How to Solve Combination Circuits Without Losing Track of Current
Learn a reliable method for solving series-parallel circuits, from simplifying equivalent resistance to calculating total and branch currents. Two worked examples, annotated circuit diagrams, and five progressively challenging questions show you how to track current, calculate voltage drops, and check your answers using conservation rules.
A 12.0 V battery feeds a 2.0 Ω resistor in series with a 6.0 Ω resistor. You add a 3.0 Ω resistor in parallel with the 6.0 Ω resistor. The total current rises from 1.5 A to 3.0 A.
Does the current through the original 6.0 Ω resistor rise too, or does it fall? Make a prediction before reading on.
It falls. The new branch allows more current to flow through the shared 2.0 Ω resistor. That increases the voltage drop across the 2.0 Ω resistor, leaving less potential difference across the parallel section. The original 6.0 Ω resistor therefore carries less current, even though the battery supplies more current overall.
More total current, less current through the original load
| BEFORE: Series circuit | Rtotal = 2.0 Ω + 6.0 Ω = 8.0 Ω Battery: 12.0 V; total current: 1.5 A 6.0 Ω load current: 1.5 A 2.0 Ω voltage drop: 3.0 V Load voltage: 9.0 V |
|---|---|
| AFTER: Parallel branch added | 6.0 Ω ∥ 3.0 Ω = 2.0 Ω; Rtotal = 4.0 Ω Battery: 12.0 V; total current: 3.0 A 2.0 Ω voltage drop rises to 6.0 V Parallel-section voltage falls to 6.0 V 6.0 Ω branch: 1.0 A; 3.0 Ω branch: 2.0 A |
This is the trap in combination circuits: the total current and the current through an individual branch do not necessarily change in the same way.
The reliable method is to simplify the circuit first, calculate the total current, and then work backwards through the original circuit. Once you understand which components share current and which share voltage, the calculations become much easier to organise.
01The two rules that stop you losing track
Imagine a road reaching a fork. The traffic arriving at the fork must equal the traffic travelling down both roads combined. Electrical current behaves similarly: at a junction, charge does not steadily pile up, so the current entering must equal the total current leaving.
Voltage works differently. It describes the energy transferred per unit of charge. Components connected between the same two electrical nodes have the same potential difference, even if their resistances are different.
A node is a continuous section of ideal connecting wire. Every point on that wire is at the same electric potential.
Series circuits: the current stays the same
Components are in series when the same current must pass through each of them. There is no alternative path at the junction between them.
For resistors in series, the equivalent resistance is the sum of their resistances:
\[ R_{\text{series}}=R_1+R_2+\cdots \]
Here, \(R_1\), \(R_2\), and the other \(R\) symbols represent resistance, measured in ohms (\(\Omega\)).
Why do the resistances add? Each resistor transfers energy from the moving charges. More resistance in the same path means a greater total opposition to current.
The current is the same through each resistor, but the potential difference across each resistor can be different.
Parallel circuits: the voltage stays the same
Components are in parallel when both ends of each component connect to the same two nodes.
The potential difference across every parallel branch is therefore equal. The current can differ between branches because their resistances differ.
For resistors in parallel,
\[ \frac{1}{R_{\text{parallel}}} = \frac{1}{R_1}+\frac{1}{R_2}+\cdots \]
For two resistors, this can be rearranged to give a particularly useful formula:
\[ R_{\text{parallel}} = \frac{R_1R_2}{R_1+R_2} \]
The total current is the sum of the branch currents:
\[ I_{\text{total}}=I_1+I_2+\cdots \]
Current is measured in amperes (A), and potential difference is measured in volts (V).
Series vs parallel: what stays the same?
| SERIES — One current path | I = I₁ = I₂; Vtotal = V₁ + V₂ Same current arrow through R₁ and R₂ Separate voltage drops: V₁ and V₂ |
|---|---|
| PARALLEL — Two shared nodes | V₁ = V₂ = VAB; Itotal = I₁ + I₂ Both resistors connect between nodes A and B; voltage is equal across both branches Incoming current splits into I₁ and I₂, then recombines |
Keep this distinction clear:
| Property | Series | Parallel |
|---|---|---|
| What identifies it? | One current path through the components | Both ends connect to the same two nodes |
| What stays the same? | Current | Potential difference |
| Equivalent resistance | Resistances add | Reciprocals add |
| What happens to current? | Same through each component | Divides between branches |
| What happens to voltage? | Divides between components | Same across each branch |
One useful check follows immediately: a parallel combination must have an equivalent resistance smaller than its smallest individual resistance. Adding another parallel branch reduces the equivalent resistance.
