the resistances and obtain \(300\ \Omega\). That calculation would apply to a series arrangement, where the same current passes through both resistors.

Question 2

An ideal \(15.0\ \mathrm{V}\) supply is connected across two parallel branches.

• Branch 1 contains a \(30\ \Omega\) resistor.
• Branch 2 contains a \(10\ \Omega\) resistor in series with an unknown resistor \(X\).

The total current supplied by the battery is \(0.750\ \mathrm{A}\).

Calculate:

a) The current through each branch.

b) The resistance of \(X\).

Assume all resistors are ohmic.

Solution 2

The current through the \(30\ \Omega\) branch is \(0.500\ \mathrm{A}\), the current through the other branch is \(0.250\ \mathrm{A}\), and the unknown resistance is \(50\ \Omega\).

The key is that both branches have the same voltage, but they do not have the same resistance. Their currents will therefore be different.

Step 1: Calculate the current through the \(30\ \Omega\) branch.

Both branches connect across the \(15.0\ \mathrm{V}\) supply.

Using KVL and Ohm’s law:

\[ I_1=\frac{15.0}{30} \]

\[ I_1=0.500\ \mathrm{A} \]

Step 2: Apply KCL at the junction.

\[ I_{\mathrm{total}}=I_1+I_2 \]

Substitute the known values:

\[ 0.750=0.500+I_2 \]

Therefore:

\[ I_2=0.250\ \mathrm{A} \]

Step 3: Apply KVL around the second branch.

The voltage drops across the \(10\ \Omega\) resistor and resistor \(X\) must add to \(15.0\ \mathrm{V}\).

\[ 15.0=I_2(10)+I_2X \]

Substitute \(I_2=0.250\ \mathrm{A}\):

\[ 15.0=0.250(10)+0.250X \]

\[ 15.0=2.50+0.250X \]

Rearrange:

\[ 12.5=0.250X \]

\[ X=50\ \Omega \]

Step 4: Check the result.

The resistance of the second branch is:

\[ R_2=10+50=60\ \Omega \]

Its current should therefore be:

\[ I_2=\frac{15.0}{60} =0.250\ \mathrm{A} \]

This agrees with our previous result.

The tempting mistake is to divide the total current equally between two branches. That only works when the branches have equal resistance. Here, the \(30\ \Omega\) branch carries twice the current of the \(60\ \Omega\) branch.

Question 3

An investigation uses two nominally identical \(82\ \Omega\) resistors, each with a stated tolerance of \(\pm5\%\).

The following readings are obtained.

a) Calculate the resistance of the first resistor using the initial readings.

b) Calculate the equivalent resistance of the two-resistor network using the second set of readings.

c) Compare the measured equivalent resistance with half the initial resistance. Does the investigation support the theoretical relationship?

d) Explain why the main ammeter alone cannot prove that the two branch currents are equal. Suggest an improvement to the investigation.

Solution 3

The first resistor has a measured resistance of approximately \(82.0\ \Omega\), and the measured equivalent resistance of the parallel combination is \(41.5\ \Omega\). This is consistent with the predicted half-resistance relationship.

Step 1: Calculate the resistance of the first resistor.

Using Ohm’s law:

\[ R_1=\frac{V}{I} \]

\[ R_1=\frac{5.00}{0.0610} \]

\[ R_1=81.97\ \Omega \]

To three significant figures:

\[ R_1=82.0\ \Omega \]

Step 2: Calculate the equivalent resistance of the parallel combination.

Use the actual measured voltage across the network, rather than assuming it remains exactly \(5.00\ \mathrm{V}\).

\[ R_{\mathrm{eq}}=\frac{4.98}{0.120} \]

\[ R_{\mathrm{eq}}=41.5\ \Omega \]

Step 3: Compare this with the theoretical prediction.

