How to Redraw a Confusing Circuit Without Changing Connections

A circuit diagram can be completely rearranged without changing how the circuit behaves. You can move resistors to different sides of the page, straighten curved wires, and spread out crowded junctions. The circuit remains electrically identical as long as every component stays connected to the same points.

The tricky part is knowing which points are actually connected.

A resistor drawn beside another resistor is not necessarily in parallel with it. Two resistors drawn end-to-end are not necessarily in series. Even a tiny junction dot can change the entire circuit.

The reliable method is to identify the nodes first, check current and voltage conservation second, and simplify resistances last.

01Redraw the connections, not the shape

Imagine a circuit where a wire leaves a resistor, loops around the battery, and eventually connects to another component. The wire might look unnecessarily complicated, but electrically, its shape does not matter.

For an ideal circuit, an uninterrupted wire has zero resistance. Every point along that wire is therefore at the same electrical potential.

All points connected by one continuous conductor belong to the same node.

What exactly is a node?

A node is a continuous region of conducting wire connecting component terminals without passing through another circuit component.

For example, imagine a wire connecting the positive terminal of a battery to one end of a resistor. The entire uninterrupted wire, including any bends and branches, is one node.

If you follow the wire around the circuit and reach another resistor, you have reached a component terminal. Passing through that resistor takes you to a different node.

This gives you the first important rule:

You can change the shape and position of a wire, but you cannot change which component terminals it connects.

What happens when wires cross?

This is one of the easiest places to make a mistake.

A crossing does not automatically mean two wires are connected.

In conventional circuit diagrams:

• A filled junction dot indicates that the wires are connected.
• A wire crossing with an explicit bridge or crossover indicates that the wires are not connected.
• A T-junction normally indicates a connection, even when a dot is not shown.

Always follow the diagram’s stated connection convention. If an unclear crossing has no reliable indication of whether the wires join, don’t invent a connection.

The five-step redraw method

Use this process whenever a circuit looks confusing.

Step 1: Find every continuous conductor.

Start at a component terminal and follow the wire until you reach another component terminal or an open connection. Include every bend and branch along the way.

Step 2: Give each node a label.

Use letters such as A, B, C, and D. Every point on the same uninterrupted conductor receives the same label.

Do not give different labels to two points simply because they are drawn far apart.

Step 3: Record which components connect each pair of nodes.

Create a small connection table.

The values are not important yet. What matters is recording the endpoints correctly.

For a battery or other voltage source, also record its positive and negative terminals. Rotating a battery symbol does not reverse its polarity.

Step 4: Draw a cleaner circuit using the same node labels.

Place the nodes wherever they make the circuit easiest to understand. Connect each component between exactly the same pair of nodes as in your table.

You can move a resistor, rotate it, or draw a wire in a completely different direction.

You cannot move one of its terminals to a different node.

Step 5: Check the connections using Kirchhoff’s laws.

Before combining resistances, check that current is conserved at every junction and that voltage changes balance around every closed loop.

This final step catches mistakes that a neat-looking drawing might hide.

02Use conservation before simplifying resistances

Why not immediately look for series and parallel resistors?

Because you might identify the wrong components as being in series or parallel if you haven’t established the actual connections.

Two conservation laws give you a way to check your redraw.

Kirchhoff’s current law: check the junctions

Kirchhoff’s current law (KCL) follows from conservation of electric charge.

Current entering a junction must equal current leaving it, assuming steady-state conditions.

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

For example, suppose 3.0 A enters a junction and divides between two branches.

If one branch carries 1.2 A, the other must carry:

\[
I_2=3.0-1.2
\]

\[
\boxed{I_2=1.8\text{ A}}
\]

The current has divided, but no charge has disappeared.

When checking a redraw, make sure the junctions you’ve identified allow the correct current paths.

Kirchhoff’s voltage law: check the loops

Kirchhoff’s voltage law (KVL) follows from conservation of energy.

The sum of all potential changes around a closed loop must equal zero.

\[
\sum \Delta V=0
\]

For a simple loop containing a battery and two resistors:

\[
\mathcal E-I R_1-I R_2=0
\]

Here:

• \(\mathcal E\) is the battery’s electromotive force (emf), measured in volts (V).
• \(I\) is the conventional current, measured in amperes (A).
• \(R_1\) and \(R_2\) are resistances, measured in ohms (Ω).

