How to Find Branch Currents in a Three-Branch Parallel Circuit

A 9.0 V power supply feeds three parallel branches containing resistors of 100 Ω, 150 Ω, and 300 Ω. An ammeter in the main wire reads 0.18 A.

How much current flows through each branch?

You cannot simply divide the total current by three. The branches have different resistances, so they carry different currents. To find each branch current, you need to establish the voltage across each branch, apply Ohm’s law, and then use conservation of current at the junction.

The reliable method is:

1. Use Kirchhoff’s voltage law to establish the potential difference across each branch.
2. Use Ohm’s law to calculate the current through each branch.
3. Use Kirchhoff’s current law to check that the branch currents add to the total current.

For three simple resistive branches connected across an ideal voltage supply:

\[ V_1=V_2=V_3=V_s \]

\[ I_1=\frac{V_s}{R_1},\quad I_2=\frac{V_s}{R_2},\quad I_3=\frac{V_s}{R_3} \]

And at the junction:

\[ \boxed{I_{\text{total}}=I_1+I_2+I_3} \]

These relationships are the foundation of finding branch currents. Let’s establish why they work, then use them to solve circuits with increasingly complicated branches.

01Start with the actual circuit

Imagine building the three-branch circuit on a laboratory bench.

The positive terminal of a DC power supply connects to a switch, followed by an ammeter in the main wire. The wire then reaches a junction, labelled A.

From junction A, the circuit divides into three separate paths. Each path contains a resistor and an ammeter connected in series. The three paths meet again at junction B.

From B, a return wire connects to the negative terminal of the power supply.

A voltmeter is connected across junctions A and B to measure the potential difference across the parallel network.

There are three important measurements to make:

• The total current entering junction A.
• The current through each individual branch.
• The potential difference between junctions A and B.

The ammeters measure current in amperes (A). The voltmeter measures potential difference in volts (V).

Notice where each instrument is connected:

• An ammeter must be connected in series with the component whose current you want to measure.
• A voltmeter must be connected in parallel across the component or section whose potential difference you want to measure.

An ammeter has very low resistance, while a voltmeter has very high resistance. This allows them to measure the circuit without significantly changing its behaviour in the idealised model.

What should the experiment show?

Before switching on the supply, predict the current in each branch.

All three resistors connect between the same two junctions. Therefore, they experience approximately the same potential difference.

For the idealised circuit, each branch has 9.0 V across it.

Using Ohm’s law:

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

The predicted currents are:

The total current should therefore be:

\[ I_{\text{total}}=0.090+0.060+0.030 \]

\[ \boxed{I_{\text{total}}=0.180\text{ A}} \]

The 100 Ω branch carries the greatest current, while the 300 Ω branch carries the smallest.

This is because all three branches have the same potential difference, but their resistances are different.

Illustrative experimental readings

Suppose the experiment produces the following readings, rounded to three significant figures.

The observations agree with two predictions:

1. The three branch currents add to approximately the total current.
2. The potential difference across each branch is approximately equal.

This supports the conservation laws used to analyse the circuit.

Real measurements will not always agree perfectly. Instrument resolution, calibration, contact resistance, and small changes in the supply voltage can introduce differences.

Why measuring only the main current is insufficient

Imagine the first experimental design only uses one ammeter in the main wire.

It measures 0.180 A, but this does not tell you how much current flows through each branch.

There are three unknown branch currents and only one equation:

\[ I_1+I_2+I_3=0.180 \]

Many combinations of currents could satisfy this equation.

To improve the investigation, measure each branch current separately by inserting the ammeter in series with that branch. Keep the supply voltage and resistor arrangement unchanged.

Alternatively, use three low-resistance ammeters simultaneously.

The revised method provides enough measurements to test the current distribution rather than simply assuming it.

02Kirchhoff’s two laws: the reasoning behind branch currents

Before simplifying resistances, establish what happens at the junctions and around the loops.

These two conservation laws determine how current and potential difference behave throughout the circuit.

Kirchhoff’s current law (KCL): what happens at a junction?

Think about a road that divides into three separate lanes.

If 60 cars enter the intersection every minute, a total of 60 cars must leave through the three lanes each minute, provided the number of cars stored at the intersection is not continually increasing.