If the resistors were exactly identical:

\[ R_{\mathrm{eq}}=\frac{81.97}{2} \]

\[ R_{\mathrm{eq}}=40.98\ \Omega \]

The difference between the predicted and measured values is:

\[ 41.5-40.98=0.52\ \Omega \]

As a percentage of the predicted value:

\[ \frac{0.52}{40.98}\times100 \approx1.3\% \]

The measured result is approximately \(1.3\%\) above the prediction.

This is consistent with the theoretical relationship, considering the stated component tolerances and possible measurement uncertainties.

However, it does not establish that the resistors have exactly equal resistance.

Step 4: Explain the limitation of the ammeter measurement.

The main ammeter measures:

\[ I_{\mathrm{total}}=I_1+I_2 \]

It does not independently measure \(I_1\) and \(I_2\).

To improve the investigation, place an ammeter in each branch, or move one ammeter between the two branches while keeping the supply voltage constant.

The ideal model predicts approximately:

\[ I_1=I_2=\frac{4.98}{82} \]

\[ I_1=I_2\approx0.0607\ \mathrm{A} \]

The two branch currents should therefore add to approximately \(0.121\ \mathrm{A}\).

The measured total of \(0.120\ \mathrm{A}\) is close to this prediction.

The important experimental distinction is that agreement with a predicted total current supports the model, while separate branch measurements provide a more direct test of equal current division.

Question 4

Two identical \(80\ \Omega\) resistors are connected in parallel across an ideal \(12.0\ \mathrm{V}\) supply.

A switch is connected in series with the second resistor. Initially, the switch is closed. It is then opened, disconnecting that branch.

Calculate and explain:

a) The equivalent resistance before and after opening the switch.

b) The total current supplied by the battery before and after opening the switch.

c) The current through the remaining resistor after the switch is opened.

d) The total power transferred to the resistors before and after opening the switch.

Assume the supply voltage and resistor temperatures remain constant.

Solution 4

Opening the switch doubles the equivalent resistance from \(40\ \Omega\) to \(80\ \Omega\), and halves the total supply current from \(0.300\ \mathrm{A}\) to \(0.150\ \mathrm{A}\). The current through the remaining resistor stays at \(0.150\ \mathrm{A}\).

The reason is that the ideal supply maintains the same voltage across the surviving branch.

Step 1: Calculate the equivalent resistance before opening the switch.

Two identical \(80\ \Omega\) resistors are connected in parallel.

\[ R_{\mathrm{eq}}=\frac{R}{2} \]

\[ R_{\mathrm{eq}}=\frac{80}{2} \]

\[ R_{\mathrm{eq}}=40\ \Omega \]

Step 2: Calculate the initial total current.

\[ I_{\mathrm{total}}=\frac{V}{R_{\mathrm{eq}}} \]

\[ I_{\mathrm{total}}=\frac{12.0}{40} \]

\[ I_{\mathrm{total}}=0.300\ \mathrm{A} \]

Each branch carries:

\[ I_1=I_2=\frac{12.0}{80} \]

\[ I_1=I_2=0.150\ \mathrm{A} \]

KCL is satisfied:

\[ 0.300=0.150+0.150 \]

Step 3: Determine what happens when the switch opens.

The second branch is now disconnected.

There is only one conducting resistor branch, so:

\[ R_{\mathrm{eq}}=80\ \Omega \]

The new total current is:

\[ I_{\mathrm{total}}=\frac{12.0}{80} \]

\[ I_{\mathrm{total}}=0.150\ \mathrm{A} \]

The current through the remaining resistor is also:

\[ I_1=0.150\ \mathrm{A} \]

It has not changed.

Step 4: Explain why the remaining current stays constant.

Before opening the switch, the surviving resistor had \(12.0\ \mathrm{V}\) across it.

After opening the switch, it still has \(12.0\ \mathrm{V}\) across it because the supply is ideal and the resistor remains connected directly across the same two supply terminals.

Using Ohm’s law:

\[ I_1=\frac{V}{R} =\frac{12.0}{80} =0.150\ \mathrm{A} \]

The branch current therefore stays constant.