Be careful, though. Components drawn next to each other on a page are not automatically in parallel. Their connections determine the relationship, not their appearance.
02A repeatable method for combination circuits
Use this method whenever a circuit contains a mixture of series and parallel resistors.
Step 1: Identify the nodes and junctions.
Trace the wires. If two resistors connect between the same pair of nodes, they are in parallel. If the same current must pass through two resistors one after the other, they are in series.
Step 2: Simplify the circuit from the inside out.
Start with the simplest identifiable group. Replace a parallel group with its equivalent resistance, then combine any resistors that are now in series.
Redraw the circuit after each reduction. This prevents you from losing track of which components remain connected.
Step 3: Find the total current.
Once the circuit has been reduced to one equivalent resistance, use Ohm’s law:
\[ V=IR \]
Here, \(V\) is potential difference in volts, \(I\) is current in amperes, and \(R\) is resistance in ohms.
For an ideal battery with fixed voltage,
\[ I_{\text{total}}=\frac{V_{\text{battery}}}{R_{\text{eq}}} \]
Step 4: Work backwards through the original circuit.
Restore the original resistors, one group at a time.
- For a resistor in series with the whole circuit, use the total current to find its voltage drop.
- For a parallel group, use the fact that every branch has the same voltage.
- For each branch, calculate its current using \(I=V/R\).
- For resistors in series within a branch, the branch current is the same through each resistor.
Step 5: Check the result before moving on.
At every junction, the currents must balance. Around every complete loop, the voltage drops must add to the battery’s potential difference.
These checks are not optional extras. They catch errors that can survive even when your equivalent resistance looks reasonable.
If a question also asks you to place meters, remember that an ammeter goes in series and a voltmeter goes in parallel. See Why Ammeters Go in Series and Voltmeters in Parallel.
03Worked example: Find the current in each branch
A 12.0 V ideal battery is connected to a 2.0 Ω resistor in series with a parallel pair of 6.0 Ω and 3.0 Ω resistors. Find the equivalent resistance of the whole circuit, the total current, and the current through each parallel branch.
Step 1: Simplify the parallel section.
The 6.0 Ω and 3.0 Ω resistors share both end nodes, so they are in parallel.
Using the two-resistor formula,
\[ \begin{aligned} R_{\text{parallel}} &=\frac{R_1R_2}{R_1+R_2}\\\\[4pt] &=\frac{(6.0)(3.0)}{6.0+3.0}\\,\Omega\\\\[4pt] &=2.0\\,\Omega \end{aligned} \]
The parallel pair behaves like a single 2.0 Ω resistor.
Step 2: Find the total resistance.
The 2.0 Ω resistor is in series with the equivalent 2.0 Ω parallel section.
\[ \begin{aligned} R_{\text{eq}} &=2.0+2.0\\\ &=4.0\\,\Omega \end{aligned} \]
Step 3: Calculate the total current.
The battery supplies 12.0 V across the equivalent 4.0 Ω circuit.
\[ \begin{aligned} I_{\text{total}} &=\frac{V_{\text{battery}}}{R_{\text{eq}}}\\\\[4pt] &=\frac{12.0}{4.0}\\\ &=3.0\\,\text{A} \end{aligned} \]
This is the current through the series resistor and the current entering the parallel section.
Step 4: Find the voltage across the parallel section.
The 2.0 Ω series resistor carries 3.0 A. Its voltage drop is
\[ \begin{aligned} V_{\text{series}}&=IR\\\ &=(3.0)(2.0)\\\ &=6.0\\,\text{V} \end{aligned} \]
The battery provides 12.0 V in total, so the remaining potential difference across the parallel section is
\[ V_{\text{parallel}}=12.0-6.0=6.0\\,\text{V} \]
Both parallel branches therefore have a potential difference of 6.0 V.
Step 5: Calculate the branch currents.
For the 6.0 Ω branch,
\[ I_6=\frac{6.0\\,\text{V}}{6.0\\,\Omega} =1.0\\,\text{A} \]
For the 3.0 Ω branch,
\[ I_3=\frac{6.0\\,\text{V}}{3.0\\,\Omega} =2.0\\,\text{A} \]
The lower-resistance branch carries more current because both branches have the same voltage across them.