The sign convention is straightforward:

• Moving through a battery from its negative to positive terminal gives a voltage rise, \(+\mathcal E\).
• Moving through a resistor in the direction of conventional current gives a voltage drop, \(-IR\).
• Moving against the current through a resistor gives a voltage rise, \(+IR\).

Choose a direction around the loop and remain consistent.

If your proposed circuit connections produce inconsistent junction currents or loop voltages, revisit the node labels and connection table.

Only then identify series and parallel resistors

Once the connections are clear, deciding whether resistors can be combined becomes much easier.

Notice the difference.

Two resistors sharing one node are not automatically in series. There must be no additional current path leaving that shared node.

Similarly, two resistors sharing one node are not automatically parallel. They must share both endpoint nodes.

For more practice calculating how current divides between parallel branches, see How to Find Branch Currents in a Three-Branch Parallel Circuit.

03Experimentally checking whether two points are connected

Sometimes you are given an actual circuit rather than a schematic. You might need to determine whether two terminals belong to the same continuous conductor.

A multimeter can help, but there is an important limitation.

Setting up the continuity test

Consider a low-voltage teaching circuit with two points, X and Y, that you suspect belong to the same node.

Disconnect the power supply before testing. Set the multimeter to resistance or continuity mode, then place one probe on X and the other on Y.

What should you predict?

If X and Y are connected by an uninterrupted conductor, their resistance should be very close to zero in the ideal-circuit model.

Consider these illustrative readings:

The estimated conductor resistance is:

\[
R_{XY}=0.3-0.2
\]

\[
\boxed{R_{XY}=0.1\ \Omega}
\]

This is consistent with a very low-resistance connection.

However, it does not prove that X and Y are the same node.

Why the original test can be misleading

A continuity tester detects a low-resistance path. That path might travel through another component or around a different branch of the circuit.

A resistor can also produce a continuity reading, depending on its resistance and the meter’s threshold.

To improve the test, isolate the suspected conductor from other possible current paths before interpreting the reading.

Never use resistance or continuity mode on an energised circuit.

The key distinction is that a continuity measurement provides evidence about electrical resistance. The actual circuit connections determine the node structure.

For more on why measurements from real circuits can differ from ideal predictions, see Why Circuit Practical Results Differ from Ideal Calculations.

04Worked examples: applying the redraw method

A common trap: two resistors that look like they are in series

Consider two resistors, \(R_1\) and \(R_2\), drawn end-to-end.

A student might immediately assume they are in series because they appear to form one continuous path.

But suppose a third resistor, \(R_3\), also connects to their shared junction B.

In the first arrangement, the current entering B divides:

\[
I_1=I_2+I_3
\]

Therefore, \(R_1\) and \(R_2\) do not carry the same current and cannot be treated as a simple series pair.

In the second arrangement, there is no additional path leaving B. The same current must pass through both resistors.

The lesson is simple: look at the number of current paths leaving a node, not the apparent position of the resistors.

Why can’t you just treat the bridge as two parallel branches?

A tempting approach is to ignore the middle resistor and calculate two independent series paths:

\[
R_{\text{upper}}=2+8=10\ \Omega
\]

\[
R_{\text{lower}}=3+2=5\ \Omega
\]

This would give:

\[
R_{\text{eq}}=\frac{10\times5}{10+5}
\]

\[
R_{\text{eq}}=3.33\ \Omega
\]

But this is incorrect because the bridge resistor provides an additional current path between B and C.

The actual equivalent resistance is 3.0 Ω.

When a bridge resistor connects two intermediate junctions, you cannot automatically treat the upper and lower paths as independent parallel branches.

You need to establish the current distribution using Kirchhoff’s laws.

Same node or just the same potential?

There is one more distinction worth remembering.

Two points connected by an uninterrupted ideal wire belong to the same node. They must have the same potential.

However, two separate nodes can also happen to have the same potential.

For example, in a balanced bridge, the two middle junctions can have equal potentials even though they are connected by a resistor rather than an uninterrupted wire.

Do not merge two nodes simply because a calculation gives them equal potentials. Node identity comes from the connections, not from the numerical voltage.