Electric current follows the same conservation principle.

Current is the rate at which electric charge passes a point. Charge cannot simply disappear when it reaches a junction.

Therefore:

\[ \text{Current entering} = \text{Current leaving} \]

For a three-branch parallel circuit:

\[ \boxed{I_{\text{total}}=I_1+I_2+I_3} \]

Here:

• \(I_{\text{total}}\) is the current entering the first junction.
• \(I_1\) is the current through branch 1.
• \(I_2\) is the current through branch 2.
• \(I_3\) is the current through branch 3.

The same currents recombine at the second junction.

An important distinction: conservation of current does not mean that the current divides equally.

For example, a circuit carrying 6.0 A through its main wire could have branch currents of:

\[ I_1=2.0\text{ A},\quad I_2=3.0\text{ A},\quad I_3=1.0\text{ A} \]

These currents are different, but their sum is still 6.0 A.

Kirchhoff’s voltage law (KVL): what happens around a loop?

Now consider a different question.

Why does each parallel branch have the same potential difference?

Think about a complete circuit loop containing a battery and one resistor branch.

The battery provides electrical energy to charges. As charges pass through the resistor, electrical potential energy is transferred to other forms, mainly thermal energy.

When the charges return to their starting point, the net change in electrical potential must be zero.

This is conservation of energy.

Kirchhoff’s voltage law states:

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

In other words, the total potential rises around a closed loop equal the total potential drops.

For a circuit with an ideal 12 V supply and one resistor in each branch:

\[ +12-V_1=0 \]

\[ +12-V_2=0 \]

\[ +12-V_3=0 \]

Therefore:

\[ \boxed{V_1=V_2=V_3=12\text{ V}} \]

Each branch connects to the same two junctions, so each experiences the same potential difference.

There is an important limitation to this statement.

If a branch contains two resistors in series, the full supply voltage is shared between those two resistors. Each resistor does not receive the entire supply voltage.

For example, if a 12 V supply connects across a branch containing two series resistors:

\[ V_1+V_2=12\text{ V} \]

The individual voltage drops depend on their resistances.

For a refresher on how voltage is distributed between series components, see Series Circuits: Current, Voltage, Resistance and Power.

Sign conventions when applying Kirchhoff’s laws

In HSC calculations, choosing a consistent sign convention makes circuit equations much easier to manage.

Use conventional current, which flows from the positive terminal of the source towards the negative terminal through the external circuit.

When moving around a loop:

• Moving through a source 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 current direction gives a positive potential change.

If your assumed current direction produces a negative answer, the actual current flows in the opposite direction.

The sign is telling you about direction, not indicating an impossible negative current magnitude.

03The method: finding all three branch currents

Now combine the two conservation laws with Ohm’s law.

For an ideal supply and fixed resistors, follow this sequence.

Step 1: Identify the three branches and two main junctions.

Label the junction where the current splits as A and the junction where it recombines as B.

Check that all three branches connect between the same two points.

Step 2: Assign current directions.

Assume the total current enters junction A and the three branch currents leave it.

Label them \(I_{\text{total}}\), \(I_1\), \(I_2\), and \(I_3\).

Step 3: Apply Kirchhoff’s current law.

Write:

\[ I_{\text{total}}=I_1+I_2+I_3 \]

This gives you the relationship between the currents.

Step 4: Apply Kirchhoff’s voltage law.

Write a loop equation for each branch.

For three branches connected directly across an ideal supply:

\[ V_1=V_2=V_3=V_s \]

If a branch contains multiple components, include the voltage drops across all of them.

Step 5: Simplify resistors within individual branches.

If a branch contains resistors in series, add their resistances:

\[ R_{\text{branch}}=R_a+R_b \]

If a branch contains resistors in parallel, use the reciprocal resistance relationship within that branch.

Do not combine the three separate branches into one equivalent resistance yet if you still need their individual currents.

Step 6: Apply Ohm’s law to each branch.

For a branch with a fixed equivalent resistance:

\[ I_k=\frac{V_s}{R_k} \]

Calculate each branch current separately.

Step 7: Check the total current.