Step 6: Check the answer.
The branch currents must add to the total current:
\[ I_6+I_3=1.0+2.0=3.0\\,\text{A} \]
The voltage drops must also add to the battery voltage along either complete path. For the path through the 6.0 Ω resistor, the drops are \(6.0+6.0=12.0\) V. The same is true for the path through the 3.0 Ω resistor.
Both checks work, so the answer is consistent.
Combination circuit: current splits, voltage is shared
| Complete circuit and shared series resistor | 12.0 V → 2.0 Ω → (6.0 Ω ∥ 3.0 Ω) Equivalent resistance: 4.0 Ω Incoming and recombined current: 3.0 A Voltage drop across 2.0 Ω resistor: 6.0 V |
|---|---|
| Parallel branches between the same two nodes | V₆Ω = V₃Ω = 6.0 V 6.0 Ω branch current: 1.0 A 3.0 Ω branch current: 2.0 A Junction check: 1.0 A + 2.0 A = 3.0 A |
04Worked example: Solve a nested combination circuit
A 24.0 V ideal battery is connected to a 2.0 Ω resistor in series with a parallel network. The upper branch contains a 4.0 Ω resistor in series with two 4.0 Ω resistors in parallel. The lower branch contains a single 3.0 Ω resistor.
Find the total current and the current through every branch and resistor.
Step 1: Reduce the innermost parallel pair.
The two 4.0 Ω resistors share the same two nodes.
\[ \begin{aligned} R_{\text{inner}} &=\frac{(4.0)(4.0)}{4.0+4.0}\\,\Omega\\\\[4pt] &=2.0\\,\Omega \end{aligned} \]
Step 2: Combine the upper branch’s series resistors.
The upper branch now contains a 4.0 Ω resistor in series with the equivalent 2.0 Ω resistance.
\[ R_{\text{upper}}=4.0+2.0=6.0\\,\Omega \]
The lower branch has a resistance of 3.0 Ω.
Step 3: Reduce the outer parallel network.
The upper and lower branches are connected between the same two nodes, so they are in parallel.
\[ \begin{aligned} R_{\text{outer}} &=\frac{(6.0)(3.0)}{6.0+3.0}\\,\Omega\\\\[4pt] &=2.0\\,\Omega \end{aligned} \]
The whole circuit is now a 2.0 Ω series resistor followed by an equivalent 2.0 Ω parallel network.
\[ R_{\text{eq}}=2.0+2.0=4.0\\,\Omega \]
Step 4: Calculate the total current.
\[ \begin{aligned} I_{\text{total}} &=\frac{24.0\\,\text{V}}{4.0\\,\Omega}\\\ &=6.0\\,\text{A} \end{aligned} \]
The 2.0 Ω series resistor therefore has a voltage drop of
\[ V_{\text{series}}=(6.0)(2.0)=12.0\\,\text{V} \]
That leaves 12.0 V across the outer parallel network.
Step 5: Calculate the outer branch currents.
The lower branch contains 3.0 Ω, so its current is
\[ I_{\text{lower}}=\frac{12.0}{3.0}=4.0\\,\text{A} \]
The upper branch has an equivalent resistance of 6.0 Ω, giving
\[ I_{\text{upper}}=\frac{12.0}{6.0}=2.0\\,\text{A} \]
The junction check works:
\[ I_{\text{total}}=4.0+2.0=6.0\\,\text{A} \]
Step 6: Work backwards through the upper branch.
The upper branch current of 2.0 A passes through its 4.0 Ω series resistor, giving a voltage drop of
\[ V_{\text{upper series}}=(2.0)(4.0)=8.0\\,\text{V} \]
The upper branch has 12.0 V across it altogether. The remaining voltage across the inner parallel pair is
\[ V_{\text{inner}}=12.0-8.0=4.0\\,\text{V} \]
Each inner 4.0 Ω resistor therefore carries
\[ I_{\text{inner}}=\frac{4.0}{4.0}=1.0\\,\text{A} \]
The two inner currents add to 2.0 A, matching the current entering that part of the upper branch.
The important habit here is to restore the circuit one layer at a time. Don’t try to calculate every current from the original diagram in one go.