05Questions and solutions

The following questions move from straightforward series-parallel identification to more demanding circuits involving junctions, changing connections, and bridge networks.

Question 1

An ideal 9.0 V battery has its positive terminal at A and negative terminal at D.

Three resistors are connected:

• \(R_1=3.0\ \Omega\) between A and B.
• \(R_2=6.0\ \Omega\) between B and D.
• \(R_3=3.0\ \Omega\) between B and D.

The original drawing places the three resistors in different positions around the page.

Redraw the circuit and calculate:

(a) The equivalent resistance.

(b) The total current supplied by the battery.

(c) The current through each parallel resistor.

Solution 1

The equivalent resistance is 5.0 Ω, the total current is 1.8 A, and the branch currents are 0.60 A through the 6.0 Ω resistor and 1.20 A through the 3.0 Ω resistor.

Part (a): Equivalent resistance

Both \(R_2\) and \(R_3\) connect between B and D.

They are therefore parallel.

\[
R_{23}=\frac{6\times3}{6+3}
\]

\[
R_{23}=\frac{18}{9}=2.0\ \Omega
\]

This parallel combination is in series with \(R_1\).

\[
R_{\text{eq}}=3+2
\]

\[
\boxed{R_{\text{eq}}=5.0\ \Omega}
\]

Part (b): Total current

Using Ohm’s law:

\[
I_T=\frac{V}{R_{\text{eq}}}
\]

\[
I_T=\frac{9.0}{5.0}
\]

\[
\boxed{I_T=1.8\text{ A}}
\]

Part (c): Branch currents

The potential difference across \(R_1\) is:

\[
V_1=IR
\]

\[
V_1=1.8\times3
\]

\[
V_1=5.4\text{ V}
\]

The remaining potential difference is across the parallel section:

\[
V_{BD}=9.0-5.4=3.6\text{ V}
\]

For the 6.0 Ω resistor:

\[
I_2=\frac{3.6}{6}
\]

\[
\boxed{I_2=0.60\text{ A}}
\]

For the 3.0 Ω resistor:

\[
I_3=\frac{3.6}{3}
\]

\[
\boxed{I_3=1.20\text{ A}}
\]

Check KCL:

\[
1.8=0.60+1.20
\]

The branch currents correctly add to the total current.

Question 2

An ideal 12.0 V battery is connected between nodes A and D.

The circuit contains:

• \(R_{AB}=2.0\ \Omega\).
• \(R_{BD}=4.0\ \Omega\).
• \(R_{BC}=2.0\ \Omega\).
• \(R_{CD}=2.0\ \Omega\).

A switch is placed between B and the \(R_{BC}\) branch.

Compare the circuit when the switch is open and when it is closed.

For each condition:

(a) Determine the equivalent resistance.

(b) Calculate the total current.

(c) Determine the current through each resistor.

(d) Explain whether \(R_{AB}\) and \(R_{BD}\) are in series.

Solution 2

With the switch open, the equivalent resistance is 6.0 Ω and the total current is 2.0 A. With the switch closed, the equivalent resistance becomes 4.0 Ω and the total current becomes 3.0 A.

Part (a): Switch open

When the switch is open, the B-C-D path is disconnected from the battery at B.

There is only one complete path for current:

\[
A\rightarrow B\rightarrow D
\]

Therefore, \(R_{AB}\) and \(R_{BD}\) are in series.

\[
R_{\text{eq}}=2+4
\]

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

Part (b): Total current with the switch open

\[
I_T=\frac{V}{R_{\text{eq}}}
\]

\[
I_T=\frac{12}{6}
\]

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

Both \(R_{AB}\) and \(R_{BD}\) carry 2.0 A.

No current flows through \(R_{BC}\) or \(R_{CD}\), because their branch is open.

Part (c): Switch closed

Closing the switch creates another complete path from B to D.

There are now two paths:

• Directly through \(R_{BD}=4.0\ \Omega\).
• Through \(R_{BC}\) and \(R_{CD}\) in series.