Add the three branch currents:

\[ I_{\text{total}}=I_1+I_2+I_3 \]

If the question also asks for equivalent resistance, you can now calculate it using:

\[ \frac{1}{R_{\text{eq}}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} \]

This gives you an independent check of the total current.

The broader relationships between voltage, current, resistance, and power are covered in Parallel Circuits: Voltage, Current, Resistance and Power.

04Worked examples

05A common misconception: opening a branch changes every current

Consider a parallel circuit with an ideal 18.0 V supply and three branches:

• Branch 1: 9.0 Ω
• Branch 2: 6.0 Ω
• Branch 3: 18.0 Ω

Initially, all three branches are connected.

The branch currents are:

\[ I_1=\frac{18.0}{9.0}=2.0\text{ A} \]

\[ I_2=\frac{18.0}{6.0}=3.0\text{ A} \]

\[ I_3=\frac{18.0}{18.0}=1.0\text{ A} \]

Therefore:

\[ I_{\text{total}}=2.0+3.0+1.0 \]

\[ I_{\text{total}}=6.0\text{ A} \]

Now imagine opening the switch in branch 3.

What happens to the currents in branches 1 and 2?

A tempting prediction is that the remaining current must redistribute equally between the two branches.

That is incorrect for an ideal voltage supply.

The voltage across branches 1 and 2 remains 18.0 V.

Therefore:

\[ I_1=\frac{18.0}{9.0}=2.0\text{ A} \]

\[ I_2=\frac{18.0}{6.0}=3.0\text{ A} \]

The open branch carries zero current:

\[ I_3=0\text{ A} \]

The new total current is:

\[ I_{\text{total}}=2.0+3.0+0 \]

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

The currents in the remaining branches do not change. Only the total current decreases.

Why?

The ideal voltage supply maintains a constant potential difference across the remaining branches. Their resistances have not changed, so their currents have not changed.

This result depends on the ideal-source assumption. A real source with internal resistance can behave differently, as the final practice question explores.

06Questions and solutions

The following questions progress from straightforward calculations to more challenging applications of Kirchhoff’s laws.

Question 1

An ideal 15.0 V battery is connected to three parallel resistors:

• \(R_1=5.0\ \Omega\)
• \(R_2=10.0\ \Omega\)
• \(R_3=15.0\ \Omega\)

Calculate:

(a) The current through each branch.

(b) The total current supplied by the battery.

(c) Explain why the three branch currents are different.

Solution 1

The branch currents are 3.0 A, 1.5 A, and 1.0 A respectively, giving a total current of 5.5 A.

All three branches experience the same potential difference because they connect across the same two junctions.

Using Ohm’s law:

(a) Branch currents

For branch 1:

\[ I_1=\frac{V}{R_1} =\frac{15.0}{5.0} \]

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

For branch 2:

\[ I_2=\frac{15.0}{10.0} \]

\[ \boxed{I_2=1.5\text{ A}} \]

For branch 3:

\[ I_3=\frac{15.0}{15.0} \]

\[ \boxed{I_3=1.0\text{ A}} \]

(b) Total current

Apply Kirchhoff’s current law:

\[ I_{\text{total}}=I_1+I_2+I_3 \]

\[ =3.0+1.5+1.0 \]

\[ \boxed{I_{\text{total}}=5.5\text{ A}} \]

(c) Explanation

The branches carry different currents because they have different resistances.

Since the potential difference is the same across all three branches, Ohm’s law shows that current is inversely proportional to resistance.

The 5.0 Ω branch carries the greatest current because it has the smallest resistance.

Question 2

An ideal 18.0 V supply is connected to three parallel branches.

• Branch A contains two resistors of 3.0 Ω and 6.0 Ω connected in series.
• Branch B contains one 12.0 Ω resistor.
• Branch C contains an unknown resistor \(R\).

An ammeter in the main supply wire reads 5.0 A.

Calculate:

(a) The current through branches A, B, and C.

(b) The unknown resistance \(R\).

(c) The potential difference across each resistor in branch A.

Solution 2

The branch currents are 2.0 A, 1.5 A, and 1.5 A. The unknown resistance is 12.0 Ω.

The key is to calculate the known branch currents first, then use conservation of current to find the unknown branch current.