Nested circuit: follow the current and voltage
| Whole circuit and outer parallel split | Rtotal = 2.0 Ω + (6.0 Ω ∥ 3.0 Ω) = 4.0 Ω Battery: 24.0 V; total current: 6.0 A Series 2.0 Ω drop: 12.0 V; outer network voltage: 12.0 V Upper branch: 2.0 A through 4.0 Ω; lower branch: 4.0 A through 3.0 Ω |
|---|---|
| Upper branch: inner parallel pair | 4.0 Ω + (4.0 Ω ∥ 4.0 Ω) = 6.0 Ω Upper series 4.0 Ω drop: 8.0 V Each inner 4.0 Ω resistor has 4.0 V across it and carries 1.0 A |
05Conservation checks before you commit to an answer
Use these checks on every combination-circuit question, even if the arithmetic seems straightforward.
1. Check current at every junction
The current entering a junction must equal the total current leaving it.
\[ I_{\text{in}}=I_1+I_2+\cdots \]
If 5.0 A enters a junction and the outgoing branch currents are 2.0 A and 4.0 A, something is wrong. The outgoing currents add to 6.0 A, not 5.0 A.
Also check where branches recombine. Their currents must add back to the current flowing through the shared series section.
2. Check voltage around a complete loop
The sum of the potential differences across the resistors in a complete path must equal the battery’s potential difference, assuming an ideal battery and connecting wires.
For a path through a series resistor and a parallel section,
\[ V_{\text{battery}} = V_{\text{series}}+V_{\text{parallel}} \]
If the battery is 18.0 V, a 7.0 V drop across the series resistor leaves 11.0 V across the parallel section.
Do not add the voltage drops of two alternative parallel branches together. They are separate paths, not consecutive parts of the same path.
3. Check equivalent resistance before calculating current
Two quick rules catch many mistakes:
- A series combination must have a resistance greater than either individual resistor.
- A parallel combination must have a resistance smaller than its smallest individual resistor.
For example, two 6.0 Ω resistors in parallel have an equivalent resistance of 3.0 Ω, not 12.0 Ω. Adding a new parallel branch reduces the equivalent resistance, rather than increasing it.
For a fixed ideal battery voltage, that reduction increases the total current.
4. Check each resistor with Ohm’s law
For every resistor, verify that
\[ I=\frac{V}{R} \]
Use the potential difference across that particular resistor, not automatically the full battery voltage.
This is the check that catches the opening mistake: a resistor in a parallel section may have less voltage across it than the battery supplies if another resistor sits in series with the entire section.
Three checks for a combination circuit
| 1. Junction current check | 3.0 A = 1.0 A + 2.0 A Incoming current equals the sum of outgoing branch currents. Recombined current = 3.0 A. |
|---|---|
| 2. Complete-loop voltage check | 12.0 V = 6.0 V + 6.0 V Series 2.0 Ω resistor drop = 6.0 V. Either loop adds the remaining 6.0 V branch drop. |
| 3. Parallel voltage check | V₆Ω = V₃Ω = 6.0 V Both branches connect to the same two nodes. Different branch currents: 1.0 A and 2.0 A. |
06Questions and solutions
Question 1
A 12.0 V ideal battery is connected directly across 6.0 Ω and 3.0 Ω resistors in parallel. Calculate the equivalent resistance, the current through each resistor, and the total current supplied by the battery.
Solution 1
The equivalent resistance is 2.0 Ω, the branch currents are 2.0 A and 4.0 A, and the total current is 6.0 A.
The equivalent resistance is
\[ \begin{aligned} R_{\text{eq}} &=\frac{(6.0)(3.0)}{6.0+3.0}\\,\Omega\\\ &=2.0\\,\Omega \end{aligned} \]
Both resistors have the full 12.0 V across them because they are connected directly across the same two nodes.
For the 6.0 Ω resistor,
\[ I_6=\frac{12.0}{6.0}=2.0\\,\text{A} \]
For the 3.0 Ω resistor,
\[ I_3=\frac{12.0}{3.0}=4.0\\,\text{A} \]
The total current is the sum of the branch currents:
\[ I_{\text{total}}=2.0+4.0=6.0\\,\text{A} \]
As a final check, \(V/I_{\text{total}}=12.0/6.0=2.0\\,\Omega\), which matches the equivalent resistance.