First combine the second path:

\[
R_{BCD}=2+2
\]

\[
R_{BCD}=4.0\ \Omega
\]

The two B-D paths are parallel:

\[
R_{BD,\text{eq}}=
\frac{4\times4}{4+4}
\]

\[
R_{BD,\text{eq}}=2.0\ \Omega
\]

This combination is in series with \(R_{AB}\).

\[
R_{\text{eq}}=2+2
\]

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

The total current is:

\[
I_T=\frac{12}{4}
\]

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

The voltage across the two parallel paths is:

\[
V_{BD}=I_T R_{BD,\text{eq}}
\]

\[
V_{BD}=3\times2
\]

\[
V_{BD}=6.0\text{ V}
\]

The direct 4.0 Ω branch carries:

\[
I_{BD}=\frac{6}{4}
\]

\[
\boxed{I_{BD}=1.5\text{ A}}
\]

The alternative path also has a total resistance of 4.0 Ω:

\[
I_{BCD}=\frac{6}{4}
\]

\[
\boxed{I_{BCD}=1.5\text{ A}}
\]

Because \(R_{BC}\) and \(R_{CD}\) are in series, each carries 1.5 A.

The current through \(R_{AB}\) is the total current:

\[
\boxed{I_{AB}=3.0\text{ A}}
\]

Part (d): Are \(R_{AB}\) and \(R_{BD}\) in series?

With the switch open, yes. They carry the same current because there is no other conducting path from B.

With the switch closed, no.

At B, the 3.0 A current divides:

\[
3.0=1.5+1.5
\]

The current through \(R_{AB}\) is therefore different from the current through \(R_{BD}\).

The physical positions of the resistors have not changed. The connection at B has changed, and that changes the circuit topology.

Question 3

A single closed loop contains:

• A 12.0 V ideal battery.
• A 6.0 V ideal battery.
• A 2.0 Ω resistor.
• A 4.0 Ω resistor.

The components are connected in series.

When travelling clockwise around the loop, you pass through the 12.0 V battery from its negative to positive terminal, then through both resistors in the direction of the assumed current, and finally through the 6.0 V battery from its positive to negative terminal.

The battery symbols are drawn at different orientations.

(a) Redraw the circuit, preserving both battery polarities.

(b) Calculate the current and state its direction.

(c) The 6.0 V battery is reversed so that both batteries now assist the clockwise current. Calculate the new current.

Solution 3

The original circuit carries 1.0 A clockwise. After reversing the 6.0 V battery, the current increases to 3.0 A clockwise.

Part (a): Redraw the circuit

Both resistors are in series because they form one uninterrupted current path.

The battery polarities must remain unchanged when the circuit is rearranged.

A battery drawn sideways is not automatically reversed. Its positive and negative terminals determine its polarity.

Part (b): Original current

The two batteries oppose each other.

The 12.0 V source produces a clockwise potential rise, while the 6.0 V source produces a clockwise potential drop.

The net emf is:

\[
\mathcal E_{\text{net}}=12-6
\]

\[
\mathcal E_{\text{net}}=6.0\text{ V}
\]

The total resistance is:

\[
R_{\text{eq}}=2+4
\]

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

Using Ohm’s law:

\[
I=\frac{\mathcal E_{\text{net}}}{R_{\text{eq}}}
\]

\[
I=\frac{6}{6}
\]

\[
\boxed{I=1.0\text{ A clockwise}}
\]

Check using KVL:

\[
12-2I-4I-6=0
\]

Substituting \(I=1\):

\[
12-2-4-6=0
\]

The loop satisfies conservation of energy.

Part (c): Reverse the 6.0 V battery

Now both batteries provide a potential rise when travelling clockwise.

The net emf becomes:

\[
\mathcal E_{\text{net}}=12+6
\]

\[
\mathcal E_{\text{net}}=18.0\text{ V}
\]

The resistors have not changed, so the total resistance remains 6.0 Ω.

\[
I=\frac{18}{6}
\]

\[
\boxed{I=3.0\text{ A clockwise}}
\]

Reversing a battery changes its polarity, not the resistance of the circuit.

This is why you must preserve source orientation as well as component connections when redrawing a circuit.

Question 4

An ideal 12.0 V battery connects nodes A (positive) and D (negative).

Four resistors are connected:

In the original drawing, the wires from B and C cross using a bridge symbol. There is no electrical connection at the crossing.

(a) Redraw the original circuit and calculate its equivalent resistance and total current.