The tempting prediction is that the surviving resistor must carry more current because it is now the only remaining path.

But current is not a fixed quantity that must be distributed between whatever branches remain. The ideal voltage source maintains the potential difference, so the surviving resistor continues drawing the same current.

Step 5: Calculate the total power before and after opening the switch.

Use:

\[ P=VI \]

Initially:

\[ P_{\mathrm{initial}}=12.0\times0.300 \]

\[ P_{\mathrm{initial}}=3.60\ \mathrm{W} \]

After opening the switch:

\[ P_{\mathrm{final}}=12.0\times0.150 \]

\[ P_{\mathrm{final}}=1.80\ \mathrm{W} \]

The total power halves.

This makes sense because one resistor has been disconnected, while the remaining resistor continues operating at the same voltage and current.

Question 5

A battery is modelled as an ideal electromotive force \(E\) in series with an internal resistance \(r\).

Two identical \(40\ \Omega\) resistors are available.

An experiment is performed in two stages:

• First, only one \(40\ \Omega\) resistor is connected to the battery. The main ammeter reads \(0.200\ \mathrm{A}\).
• Next, both \(40\ \Omega\) resistors are connected in parallel. The main ammeter reads \(0.300\ \mathrm{A}\).

Assume:

• The battery’s emf and internal resistance remain constant.
• The resistors are ohmic and remain at constant temperature.
• The ammeter has negligible resistance.

Determine:

a) The battery’s internal resistance \(r\).

b) The battery’s emf \(E\).

c) The equivalent resistance of the external resistor network in each arrangement.

d) Why does the total current increase by only \(50\%\), rather than doubling, even though the external equivalent resistance halves?

A student argues that the equivalent resistances must be \(60\ \Omega\) and \(40\ \Omega\), because \(E/I\) gives those values. Explain precisely why this reasoning does not give the resistance of the external resistor network.

Solution 5

The battery has an internal resistance of \(20\ \Omega\) and an emf of \(12.0\ \mathrm{V}\). The external equivalent resistance decreases from \(40\ \Omega\) to \(20\ \Omega\), but the battery’s internal resistance prevents the total current from doubling.

This question tests an important distinction: the resistance of the external parallel network is not necessarily the same as the total resistance seen by the battery.

Step 1: Analyse the circuit with one resistor.

The external resistor has a resistance of \(40\ \Omega\), and the measured current is \(0.200\ \mathrm{A}\).

Using Ohm’s law, the terminal voltage across the external resistor is:

\[ V_1=IR \]

\[ V_1=0.200\times40 \]

\[ V_1=8.0\ \mathrm{V} \]

The battery’s internal resistance also produces a voltage drop.

Applying KVL around the complete circuit:

\[ E=V_1+I_1r \]

Substituting:

\[ E=8.0+0.200r \]

Call this equation (1).

Step 2: Analyse the circuit with both resistors connected.

The two identical resistors are in parallel.

They have the same voltage across them and therefore carry equal currents.

Using KCL:

\[ I_{\mathrm{total}}=I_1+I_2 \]

Since the total current is \(0.300\ \mathrm{A}\):

\[ 0.300=I_1+I_2 \]

The branch currents must be equal, so:

\[ I_1=I_2=\frac{0.300}{2} \]

\[ I_1=I_2=0.150\ \mathrm{A} \]

Step 3: Calculate the battery’s terminal voltage in the second arrangement.

Each resistor has \(40\ \Omega\) resistance and carries \(0.150\ \mathrm{A}\).

Using Ohm’s law:

\[ V_2=IR \]

\[ V_2=0.150\times40 \]

\[ V_2=6.0\ \mathrm{V} \]

Apply KVL around the complete circuit:

\[ E=V_2+I_{\mathrm{total}}r \]

Substituting:

\[ E=6.0+0.300r \]

Call this equation (2).

Step 4: Solve for the internal resistance.

Both equations describe the same battery, so the emf \(E\) must be equal.