Question 2
An ideal 24.0 V battery is connected to a 3.0 Ω resistor in series with a parallel network. The upper branch contains a 6.0 Ω resistor. The lower branch contains 2.0 Ω and 4.0 Ω resistors in series.
Calculate the total resistance, total current, current through every resistor, and potential difference across every resistor.
Solution 2
The total resistance is 6.0 Ω, the total current is 4.0 A, and each outer branch carries 2.0 A.
First, combine the two resistors in the lower branch:
\[ R_{\text{lower}}=2.0+4.0=6.0\\,\Omega \]
The upper and lower branches are therefore both 6.0 Ω. Their parallel equivalent is
\[ \begin{aligned} R_{\text{parallel}} &=\frac{(6.0)(6.0)}{6.0+6.0}\\,\Omega\\\ &=3.0\\,\Omega \end{aligned} \]
Adding the 3.0 Ω series resistor gives
\[ R_{\text{eq}}=3.0+3.0=6.0\\,\Omega \]
The total current is
\[ I_{\text{total}}=\frac{24.0}{6.0}=4.0\\,\text{A} \]
The series resistor has a voltage drop of
\[ V_{\text{series}}=(4.0)(3.0)=12.0\\,\text{V} \]
The parallel network receives the remaining 12.0 V. Each outer branch therefore carries
\[ I_{\text{branch}}=\frac{12.0}{6.0}=2.0\\,\text{A} \]
The upper 6.0 Ω resistor has 12.0 V across it. In the lower branch, the same 2.0 A passes through both series resistors:
\[ V_{2\Omega}=(2.0)(2.0)=4.0\\,\text{V} \]
\[ V_{4\Omega}=(2.0)(4.0)=8.0\\,\text{V} \]
The lower branch’s voltage drops add to 12.0 V, and the two outer branch currents add to 4.0 A. Both conservation checks are satisfied.
Question 3
A 12.0 V ideal battery is connected to a 2.0 Ω resistor in series with a parallel network. One branch contains a 4.0 Ω resistor, while the other contains an unknown resistance \(R\). The measured total current is 4.0 A.
Determine \(R\) and the current through each parallel branch.
Solution 3
The unknown resistance is \(\frac{4}{3}\\,\Omega\), the known branch carries 1.0 A, and the unknown branch carries 3.0 A.
Start by finding the equivalent resistance of the whole circuit:
\[ R_{\text{eq}}=\frac{12.0}{4.0}=3.0\\,\Omega \]
The 2.0 Ω resistor is in series with the parallel network, so the network’s equivalent resistance must be
\[ R_{\text{parallel}}=3.0-2.0=1.0\\,\Omega \]
Next, find the voltage across the parallel network. The series resistor carries the total current of 4.0 A, so its voltage drop is
\[ V_{\text{series}}=(4.0)(2.0)=8.0\\,\text{V} \]
The remaining voltage is
\[ V_{\text{parallel}}=12.0-8.0=4.0\\,\text{V} \]
The known 4.0 Ω branch carries
\[ I_4=\frac{4.0}{4.0}=1.0\\,\text{A} \]
The unknown branch must carry the remaining current:
\[ I_R=4.0-1.0=3.0\\,\text{A} \]
Using Ohm’s law,
\[ R=\frac{4.0\\,\text{V}}{3.0\\,\text{A}} =\frac{4}{3}\\,\Omega \approx1.33\\,\Omega \]
Notice that the unknown resistance is smaller than 4.0 Ω, as expected: the lower-resistance branch carries more current when the branch voltages are equal.
The equivalent-resistance check also works:
\[ \frac{1}{R_{\text{parallel}}} =\frac{1}{4.0}+\frac{1}{4/3} =1.0\\,\Omega^{-1} \]
Thus \(R_{\text{parallel}}=1.0\\,\Omega\), consistent with the original circuit.
Question 4
An ideal 24.0 V battery feeds a 4.0 Ω series resistor, followed by a parallel section. One branch contains an 8.0 Ω resistor. The other contains 6.0 Ω and 2.0 Ω resistors in series.
A student reports the following readings:
- Total current: 3.0 A
- Voltage across the 4.0 Ω series resistor: 12.0 V
- Voltage across the 8.0 Ω branch: 12.0 V
- Voltage across the 6.0 Ω resistor: 9.0 V
- Voltage across the 2.0 Ω resistor: 4.0 V
Exactly one reading is wrong. Identify it, calculate the correct value, and explain which conservation rules expose the mistake.