(a) Branch currents

First, calculate the equivalent resistance of branch A.

The two resistors are in series:

\[ R_A=3.0+6.0 \]

\[ R_A=9.0\ \Omega \]

The entire branch has 18.0 V across it.

Therefore:

\[ I_A=\frac{18.0}{9.0} \]

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

For branch B:

\[ I_B=\frac{18.0}{12.0} \]

\[ \boxed{I_B=1.5\text{ A}} \]

(b) Unknown resistance

Apply Kirchhoff’s current law:

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

Substitute the known values:

\[ 5.0=2.0+1.5+I_C \]

Rearrange:

\[ I_C=5.0-3.5 \]

\[ \boxed{I_C=1.5\text{ A}} \]

Now use Ohm’s law to calculate the unknown resistance:

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

\[ R=\frac{18.0}{1.5} \]

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

(c) Voltage drops in branch A

The current through both series resistors is 2.0 A.

For the 3.0 Ω resistor:

\[ V_1=IR \]

\[ V_1=(2.0)(3.0) \]

\[ \boxed{V_1=6.0\text{ V}} \]

For the 6.0 Ω resistor:

\[ V_2=(2.0)(6.0) \]

\[ \boxed{V_2=12.0\text{ V}} \]

Check using Kirchhoff’s voltage law:

\[ 6.0+12.0=18.0\text{ V} \]

The voltage drops add to the supply voltage, confirming the result.

Question 3

An ideal 24.0 V supply connects to a circuit with three main parallel branches.

• Branch A contains a 4.0 Ω resistor and an 8.0 Ω resistor in series.
• Branch B splits into two smaller parallel paths containing a 12.0 Ω resistor and a 6.0 Ω resistor.
• Branch C contains a single 8.0 Ω resistor.

Calculate:

(a) The total current through each of the three main branches.

(b) The individual currents through the 12.0 Ω and 6.0 Ω resistors inside branch B.

(c) The total current supplied by the battery.

Solution 3

The three main branch currents are 2.0 A, 6.0 A, and 3.0 A. The two smaller paths inside branch B carry 2.0 A and 4.0 A respectively. The total supply current is 11.0 A.

This question requires applying Kirchhoff’s laws at two different levels of the circuit.

(a) Current through the three main branches

Start with branch A.

Its two resistors are in series:

\[ R_A=4.0+8.0=12.0\ \Omega \]

Therefore:

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

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

Next, consider branch B.

Its two resistors are connected in parallel. Calculate their equivalent resistance:

\[ \frac{1}{R_B} = \frac{1}{12.0} + \frac{1}{6.0} \]

\[ \frac{1}{R_B} = \frac{1+2}{12.0} \]

\[ R_B=4.0\ \Omega \]

Therefore:

\[ I_B=\frac{24.0}{4.0} \]

\[ \boxed{I_B=6.0\text{ A}} \]

Finally, branch C contains one 8.0 Ω resistor:

\[ I_C=\frac{24.0}{8.0} \]

\[ \boxed{I_C=3.0\text{ A}} \]

(b) Individual currents inside branch B

Both smaller parallel paths experience the same 24.0 V potential difference.

For the 12.0 Ω resistor:

\[ I_{12\Omega}=\frac{24.0}{12.0} \]

\[ \boxed{I_{12\Omega}=2.0\text{ A}} \]

For the 6.0 Ω resistor:

\[ I_{6\Omega}=\frac{24.0}{6.0} \]

\[ \boxed{I_{6\Omega}=4.0\text{ A}} \]

Check at the internal junction using Kirchhoff’s current law:

\[ I_B=I_{12\Omega}+I_{6\Omega} \]

\[ I_B=2.0+4.0 \]

\[ I_B=6.0\text{ A} \]

This agrees with the main branch current calculated earlier.

(c) Total supply current

Now apply Kirchhoff’s current law at the main junction:

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

\[ I_{\text{total}}=2.0+6.0+3.0 \]

\[ \boxed{I_{\text{total}}=11.0\text{ A}} \]

The important idea is that current conservation applies at every junction, not just the first junction where the supply current divides.

Question 4

A student investigates a three-branch parallel circuit using a 9.0 V supply.

