Why Adding a Parallel Branch Increases Total Current

A \(12.0\ \mathrm{V}\) ideal supply is connected to a \(12.0\ \Omega\) resistor. The current is \(1.00\ \mathrm{A}\). Now connect a second resistor, \(6.0\ \Omega\), across the same two points.

Does the original resistor lose some current because there is now another path?

No. The original branch still carries \(1.00\ \mathrm{A}\). The new branch carries \(2.00\ \mathrm{A}\), so the total current supplied by the battery increases to \(3.00\ \mathrm{A}\).

Adding a parallel branch increases total current because an ideal voltage supply maintains the same potential difference while providing an additional path for current. The current entering the junction must equal the sum of the currents leaving it.

The important detail is that the battery does not supply a fixed amount of current that must be divided between the branches. It supplies the voltage, and the circuit determines how much current flows.

01What the switch experiment is testing

Imagine setting up a simple investigation to find out what actually happens when another parallel branch is connected.

The question is straightforward:

Does the original branch current decrease, or does the power supply deliver more total current?

Setting up the circuit

Use the following equipment:

• A regulated \(12.0\ \mathrm{V}\) DC power supply.
• A \(12.0\ \Omega\) resistor in the original branch.
• A \(6.0\ \Omega\) resistor in a second branch.
• A switch to connect or disconnect the second branch.
• Three ammeters to measure the total current and the two branch currents.
• A voltmeter to measure the potential difference across the parallel network.

The positive terminal of the power supply connects to the main ammeter. Its output then reaches junction A, where the circuit divides into two branches.

The first branch contains the \(12.0\ \Omega\) resistor and an ammeter. The second contains the switch, the \(6.0\ \Omega\) resistor, and another ammeter.

Both branches meet again at junction B. A return wire connects B to the negative terminal of the supply.

The voltmeter connects across A and B.

An ammeter is connected in series with the component whose current it measures. A voltmeter is connected in parallel across the section whose potential difference it measures.

What should happen when the switch closes?

Start with the switch open. Only the \(12.0\ \Omega\) resistor is connected.

Using Ohm’s law:

\[ I_1=\frac{V}{R_1} \]

\[ I_1=\frac{12.0}{12.0} =1.00\ \mathrm{A} \]

The main ammeter also reads \(1.00\ \mathrm{A}\).

Now close the switch.

The \(6.0\ \Omega\) resistor creates a second path between the same two junctions. Both resistors are connected directly across the supply, so each has \(12.0\ \mathrm{V}\) across it.

The new branch current is:

\[ I_2=\frac{12.0}{6.0} =2.00\ \mathrm{A} \]

The total current is therefore:

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

\[ =1.00+2.00 \]

\[ \boxed{I_{\text{total}}=3.00\ \mathrm{A}} \]

Notice what has happened:

• The original branch still carries \(1.00\ \mathrm{A}\).
• The new branch contributes another \(2.00\ \mathrm{A}\).
• The power supply now delivers \(3.00\ \mathrm{A}\).

Illustrative experimental readings

The following table gives idealised readings predicted for this investigation. They are illustrative values, not claimed measurements from an actual experiment.

The pattern supports two important predictions:

1. The original branch current remains unchanged because its resistance and potential difference remain unchanged.
2. The total current increases by the amount of current drawn by the new branch.

Why the first experimental design needs improving

Suppose the investigation initially uses only one ammeter, positioned before the junction.

It records the increase in total current, but it cannot establish what happened to the current in the original branch.

The original branch might have stayed at \(1.00\ \mathrm{A}\), or its current might have changed. The main ammeter alone cannot distinguish these possibilities.

A better design uses an ammeter in each branch and a voltmeter across the parallel network.

The supply voltage must also be controlled. A regulated supply with a current limit above the expected \(3.00\ \mathrm{A}\) is suitable for this idealised test.