Check the voltage readings in a combination circuit
| Circuit topology and reported readings | 24.0 V → 4.0 Ω → [8.0 Ω ∥ (6.0 Ω + 2.0 Ω)] Total current: 3.0 A; series 4.0 Ω reading: 12.0 V Upper 8.0 Ω branch reading: 12.0 V Lower branch: 6.0 Ω reading: 9.0 V; 2.0 Ω reading: 4.0 V |
|---|---|
| Conservation checks | Vparallel = 12.0 V; Ibranch = 1.5 A Correct 6.0 Ω drop: 1.5 A × 6.0 Ω = 9.0 V Correct 2.0 Ω drop: 1.5 A × 2.0 Ω = 3.0 V Loop check: 12.0 V + 9.0 V + 3.0 V = 24.0 V |
Solution 4
The reported 4.0 V across the 2.0 Ω resistor is wrong; the correct voltage is 3.0 V.
The total current of 3.0 A passes through the 4.0 Ω series resistor. Its voltage drop is
\[ V_{\text{series}}=(3.0)(4.0)=12.0\\,\text{V} \]
This leaves 12.0 V across the parallel section, consistent with the reported voltage across the 8.0 Ω branch.
The current in that branch is
\[ I_8=\frac{12.0}{8.0}=1.5\\,\text{A} \]
The other branch has a total resistance of
\[ R_{\text{other}}=6.0+2.0=8.0\\,\Omega \]
It also carries 1.5 A because it has the same 12.0 V across its 8.0 Ω total resistance.
The correct voltage across the 2.0 Ω resistor is therefore
\[ V_{2\Omega}=(1.5)(2.0)=3.0\\,\text{V} \]
The 6.0 Ω resistor’s reported 9.0 V drop is correct:
\[ V_{6\Omega}=(1.5)(6.0)=9.0\\,\text{V} \]
The two drops in that branch must add to the 12.0 V across the branch. The correct values give \(9.0+3.0=12.0\) V, while the reported values give \(9.0+4.0=13.0\) V.
That extra volt cannot appear from nowhere. The incorrect reading violates the voltage rule for a complete loop, while the current readings remain consistent with conservation of current at the junction.
Question 5
A 12.0 V ideal battery feeds a 2.0 Ω resistor in series with a two-branch network.
- Branch A contains a single 4.0 Ω resistor.
- Branch B contains a 2.0 Ω resistor in series with an original 4.0 Ω resistor.
A switch then connects a second 4.0 Ω resistor directly in parallel with the original 4.0 Ω resistor in branch B.
Calculate the total current, the current in branch A, and the current through the original 4.0 Ω resistor in branch B, both before and after the switch closes. Determine which currents rise or fall, and explain how the total current can increase even if some individual currents decrease.
Closing a switch changes branch currents
| BEFORE — Switch open | 12.0 V → shared 2.0 Ω → outer nodes A–B; A: 4.0 Ω ∥ B: (2.0 Ω + original 4.0 Ω) Total: 2.73 A; branch A: 1.64 A; branch B: 1.09 A. Original 4.0 Ω current: 1.09 A; outer junctions: A–B. |
|---|---|
| AFTER — Switch closed | Outer nodes A–B; inner nodes C–D connect original and added 4.0 Ω resistors in parallel. Total: 3.00 A; branch A: 1.50 A; branch B: 1.50 A. Original 4.0 Ω current: 0.75 A; added 4.0 Ω current: 0.75 A. |
Solution 5
The total current rises from approximately 2.73 A to 3.00 A. However, the current in branch A falls from approximately 1.64 A to 1.50 A, and the current through the original 4.0 Ω resistor in branch B falls from approximately 1.09 A to 0.75 A.
The key is to track both levels of the circuit: the outer parallel network and the inner parallel pair created by the switch.