The following current readings are recorded:

Each ammeter has an uncertainty of ±0.002 A.

The student claims that Kirchhoff’s current law is incorrect because the sum of the branch currents is not exactly equal to the main supply current.

(a) Calculate the sum of the three branch currents.

(b) Determine the difference between the measured total current and the sum of the branch currents.

(c) Assess whether the measurements disprove Kirchhoff’s current law.

(d) Suggest two improvements to the experimental procedure.

Solution 4

The branch currents sum to 0.180 A, giving a difference of only 0.001 A from the main current. This does not disprove Kirchhoff’s current law because the difference is within the measurement uncertainty.

(a) Sum of the branch currents

Using Kirchhoff’s current law:

\[ I_{\text{branches}}=I_1+I_2+I_3 \]

\[ =0.090+0.060+0.030 \]

\[ \boxed{I_{\text{branches}}=0.180\text{ A}} \]

(b) Difference between measurements

\[ \Delta I=0.181-0.180 \]

\[ \boxed{\Delta I=0.001\text{ A}} \]

The difference is 1 mA.

(c) Assessing the claim

The student’s conclusion is incorrect.

Each ammeter has an uncertainty of ±0.002 A. The uncertainty in the sum of the three branch readings could be as large as ±0.006 A using a conservative addition of the individual uncertainties.

The main current also has an uncertainty of ±0.002 A.

The observed difference of 0.001 A is smaller than these uncertainties.

Therefore, the measurements are consistent with Kirchhoff’s current law.

Experimental results do not need to match theoretical predictions perfectly. They need to agree within the limitations of the measurement process.

(d) Improvements

Two suitable improvements are:

1. Use calibrated ammeters with smaller measurement uncertainties to obtain more precise current readings.
2. Use a regulated power supply and maintain stable electrical connections to reduce changes in voltage and contact resistance during measurements.

Repeating the measurements and calculating an average would also improve reliability.

A single ammeter in the main supply wire would not be enough to determine all three branch currents. Additional branch measurements are necessary.

Question 5

Extension: A battery with internal resistance

A battery has an electromotive force (emf) of 18.0 V and an internal resistance of 1.0 Ω.

It supplies a three-branch parallel circuit containing:

• Branch A: 6.0 Ω
• Branch B: 3.0 Ω
• Branch C: 2.0 Ω

A switch is initially closed in branch C.

The switch is then opened, disconnecting branch C.

A student predicts that the currents in branches A and B must decrease because the total current supplied by the battery decreases.

Is the student correct?

Calculate:

(a) The initial total current supplied by the battery.

(b) The initial current through each branch and the initial terminal voltage of the battery.

(c) The new total current after branch C is disconnected.

(d) The new currents through branches A and B.

(e) Explain why the currents in the remaining branches change even though their resistances remain constant.

Assume all external resistors are fixed and ohmic.

Solution 5

The student is incorrect.

When branch C is disconnected, the total current decreases from 9.0 A to 6.0 A. However, the currents through branches A and B actually increase.

The reason is that the battery has internal resistance, so its terminal voltage changes when the total current changes.

This is a more advanced situation than the ideal-source examples above.

(a) Initial total current

First, calculate the equivalent resistance of the three external parallel branches.

\[ \frac{1}{R_p} = \frac{1}{6.0} + \frac{1}{3.0} + \frac{1}{2.0} \]

\[ \frac{1}{R_p} = \frac{1+2+3}{6} \]

\[ R_p=1.0\ \Omega \]

The battery’s internal resistance is in series with the external parallel network.

Therefore, the total circuit resistance is:

\[ R_{\text{total}}=r+R_p \]

\[ R_{\text{total}}=1.0+1.0 \]

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

Using the battery emf:

\[ I_{\text{total}}=\frac{E}{R_{\text{total}}} \]

\[ I_{\text{total}}=\frac{18.0}{2.0} \]

\[ \boxed{I_{\text{total}}=9.0\text{ A}} \]

(b) Initial branch currents and terminal voltage

The battery’s terminal voltage is the potential difference across the external parallel network.