This matters because a real battery or power supply might experience a voltage drop when the load increases. Measuring the actual voltage across A and B helps identify whether the original branch current changed because of a voltage change.

The investigation is now testing the correct idea: what happens to the current distribution when an additional path is connected while the branch voltage is maintained?

02The physics: conservation at junctions and around loops

Before using equivalent resistance, establish two things:

• What happens to current at a junction?
• What happens to potential difference around a complete circuit loop?

These are described by Kirchhoff’s current law and Kirchhoff’s voltage law.

1. Kirchhoff’s current law: charge is conserved

Current measures the rate at which electric charge passes a point.

\[ I=\frac{\Delta Q}{\Delta t} \]

Here:

• \(I\) is current, measured in amperes (\(\mathrm{A}\)).
• \(\Delta Q\) is the charge passing a point, measured in coulombs (\(\mathrm{C}\)).
• \(\Delta t\) is the time interval, measured in seconds (\(\mathrm{s}\)).

At a junction, charge cannot continuously disappear or accumulate in a steady-state circuit.

Therefore, the total current entering a junction must equal the total current leaving it.

This is Kirchhoff’s current law (KCL):

\[ \boxed{\sum I_{\text{in}}=\sum I_{\text{out}}} \]

For the two-branch circuit:

\[ \boxed{I_{\text{total}}=I_1+I_2} \]

The same relationship applies where the branches recombine.

A useful way to picture this is a pump supplying two parallel pipes. If the pump maintains the same pressure difference, opening another pipe allows a greater combined flow.

The analogy has limits. Electric charge is not water, and the electric field establishes the current throughout the circuit. However, the idea of an additional path under the same driving difference is useful.

The key misconception is thinking that current is a fixed quantity that gets divided between branches.

KCL does not say that. It says that whatever current enters a junction must leave it.

2. Kirchhoff’s voltage law: energy is conserved

Now consider why both branches experience the same potential difference.

Kirchhoff’s voltage law (KVL) states that the sum of all potential changes around a complete closed loop is zero:

\[ \boxed{\sum \Delta V=0} \]

This follows from conservation of energy. A charge that travels around a complete circuit loop and returns to its starting point has no net change in electrical potential energy.

Use the following sign convention:

• Moving through a supply from its negative terminal to its positive terminal gives a positive potential change.
• Moving through a resistor in the direction of conventional current gives a negative potential change.
• Moving through a resistor against the conventional current gives a positive potential change.

For the original branch, the loop equation is:

\[ +V_s-V_1=0 \]

Therefore:

\[ V_1=V_s \]

For the second branch:

\[ +V_s-V_2=0 \]

Therefore:

\[ V_2=V_s \]

Combining these results:

\[ \boxed{V_1=V_2=V_s} \]

This applies because both branches connect directly between the same two supply terminals, with ideal connecting wires.

This is the second essential part of the explanation.

KCL tells us that branch currents add together. KVL tells us that each directly connected parallel branch has the same potential difference.

Together, they explain why adding another branch increases the total current.

3. Apply Ohm’s law to find the new total current

Only after establishing the junction and loop relationships should you apply Ohm’s law.

For an ohmic resistor:

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

Since both branches have the same voltage:

\[ I_1=\frac{V_s}{R_1} \]

\[ I_2=\frac{V_s}{R_2} \]

Using Kirchhoff’s current law:

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

Substituting:

\[ \boxed{ I_{\text{total}} = \frac{V_s}{R_1} + \frac{V_s}{R_2} } \]

Now the reason for the increase is clear.

Adding another resistor in parallel introduces another positive current contribution. The original branch does not lose current when its voltage remains unchanged.

4. Why equivalent resistance decreases

We can now use the equivalent resistance equation as a shortcut.

The equivalent resistance is the resistance of a single hypothetical resistor that would draw the same total current from the supply.