(b) Calculate the potentials at B and C relative to D.

(c) A filled junction dot is added at the crossing, connecting B and C. Calculate the new equivalent resistance and total current.

Explain why this small drawing change produces a different result.

Solution 4

Without the junction dot, the equivalent resistance is 4.5 Ω and the total current is 2.67 A. With the junction dot, the equivalent resistance becomes 4.0 Ω and the total current becomes 3.0 A.

Part (a): Original circuit without the junction

The crossing does not connect B and C.

There are two separate branches between A and D.

The upper branch contains:

\[
R_{\text{upper}}=3+6
\]

\[
R_{\text{upper}}=9.0\ \Omega
\]

The lower branch contains:

\[
R_{\text{lower}}=6+3
\]

\[
R_{\text{lower}}=9.0\ \Omega
\]

These two branches are parallel.

\[
R_{\text{eq}}=
\frac{9\times9}{9+9}
\]

\[
\boxed{R_{\text{eq}}=4.5\ \Omega}
\]

The total current is:

\[
I_T=\frac{12}{4.5}
\]

\[
\boxed{I_T=2.67\text{ A}}
\]

Part (b): Potentials at B and C

Each branch has 12.0 V across it.

The current in each branch is:

\[
I_{\text{branch}}=\frac{12}{9}
\]

\[
I_{\text{branch}}=1.33\text{ A}
\]

Take D as the zero-potential reference.

The potential at B is the voltage across \(R_{BD}\):

\[
V_B=1.33\times6
\]

\[
\boxed{V_B=8.0\text{ V}}
\]

The potential at C is the voltage across \(R_{CD}\):

\[
V_C=1.33\times3
\]

\[
\boxed{V_C=4.0\text{ V}}
\]

Notice that B and C have different potentials.

This confirms why they cannot be treated as one node in the original circuit.

Part (c): Add the junction dot

The dot connects B and C with an ideal conductor.

They now become one node, which we can label X.

The two resistors between A and X are parallel:

\[
R_{AX}=\frac{3\times6}{3+6}
\]

\[
R_{AX}=2.0\ \Omega
\]

The two resistors between X and D are also parallel:

\[
R_{XD}=\frac{6\times3}{6+3}
\]

\[
R_{XD}=2.0\ \Omega
\]

These two equivalent resistances are in series:

\[
R_{\text{eq}}=2+2
\]

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

The new total current is:

\[
I_T=\frac{12}{4}
\]

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

The dot changes the circuit because it creates an additional conducting connection between B and C.

It is not a cosmetic change to the drawing. It changes the number of nodes and the available current paths.

Question 5

An ideal 12.0 V battery is connected between nodes A (positive) and D (negative).

A bridge circuit contains the following resistors:

The bridge resistor connects the two intermediate junctions B and C.

(a) Determine whether current flows through the 5.0 Ω bridge resistor. Calculate the equivalent resistance and total battery current.

(b) The battery voltage is doubled to 24.0 V, with all resistances unchanged. Does current now flow through the bridge resistor? Calculate the new total current.

(c) The battery remains at 24.0 V, but \(R_{AB}\) is increased from 4.0 Ω to 8.0 Ω. The bridge resistor remains connected and has a resistance of 5.0 Ω.

Calculate:

• The actual potentials at B and C.
• The current and direction through the bridge resistor.
• The total current supplied by the battery.
• The new equivalent resistance.

Explain why calculating the two middle-node potentials as though the bridge resistor were disconnected can predict the direction of the bridge current, but cannot give its actual magnitude.

Solution 5

Initially, the bridge carries zero current and the equivalent resistance is 3.6 Ω.

Doubling the battery voltage does not produce a bridge current because the two middle junctions remain at equal potentials.

After increasing \(R_{AB}\) to 8.0 Ω, the bridge carries approximately 0.372 A from C to B. The new equivalent resistance is approximately 4.69 Ω.

Part (a): Determine whether the original bridge carries current

The upper branch contains two series resistors:

\[
R_{\text{upper}}=4+2
\]

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

The lower branch also contains two series resistors:

\[
R_{\text{lower}}=6+3
\]

\[
R_{\text{lower}}=9.0\ \Omega
\]

Calculate the current in each branch.