Equate equations (1) and (2):

\[ 8.0+0.200r=6.0+0.300r \]

Rearrange:

\[ 2.0=0.100r \]

Therefore:

\[ r=20\ \Omega \]

Step 5: Calculate the battery’s emf.

Substitute \(r=20\ \Omega\) into equation (1):

\[ E=8.0+0.200(20) \]

\[ E=8.0+4.0 \]

\[ E=12.0\ \mathrm{V} \]

The battery therefore has an emf of \(12.0\ \mathrm{V}\) and an internal resistance of \(20\ \Omega\).

Step 6: Calculate the external equivalent resistance in each arrangement.

With one resistor:

\[ R_{\mathrm{eq,1}}=40\ \Omega \]

With two identical resistors in parallel:

\[ R_{\mathrm{eq,2}}=\frac{40}{2} \]

\[ R_{\mathrm{eq,2}}=20\ \Omega \]

The external equivalent resistance has halved, exactly as predicted.

Step 7: Explain why the battery current does not double.

This is the important conceptual part.

The battery has an internal resistance of \(20\ \Omega\), which remains in series with the external resistor network.

With one resistor, the total circuit resistance is:

\[ R_{\mathrm{total,1}}=20+40 \]

\[ R_{\mathrm{total,1}}=60\ \Omega \]

With both resistors in parallel:

\[ R_{\mathrm{total,2}}=20+20 \]

\[ R_{\mathrm{total,2}}=40\ \Omega \]

The total current is therefore:

For one resistor:

\[ I_1=\frac{12.0}{60} =0.200\ \mathrm{A} \]

For two parallel resistors:

\[ I_2=\frac{12.0}{40} =0.300\ \mathrm{A} \]

The current increases by:

\[ \frac{0.300-0.200}{0.200}\times100 =50\% \]

It does not double because the internal resistance remains part of the complete circuit.

The terminal voltage also falls from \(8.0\ \mathrm{V}\) to \(6.0\ \mathrm{V}\), as the greater total current produces a larger voltage drop across the battery’s internal resistance.

Step 8: Identify the student’s mistake.

The student calculates:

\[ \frac{E}{I_1}=\frac{12.0}{0.200} =60\ \Omega \]

and:

\[ \frac{E}{I_2}=\frac{12.0}{0.300} =40\ \Omega \]

Both calculations are mathematically correct.

However, these values represent the total circuit resistance, including the battery’s internal resistance.

They are not the equivalent resistances of the external resistor network.

To calculate the external equivalent resistance, use the terminal voltage across the external network:

\[ R_{\mathrm{eq}}=\frac{V_{\mathrm{terminal}}}{I_{\mathrm{total}}} \]

For the first arrangement:

\[ R_{\mathrm{eq,1}}=\frac{8.0}{0.200} =40\ \Omega \]

For the second arrangement:

\[ R_{\mathrm{eq,2}}=\frac{6.0}{0.300} =20\ \Omega \]

The external resistance has halved, just as it should.

The student’s mistake was including the battery’s internal resistance when calculating the resistance of the external load.

08What to remember when solving HSC parallel-resistance questions

Before using the shortcut for two identical parallel resistors, check the actual circuit.

1. Both resistors must connect between the same two nodes.
2. KVL establishes that the voltage across each resistor is the same.
3. KCL establishes that the total current is the sum of the branch currents.
4. Ohm’s law establishes the current through each resistor.
5. The equivalent resistance is calculated using the total current.

For two identical resistors:

\[ R_{\mathrm{eq}}=\frac{R}{2} \]

For \(n\) identical resistors:

\[ R_{\mathrm{eq}}=\frac{R}{n} \]

The shortcut is useful, but understanding where it comes from is more important. It allows you to recognise when it applies, when it does not, and how to handle circuits containing additional components.

The natural next step is applying the same conservation-first method to three or more parallel branches, particularly when the resistors have different values. See How to Find Branch Currents in a Three-Branch Parallel Circuit.