Since:

\[ I_{\text{total}}=\frac{V_s}{R_{\text{eq}}} \]

and:

\[ I_{\text{total}} = \frac{V_s}{R_1} + \frac{V_s}{R_2} \]

we obtain:

\[ \boxed{ \frac{1}{R_{\text{eq}}} = \frac{1}{R_1} + \frac{1}{R_2} } \]

Adding another parallel branch increases the sum of the reciprocals, so the equivalent resistance decreases.

For a constant-voltage supply:

\[ I_{\text{total}}=\frac{V_s}{R_{\text{eq}}} \]

A smaller equivalent resistance produces a larger total current.

This is the mathematical version of the same physical explanation.

03An important contrast: when the original branch current does change

There is one assumption in the earlier explanation that you must not overlook.

The original branch current remains unchanged when the potential difference across that branch remains unchanged.

But what if the parallel network is connected to the supply through another resistor?

The parallel branches still have the same voltage as each other. However, they might not have the full supply voltage.

Consider a \(24.0\ \mathrm{V}\) ideal supply connected to a \(2.0\ \Omega\) resistor in series with a parallel network containing \(6.0\ \Omega\) and \(12.0\ \Omega\) branches.

Now add a \(4.0\ \Omega\) branch in parallel with the other two.

The total current increases, but the voltage across the parallel network decreases. Consequently, the original branch currents also decrease.

Why?

The additional branch reduces the equivalent resistance of the parallel network. This increases the current through the \(2.0\ \Omega\) series resistor.

A larger current through that resistor creates a larger voltage drop across it. Less voltage is therefore available across the parallel section.

The distinction is:

The additional branch still increases the total source current in the series-resistor example. It simply changes the voltage available to the existing branches.

For a deeper explanation of how to identify which components actually share the same two nodes, read How to Recognise Nodes That Are Electrically the Same Point.

For the broader relationships between voltage, current, resistance, and power, see Parallel Circuits: Voltage, Current, Resistance and Power.

04Worked examples

05A reliable method for these questions

When a question asks what happens after another branch is added, use this sequence:

1. Identify the nodes. Check whether the new branch actually connects between the same two points as the existing branches.
2. Apply Kirchhoff’s current law. Write the relationship between the total current and the individual branch currents.
3. Apply Kirchhoff’s voltage law. Establish the voltage across each branch using complete circuit loops.
4. Use Ohm’s law. Calculate each branch current using its actual potential difference.
5. Calculate equivalent resistance. Use resistance reductions as a shortcut or a final check, not as a replacement for understanding the circuit.

The most important decision is whether the voltage across the existing branches stays constant when the new branch is added.

For an ideal voltage supply connected directly across the parallel network, it does. If there is another component in series, it might not.

06Questions and solutions

Question 1

An ideal \(12.0\ \mathrm{V}\) supply is connected to a \(4.0\ \Omega\) resistor.

A second \(6.0\ \Omega\) resistor is connected in parallel with the original resistor.

Calculate:

(a) The current through the original resistor before the second branch is connected.

(b) The current through each branch after the second resistor is connected.

(c) The new total current supplied by the battery.

Solution 1

The total current increases from \(3.0\ \mathrm{A}\) to \(5.0\ \mathrm{A}\), while the original branch continues carrying \(3.0\ \mathrm{A}\).

(a) Original branch current

Before adding the second resistor, only the \(4.0\ \Omega\) resistor is connected.

Using Ohm’s law:

\[ I_1=\frac{V}{R_1} \]

\[ I_1=\frac{12.0}{4.0} \]

\[ \boxed{I_1=3.0\ \mathrm{A}} \]

(b) Currents after adding the second branch

Both resistors connect directly across the ideal supply.

Therefore, each receives \(12.0\ \mathrm{V}\).