The voltage lost across the internal resistance is:

\[ V_r=Ir \]

\[ V_r=(9.0)(1.0)=9.0\text{ V} \]

Therefore, the terminal voltage is:

\[ V_t=E-V_r \]

\[ V_t=18.0-9.0 \]

\[ \boxed{V_t=9.0\text{ V}} \]

Now calculate each branch current.

Branch A:

\[ I_A=\frac{9.0}{6.0} \]

\[ \boxed{I_A=1.5\text{ A}} \]

Branch B:

\[ I_B=\frac{9.0}{3.0} \]

\[ \boxed{I_B=3.0\text{ A}} \]

Branch C:

\[ I_C=\frac{9.0}{2.0} \]

\[ \boxed{I_C=4.5\text{ A}} \]

Check using Kirchhoff’s current law:

\[ 1.5+3.0+4.5=9.0\text{ A} \]

The branch currents correctly add to the total current.

(c) New total current after opening branch C

With branch C disconnected, only branches A and B remain.

Their equivalent resistance is:

\[ \frac{1}{R_p'} = \frac{1}{6.0} + \frac{1}{3.0} \]

\[ \frac{1}{R_p'}=\frac{3}{6} \]

\[ R_p'=2.0\ \Omega \]

The total resistance, including internal resistance, becomes:

\[ R_{\text{total}}'=1.0+2.0 \]

\[ R_{\text{total}}'=3.0\ \Omega \]

Therefore:

\[ I_{\text{total}}'=\frac{18.0}{3.0} \]

\[ \boxed{I_{\text{total}}'=6.0\text{ A}} \]

The total current has decreased from 9.0 A to 6.0 A.

(d) New currents through branches A and B

The current through the internal resistance is now 6.0 A.

Therefore, the voltage drop across the internal resistance is:

\[ V_r'=I_{\text{total}}'r \]

\[ V_r'=(6.0)(1.0) \]

\[ V_r'=6.0\text{ V} \]

The new terminal voltage is:

\[ V_t'=E-V_r' \]

\[ V_t'=18.0-6.0 \]

\[ \boxed{V_t'=12.0\text{ V}} \]

The remaining branches now experience 12.0 V rather than 9.0 V.

For branch A:

\[ I_A'=\frac{12.0}{6.0} \]

\[ \boxed{I_A'=2.0\text{ A}} \]

For branch B:

\[ I_B'=\frac{12.0}{3.0} \]

\[ \boxed{I_B'=4.0\text{ A}} \]

Check:

\[ I_{\text{total}}'=2.0+4.0 \]

\[ I_{\text{total}}'=6.0\text{ A} \]

The results are consistent.

(e) Why do the remaining branch currents increase?

This is the central idea of the question.

When branch C is disconnected, the external equivalent resistance increases from 1.0 Ω to 2.0 Ω.

Consequently, the total current supplied by the battery decreases.

However, a smaller current through the internal resistance also means a smaller voltage drop inside the battery.

The battery’s terminal voltage therefore increases from 9.0 V to 12.0 V.

This higher terminal voltage increases the current through both remaining branches.

The changes can be summarised as follows:

The surprising result is that the total current decreases while the individual currents in the remaining branches increase.

There is no contradiction. The removal of branch C eliminates 4.5 A of current, while the other two branch currents increase by a combined 1.5 A.

This is why assumptions about the power supply matter. With an ideal voltage supply, the remaining branch currents would stay constant. With internal resistance, the terminal voltage can change.

For ordinary three-branch HSC calculations, use the ideal-source model unless the question supplies internal resistance or other information indicating that the source voltage changes.

07What to remember when solving three-branch circuits

When you encounter a three-branch parallel circuit, don’t immediately reach for the equivalent resistance formula.

Start with the physical arrangement and the conservation laws.

The most useful order is:

1. Identify the junctions and branches.
2. Apply Kirchhoff’s current law.
3. Apply Kirchhoff’s voltage law.
4. Simplify components within individual branches where necessary.
5. Calculate each branch current using Ohm’s law.
6. Add the branch currents and check the result.
7. Calculate equivalent resistance only when required.

Once you can handle three branches, the next useful step is analysing circuits where parallel networks are combined with series components, multiple junctions, and more than one voltage source. The same conservation laws still apply; you simply need to keep track of each loop and junction carefully.