Upper branch:

\[
I_{\text{upper}}=\frac{12}{6}
\]

\[
I_{\text{upper}}=2.0\text{ A}
\]

Lower branch:

\[
I_{\text{lower}}=\frac{12}{9}
\]

\[
I_{\text{lower}}=1.33\text{ A}
\]

Now calculate the potential at each middle junction relative to D.

For B:

\[
V_B=I_{\text{upper}}R_{BD}
\]

\[
V_B=2\times2
\]

\[
V_B=4.0\text{ V}
\]

For C:

\[
V_C=I_{\text{lower}}R_{CD}
\]

\[
V_C=\frac{12}{9}\times3
\]

\[
V_C=4.0\text{ V}
\]

Therefore:

\[
V_B=V_C
\]

The potential difference across the bridge resistor is:

\[
V_{BC}=V_B-V_C
\]

\[
V_{BC}=0\text{ V}
\]

Using Ohm’s law:

\[
I_{BC}=\frac{0}{5}
\]

\[
\boxed{I_{BC}=0\text{ A}}
\]

The bridge resistor carries no current because its two terminals have equal potentials.

This is called a balanced bridge.

Notice that B and C are still different nodes. They are connected through a 5.0 Ω resistor, not an uninterrupted wire.

The total battery current is:

\[
I_T=2+\frac{12}{9}
\]

\[
I_T=3.33\text{ A}
\]

The equivalent resistance is:

\[
R_{\text{eq}}=\frac{12}{3.33}
\]

Using the unrounded current:

\[
R_{\text{eq}}=\frac{12}{10/3}
\]

\[
\boxed{R_{\text{eq}}=3.6\ \Omega}
\]

Part (b): Double the battery voltage

The battery now supplies 24.0 V.

The resistor values and their connections have not changed.

The upper branch current becomes:

\[
I_{\text{upper}}=\frac{24}{6}
\]

\[
I_{\text{upper}}=4.0\text{ A}
\]

The lower branch current becomes:

\[
I_{\text{lower}}=\frac{24}{9}
\]

\[
I_{\text{lower}}=2.67\text{ A}
\]

Calculate the middle-node potentials again.

For B:

\[
V_B=4\times2
\]

\[
V_B=8.0\text{ V}
\]

For C:

\[
V_C=\frac{24}{9}\times3
\]

\[
V_C=8.0\text{ V}
\]

The two potentials are still equal.

Therefore:

\[
\boxed{I_{BC}=0\text{ A}}
\]

The new total current is:

\[
I_T=4+\frac{24}{9}
\]

\[
\boxed{I_T=6.67\text{ A}}
\]

The equivalent resistance remains:

\[
R_{\text{eq}}=\frac{24}{6.67}
\]

\[
\boxed{R_{\text{eq}}=3.6\ \Omega}
\]

Doubling the source voltage doubles the branch currents, but it does not change the balance of the bridge.

The reason is that the balance depends on the resistance ratios, not the absolute supply voltage.

Part (c): Increase \(R_{AB}\) to 8.0 Ω

This changes the resistance ratio of the upper branch.

The bridge is no longer balanced, so we expect a current through \(R_{BC}\).

However, we cannot simply calculate the two independent branch currents and assume they remain unchanged. The bridge resistor now provides a current path between B and C.

We must use KCL at both junctions.

Take:

\[
V_A=24\text{ V}
\]

\[
V_D=0\text{ V}
\]

Let the unknown middle-node potentials be \(V_B\) and \(V_C\).

Step 1: Write KCL at junction B.

Current enters B from A through the 8.0 Ω resistor.

It leaves through the 2.0 Ω resistor and the 5.0 Ω bridge resistor.

Therefore:

\[
\frac{24-V_B}{8}
=
\frac{V_B}{2}
+
\frac{V_B-V_C}{5}
\]

Multiply through by 40:

\[
5(24-V_B)=20V_B+8(V_B-V_C)
\]

Simplifying:

\[
\boxed{33V_B-8V_C=120}
\]

Step 2: Write KCL at junction C.

Current enters C from A and through the bridge resistor.

It leaves through the 3.0 Ω resistor.