The original branch remains:

\[ I_1=\frac{12.0}{4.0} =3.0\ \mathrm{A} \]

The new branch carries:

\[ I_2=\frac{12.0}{6.0} =2.0\ \mathrm{A} \]

(c) Total current

Apply Kirchhoff’s current law:

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

\[ =3.0+2.0 \]

\[ \boxed{I_{\text{total}}=5.0\ \mathrm{A}} \]

The original current does not decrease because the supply maintains the same voltage across that resistor. The additional branch contributes another \(2.0\ \mathrm{A}\).

The tempting mistake is to divide the original \(3.0\ \mathrm{A}\) between the two branches. That assumes a fixed total current, which is not how an ideal voltage supply operates.

Question 2

An ideal \(24.0\ \mathrm{V}\) supply is connected to three parallel branches.

• Branch A contains a \(4.0\ \Omega\) resistor in series with an \(8.0\ \Omega\) resistor.
• Branch B contains a \(6.0\ \Omega\) resistor.
• Branch C contains an unknown resistor, \(R_C\).

After all three branches are connected, an ammeter in the main supply wire reads \(8.0\ \mathrm{A}\).

Calculate:

(a) The current through each branch.

(b) The resistance of branch C.

(c) The potential difference across the \(8.0\ \Omega\) resistor in branch A.

Explain why the full supply voltage cannot be applied separately to both resistors in branch A.

Solution 2

The branch currents are \(2.0\ \mathrm{A}\), \(4.0\ \mathrm{A}\), and \(2.0\ \mathrm{A}\), respectively. The unknown resistance is \(12.0\ \Omega\), and the voltage across the \(8.0\ \Omega\) resistor is \(16.0\ \mathrm{V}\).

Step 1: Calculate the equivalent resistance of branch A.

The two resistors in branch A are in series, so their resistances add:

\[ R_A=4.0+8.0 \]

\[ R_A=12.0\ \Omega \]

Step 2: Apply Kirchhoff’s voltage law.

All three complete branches connect across the same two supply nodes.

Therefore:

\[ V_A=V_B=V_C=24.0\ \mathrm{V} \]

Step 3: Calculate the current through branches A and B.

For branch A:

\[ I_A=\frac{24.0}{12.0} \]

\[ \boxed{I_A=2.0\ \mathrm{A}} \]

For branch B:

\[ I_B=\frac{24.0}{6.0} \]

\[ \boxed{I_B=4.0\ \mathrm{A}} \]

Step 4: Use Kirchhoff’s current law to find branch C current.

The total current is the sum of all three branch currents:

\[ I_{\text{total}}=I_A+I_B+I_C \]

Substitute the known values:

\[ 8.0=2.0+4.0+I_C \]

\[ \boxed{I_C=2.0\ \mathrm{A}} \]

Step 5: Calculate the unknown resistance.

Using Ohm’s law:

\[ R_C=\frac{V_C}{I_C} \]

\[ R_C=\frac{24.0}{2.0} \]

\[ \boxed{R_C=12.0\ \Omega} \]

Step 6: Find the voltage across the \(8.0\ \Omega\) resistor.

The two resistors in branch A share the branch current of \(2.0\ \mathrm{A}\).

Therefore:

\[ V_{8\Omega}=IR \]

\[ V_{8\Omega}=2.0\times8.0 \]

\[ \boxed{V_{8\Omega}=16.0\ \mathrm{V}} \]

The remaining voltage across the \(4.0\ \Omega\) resistor is:

\[ V_{4\Omega}=2.0\times4.0 =8.0\ \mathrm{V} \]

Check using Kirchhoff’s voltage law:

\[ 16.0+8.0=24.0\ \mathrm{V} \]

The tempting mistake is to give each resistor in branch A the full \(24.0\ \mathrm{V}\). Only the complete branch has the full supply voltage. The two series resistors divide that voltage between themselves.

Question 3

An ideal \(12.0\ \mathrm{V}\) supply is connected to a \(2.0\ \Omega\) resistor in series with a parallel network.

The parallel network initially contains:

• A \(4.0\ \Omega\) resistor.
• A \(12.0\ \Omega\) resistor.