Before the switch closes
Branch B contains two series resistors, so its resistance is
\[ R_B=2.0+4.0=6.0\\,\Omega \]
Branch A has a resistance of 4.0 Ω. The equivalent resistance of the outer parallel network is
\[ \begin{aligned} R_{\text{parallel}} &=\frac{(4.0)(6.0)}{4.0+6.0}\\,\Omega\\\ &=2.4\\,\Omega \end{aligned} \]
Including the shared 2.0 Ω resistor,
\[ R_{\text{eq}}=2.0+2.4=4.4\\,\Omega \]
The total current is
\[ \begin{aligned} I_{\text{total}} &=\frac{12.0}{4.4}\\\ &=\frac{30}{11}\\,\text{A}\\\ &\approx2.73\\,\text{A} \end{aligned} \]
The voltage across the outer parallel network is
\[ \begin{aligned} V_{\text{parallel}} &=I_{\text{total}}R_{\text{parallel}}\\\ &=\frac{30}{11}\times2.4\\\ &=\frac{72}{11}\\,\text{V}\\\ &\approx6.55\\,\text{V} \end{aligned} \]
The current in branch A is
\[ I_A=\frac{72/11}{4.0} =\frac{18}{11}\\,\text{A} \approx1.64\\,\text{A} \]
The current in branch B is
\[ I_B=\frac{72/11}{6.0} =\frac{12}{11}\\,\text{A} \approx1.09\\,\text{A} \]
Before the switch closes, the original 4.0 Ω resistor is in series with the 2.0 Ω resistor in branch B. It therefore carries the same 1.09 A current as branch B.
After the switch closes
The original 4.0 Ω resistor now shares its two end nodes with another 4.0 Ω resistor. The inner parallel pair has an equivalent resistance of
\[ R_{\text{inner}} =\frac{(4.0)(4.0)}{4.0+4.0} =2.0\\,\Omega \]
Branch B now has a total resistance of
\[ R_B=2.0+2.0=4.0\\,\Omega \]
Both outer branches now have a resistance of 4.0 Ω. Their parallel equivalent is
\[ R_{\text{parallel}} =\frac{(4.0)(4.0)}{4.0+4.0} =2.0\\,\Omega \]
The total circuit resistance is
\[ R_{\text{eq}}=2.0+2.0=4.0\\,\Omega \]
The total current therefore rises to
\[ I_{\text{total}}=\frac{12.0}{4.0}=3.00\\,\text{A} \]
The shared series resistor has a voltage drop of
\[ V_{\text{series}}=(3.00)(2.0)=6.0\\,\text{V} \]
This leaves 6.0 V across the outer parallel network. Each outer branch carries
\[ I_A=I_B=\frac{6.0}{4.0}=1.50\\,\text{A} \]
Branch A’s current has fallen from 1.64 A to 1.50 A, even though the total current has risen.
Now look inside branch B. Its 2.0 Ω series resistor carries 1.50 A, so its voltage drop is
\[ V_{2\Omega}=(1.50)(2.0)=3.0\\,\text{V} \]
The inner parallel pair receives the remaining 3.0 V. Each 4.0 Ω resistor therefore carries
\[ I_{4\Omega}=\frac{3.0}{4.0}=0.75\\,\text{A} \]
The original 4.0 Ω resistor’s current has fallen from approximately 1.09 A to 0.75 A.
Why the tempting prediction fails
It is tempting to think that adding a parallel resistor must increase the current through every existing resistor. Adding a parallel path does lower the equivalent resistance of the network, and the total current does increase. But that does not mean every branch current increases.
The shared 2.0 Ω resistor now carries more current, so its voltage drop rises from approximately 5.45 V to 6.0 V. The voltage left across the outer parallel network consequently falls from approximately 6.55 V to 6.0 V. That reduces the current in branch A.
Inside branch B, the branch current increases, but it now splits between two equal 4.0 Ω resistors. Each receives only half the branch current, and the voltage across the pair has also fallen.
The current-conservation check confirms the result. Before the switch closes,
\[ \frac{18}{11}+\frac{12}{11} =\frac{30}{11}\\,\text{A} \]
After it closes,
\[ 1.50+1.50=3.00\\,\text{A} \]
Branch A loses \(\frac{3}{22}\) A, but branch B gains \(\frac{9}{22}\) A. The increase in branch B is larger than the decrease in branch A, so the total current rises by \(\frac{3}{11}\) A.
The voltage checks work as well: after the switch closes, the 3.0 V drop across the 2.0 Ω resistor in branch B plus the 3.0 V across the inner parallel pair gives the full 6.0 V across that branch.
These are the two conservation checks to trust when a result looks surprising: currents must balance at every junction, and voltage drops must balance around every complete loop. Once you can apply both reliably, the next step is using Kirchhoff’s junction and loop rules to analyse circuits that cannot be simplified into straightforward series and parallel groups.