Therefore:

\[
\frac{24-V_C}{6}
+
\frac{V_B-V_C}{5}
=
\frac{V_C}{3}
\]

Multiply through by 30:

\[
5(24-V_C)+6(V_B-V_C)=10V_C
\]

Simplifying:

\[
\boxed{6V_B-21V_C=-120}
\]

Step 3: Solve the simultaneous equations.

We have:

\[
33V_B-8V_C=120
\]

\[
6V_B-21V_C=-120
\]

Solving these equations gives:

\[
V_B=\frac{232}{43}
\]

\[
\boxed{V_B=5.40\text{ V}}
\]

And:

\[
V_C=\frac{312}{43}
\]

\[
\boxed{V_C=7.26\text{ V}}
\]

Step 4: Calculate the bridge current.

The potential difference across the bridge resistor is:

\[
V_C-V_B
=
7.26-5.40
\]

\[
V_C-V_B=1.86\text{ V}
\]

Using Ohm’s law:

\[
I_{CB}=\frac{V_C-V_B}{5}
\]

Using the exact values:

\[
I_{CB}=\frac{80}{43\times5}
\]

\[
\boxed{I_{CB}=0.372\text{ A}}
\]

Because C is at a higher potential than B, conventional current flows from C to B.

Step 5: Calculate the total battery current.

Current leaves the positive terminal through two branches.

Through the upper-left resistor:

\[
I_{AB}=\frac{24-V_B}{8}
\]

\[
I_{AB}=\frac{100}{43}
\]

\[
I_{AB}=2.33\text{ A}
\]

Through the lower-left resistor:

\[
I_{AC}=\frac{24-V_C}{6}
\]

\[
I_{AC}=\frac{120}{43}
\]

\[
I_{AC}=2.79\text{ A}
\]

Therefore:

\[
I_T=I_{AB}+I_{AC}
\]

\[
I_T=\frac{100}{43}+\frac{120}{43}
\]

\[
\boxed{I_T=5.12\text{ A}}
\]

Step 6: Calculate the new equivalent resistance.

Using the battery voltage and total current:

\[
R_{\text{eq}}=\frac{V}{I_T}
\]

\[
R_{\text{eq}}=
\frac{24}{220/43}
\]

\[
\boxed{R_{\text{eq}}=4.69\ \Omega}
\]

Step 7: Check the current at both junctions.

At B, current enters from A and C:

\[
I_{AB}+I_{CB}
=
\frac{100}{43}+\frac{16}{43}
\]

\[
=\frac{116}{43}\text{ A}
\]

This equals the current leaving B through the 2.0 Ω resistor:

\[
I_{BD}=\frac{V_B}{2}
=\frac{116}{43}\text{ A}
\]

At C, current entering from A equals the currents leaving towards B and D:

\[
\frac{120}{43}
=
\frac{16}{43}+\frac{104}{43}
\]

Both junctions satisfy KCL.

The important final distinction: predicted direction versus actual current

There is a subtle trap in the final part.

You might remove the bridge resistor temporarily and calculate the two middle-node potentials as though the upper and lower branches were independent.

With the modified 8.0 Ω upper-left resistor:

\[
V_{B,\text{open}}=
24\times\frac{2}{8+2}
\]

\[
V_{B,\text{open}}=4.8\text{ V}
\]

For the lower branch:

\[
V_{C,\text{open}}=
24\times\frac{3}{6+3}
\]

\[
V_{C,\text{open}}=8.0\text{ V}
\]

This tells you that C would be at a higher potential than B if the bridge were disconnected.

Therefore, when the bridge resistor is connected, current initially flows from C towards B.

That correctly predicts the direction.

But these open-circuit potentials are not the actual potentials once the 5.0 Ω resistor is connected. Current through the bridge changes the potential distribution in both branches.

The actual values must be found using KCL, which is why the final answer is 0.372 A rather than simply dividing the open-circuit potential difference by 5.0 Ω.

This is the difference between identifying the direction of a current and calculating its actual magnitude.

When you encounter an unfamiliar circuit in an HSC Physics question, resist the urge to combine resistors immediately. First establish the nodes, preserve every component connection and source polarity, and check the junctions and loops. Once the topology is correct, you can decide whether ordinary series-parallel reductions are sufficient or whether you need simultaneous equations using Kirchhoff’s laws.