A third \(6.0\ \Omega\) resistor is then added in parallel.

Calculate:

(a) The initial total current supplied.

(b) The initial current through both parallel branches.

(c) The new total current after the third branch is added.

(d) The new current through all three branches.

(e) Explain why the original branch currents change even though their resistances remain constant.

Solution 3

The total current increases from \(2.4\ \mathrm{A}\) to \(3.0\ \mathrm{A}\), but both original branch currents decrease because the voltage across the parallel network falls.

(a) Initial total current

First, calculate the equivalent resistance of the original parallel network:

\[ \frac{1}{R_p} = \frac{1}{4.0} + \frac{1}{12.0} \]

\[ \frac{1}{R_p}=\frac{4}{12} \]

\[ R_p=3.0\ \Omega \]

The \(2.0\ \Omega\) resistor is in series with the parallel network:

\[ R_{\text{total}}=2.0+3.0 \]

\[ R_{\text{total}}=5.0\ \Omega \]

Using Ohm’s law:

\[ I_{\text{total}}=\frac{12.0}{5.0} \]

\[ \boxed{I_{\text{total}}=2.4\ \mathrm{A}} \]

(b) Initial branch currents

The voltage across the \(2.0\ \Omega\) series resistor is:

\[ V_s=IR \]

\[ V_s=2.4\times2.0 =4.8\ \mathrm{V} \]

The remaining voltage is across the parallel network:

\[ V_p=12.0-4.8 \]

\[ V_p=7.2\ \mathrm{V} \]

For the \(4.0\ \Omega\) branch:

\[ I_1=\frac{7.2}{4.0} \]

\[ \boxed{I_1=1.8\ \mathrm{A}} \]

For the \(12.0\ \Omega\) branch:

\[ I_2=\frac{7.2}{12.0} \]

\[ \boxed{I_2=0.6\ \mathrm{A}} \]

Their sum is \(2.4\ \mathrm{A}\), as required by Kirchhoff’s current law.

(c) New total current

After adding the \(6.0\ \Omega\) resistor:

\[ \frac{1}{R_p} = \frac{1}{4.0} + \frac{1}{12.0} + \frac{1}{6.0} \]

\[ \frac{1}{R_p}=\frac{6}{12} \]

\[ R_p=2.0\ \Omega \]

The total resistance is now:

\[ R_{\text{total}}=2.0+2.0 \]

\[ R_{\text{total}}=4.0\ \Omega \]

Therefore:

\[ I_{\text{total}}=\frac{12.0}{4.0} \]

\[ \boxed{I_{\text{total}}=3.0\ \mathrm{A}} \]

(d) New branch currents

The voltage drop across the series resistor is:

\[ V_s=3.0\times2.0 =6.0\ \mathrm{V} \]

The voltage across the parallel network becomes:

\[ V_p=12.0-6.0 \]

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

The three branch currents are:

\[ I_1=\frac{6.0}{4.0} =1.5\ \mathrm{A} \]

\[ I_2=\frac{6.0}{12.0} =0.5\ \mathrm{A} \]

\[ I_3=\frac{6.0}{6.0} =1.0\ \mathrm{A} \]

Check using KCL:

\[ I_{\text{total}}=1.5+0.5+1.0 \]

\[ \boxed{I_{\text{total}}=3.0\ \mathrm{A}} \]

(e) Explain the changes

The original branch currents decrease because the voltage across their common parallel section falls from \(7.2\ \mathrm{V}\) to \(6.0\ \mathrm{V}\).

The current through the series resistor increases, so its voltage drop increases. This leaves less voltage across the parallel network.

The tempting answer is that parallel branches always retain their original current. That is only true when their potential difference remains unchanged.

The new branch still increases the total source current, but it also changes the voltage distribution throughout the circuit.

Question 4

A student investigates the effect of adding a parallel branch using a regulated supply set to \(9.0\ \mathrm{V}\).

The original branch contains an \(18\ \Omega\) resistor. A second branch containing a \(9.0\ \Omega\) resistor can be connected using a switch.

The student records the following illustrative measurements:

The student makes two claims:

• The original branch current has decreased.
• Kirchhoff’s current law is incorrect because the two branch readings add to \(1.49\ \mathrm{A}\), not exactly \(1.50\ \mathrm{A}\).

Evaluate both claims. Use Ohm’s law, Kirchhoff’s current law, and the measurement uncertainties in your explanation.

Solution 4

Neither claim is supported by the measurements. The results are consistent with the original branch current remaining approximately constant and with Kirchhoff’s current law being satisfied.

Step 1: Check the original branch current using Ohm’s law.

Before adding the new branch:

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

\[ I_1=\frac{8.98}{18} \]

\[ I_1=0.499\ \mathrm{A} \]

This agrees with the measured value of \(0.50\ \mathrm{A}\).

After adding the branch:

\[ I_1=\frac{8.96}{18} \]

\[ I_1=0.498\ \mathrm{A} \]

This is also approximately \(0.50\ \mathrm{A}\).

The measured change from \(0.50\ \mathrm{A}\) to \(0.49\ \mathrm{A}\) is only \(0.01\ \mathrm{A}\).

The combined uncertainty of the two current readings is greater than this difference. Therefore, the measurements do not establish that the original branch current has decreased.

Step 2: Check the new branch current.

Using the measured voltage:

\[ I_2=\frac{8.96}{9.0} \]

\[ I_2=0.996\ \mathrm{A} \]

This agrees with the measured value of \(1.00\ \mathrm{A}\).

Step 3: Check Kirchhoff’s current law.

After the new branch is connected:

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

Using the measured branch currents:

\[ I_{\text{total}}=0.49+1.00 \]

\[ I_{\text{total}}=1.49\ \mathrm{A} \]

The main ammeter reads \(1.50\ \mathrm{A}\).

The difference is only \(0.01\ \mathrm{A}\), which is consistent with the stated measurement uncertainties.

There is no evidence that Kirchhoff’s current law has failed.

Step 4: Evaluate the overall experiment.

The supply current increases from approximately \(0.50\ \mathrm{A}\) to \(1.50\ \mathrm{A}\).

The new branch contributes approximately \(1.00\ \mathrm{A}\), accounting for almost the entire increase.

The small voltage difference between the two measurements is also within the voltmeter’s uncertainty.

Therefore, the results are consistent with the constant-voltage model: the original branch current remains approximately unchanged, while the new branch increases the total current.

The important experimental lesson is that small differences between readings do not automatically indicate a failure of a physical law. Measurement uncertainty must be considered before drawing that conclusion.

Question 5

Extension: When a power supply reaches its current limit

A laboratory power supply operates in the following way:

• It maintains a constant output voltage of \(12.0\ \mathrm{V}\) while the required output current is no greater than \(2.0\ \mathrm{A}\).
• If the circuit would draw more than \(2.0\ \mathrm{A}\) at \(12.0\ \mathrm{V}\), the supply enters constant-current mode.
• In constant-current mode, it maintains \(2.0\ \mathrm{A}\) and reduces its output voltage as necessary.

Three identical \(12.0\ \Omega\) resistors are connected in parallel across the supply.

Initially, only the first resistor is connected. The second and third resistors are then switched on, one at a time.

Determine the following for each stage:

(a) The total current supplied.

(b) The voltage across each resistor.

(c) The current through each connected branch.

(d) Explain why adding the third branch does not increase the total current beyond \(2.0\ \mathrm{A}\).

A student argues:

“Three identical resistors are connected across a \(12.0\ \mathrm{V}\) supply, so each must carry \(1.0\ \mathrm{A}\). The total current must therefore be \(3.0\ \mathrm{A}\).”

Explain precisely why this prediction fails.

Solution 5

The total current increases from \(1.0\ \mathrm{A}\) to \(2.0\ \mathrm{A}\) when the second resistor is added, but remains at \(2.0\ \mathrm{A}\) when the third resistor is added. The supply reduces its voltage to \(8.0\ \mathrm{V}\), so each branch carries approximately \(0.67\ \mathrm{A}\).

This question tests an important assumption: does the supply continue maintaining \(12.0\ \mathrm{V}\) after another branch is connected?

Not necessarily.

Step 1: Consider the first connected resistor.

Only one \(12.0\ \Omega\) resistor is connected.

The supply maintains \(12.0\ \mathrm{V}\), so:

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

\[ \boxed{I_{\text{total}}=1.0\ \mathrm{A}} \]

The supply is operating below its \(2.0\ \mathrm{A}\) current limit.

Step 2: Connect the second resistor.

Two identical \(12.0\ \Omega\) resistors are now in parallel.

Their equivalent resistance is:

\[ R_{\text{eq}}=\frac{12.0}{2} \]

\[ R_{\text{eq}}=6.0\ \Omega \]

The current demanded at \(12.0\ \mathrm{V}\) is:

\[ I_{\text{total}}=\frac{12.0}{6.0} \]

\[ \boxed{I_{\text{total}}=2.0\ \mathrm{A}} \]

The supply has reached its current limit, but it can still maintain \(12.0\ \mathrm{V}\) at this exact boundary.

Each branch carries:

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

\[ \boxed{I_1=I_2=1.0\ \mathrm{A}} \]

Step 3: Connect the third resistor.

Three identical resistors are now connected in parallel.

Their equivalent resistance is:

\[ R_{\text{eq}}=\frac{12.0}{3} \]

\[ R_{\text{eq}}=4.0\ \Omega \]

If the supply could maintain \(12.0\ \mathrm{V}\), the circuit would demand:

\[ I_{\text{demand}}=\frac{12.0}{4.0} \]

\[ I_{\text{demand}}=3.0\ \mathrm{A} \]

But the supply cannot deliver \(3.0\ \mathrm{A}\). Its maximum current is \(2.0\ \mathrm{A}\).

It therefore enters constant-current mode.

Step 4: Calculate the new output voltage.

The supply maintains:

\[ I_{\text{total}}=2.0\ \mathrm{A} \]

The load equivalent resistance is still \(4.0\ \Omega\).

Using Ohm’s law:

\[ V=IR \]

\[ V=2.0\times4.0 \]

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

Step 5: Calculate the current through each branch.

All three resistors remain connected across the same two supply nodes.

Each now has \(8.0\ \mathrm{V}\) across it.

Therefore:

\[ I_1=I_2=I_3 \]

\[ I=\frac{8.0}{12.0} \]

\[ \boxed{I=0.667\ \mathrm{A}} \]

Check using Kirchhoff’s current law:

\[ I_{\text{total}}=0.667+0.667+0.667 \]

\[ \boxed{I_{\text{total}}\approx2.0\ \mathrm{A}} \]

The student’s prediction fails because it assumes the supply continues providing \(12.0\ \mathrm{V}\).

Once the third branch is connected, the supply reaches its current limit and reduces its output voltage. The voltage across every branch consequently falls, reducing each branch current.

This also reveals an important distinction between Kirchhoff’s laws and the behaviour of the power supply.

KCL tells us that the total current equals the sum of the branch currents. It does not independently determine what that total current must be. The supply’s operating conditions and the circuit resistance determine the actual values.

07Where to go next

The central skill is recognising what stays constant when a circuit changes.

With an ideal voltage supply directly across a parallel network, adding a branch increases total current while leaving the original branch currents unchanged.

With additional series resistance or a supply limitation, the voltage across the parallel network can change, so the original branch currents may change too.

The next useful step is applying the same reasoning to mixed series-parallel circuits, where identifying nodes, writing loop equations, and tracking voltage drops become essential for comparing component currents and bulb brightness.