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A \(6.0\ \Omega\) resistor is connected between points A and B. A separate wire leaves A, loops around the edge of the circuit diagram, and connects directly to B. What is the potential difference across the resistor?

It is \(0\ \mathrm{V}\), assuming an ideal wire. The wire makes A and B the same electrical node, so both ends of the resistor are at the same potential.

This is the key to recognising nodes in a circuit: two points are the same node if an uninterrupted path of ideal conducting wire connects them. The wire can bend, loop around the page, or connect points that look nowhere near each other. Its shape and length do not matter.

There is an important catch, though. Two points can have the same voltage without being the same node. Confusing these two ideas can lead you to simplify an entire circuit incorrectly.

01The wire-tracing rule

A node is a continuous region of an ideal conductor. Every point along that connected conductor has the same electric potential.

Think of a node as one uninterrupted piece of metal. Bending the metal does not create another piece. However, placing a resistor, battery, or other circuit component between two sections separates their nodes.

For example, suppose you follow a wire from A towards B.

• If you can reach B using only uninterrupted wire, A and B are the same node.
• If you must pass through a resistor, they are different nodes.
• If you encounter an open switch, there is no conducting connection through that switch.
• If you pass through an ideal closed switch, the two sides are connected and belong to the same node.

The distinction is summarised below.

A resistor separates its two terminals into different nodes, unless another conducting path connects those terminals.

An ideal ammeter has zero resistance, so it does not create a potential difference between its terminals. An ideal voltmeter has effectively infinite resistance and does not connect its measurement terminals together.

Why the drawing can be misleading

Circuit diagrams show electrical connections, not necessarily the physical layout of the wires.

A wire might travel halfway around a circuit before reaching another component. It is still the same node throughout that uninterrupted connection.

Wire crossings need particular attention. In the usual circuit-diagram convention:

• A junction dot indicates that the wires connect.
• A bridge or crossing without a junction indicates that the wires pass over each other without connecting.

If a diagram is ambiguous, redraw the connections before doing any calculations. Don’t decide that two wires connect just because they appear to touch on the page.

02Use conservation before reducing resistors

Recognising nodes is only the beginning. Before combining resistors, use conservation of charge at junctions and conservation of energy around loops to check your circuit interpretation.

These are Kirchhoff’s two laws.

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

Imagine a wire carrying \(3.0\ \mathrm{A}\) of conventional current towards a junction. At the junction, the wire splits into two branches.

Suppose one branch carries \(1.0\ \mathrm{A}\). How much current must enter the other branch?

It must carry \(2.0\ \mathrm{A}\).

Charge cannot continuously accumulate at a junction in a steady DC circuit. The amount of charge entering each second must equal the amount leaving each second.

This gives Kirchhoff’s current law:

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

Here, \(I\) represents current, measured in amperes (\(\mathrm{A}\)).

For a junction with three branches:

\[ I_1=I_2+I_3 \]

provided \(I_1\) enters and \(I_2\) and \(I_3\) leave.

One subtle point: two locations can belong to the same node without carrying the same current.

Imagine a long horizontal wire with branches joining it at J and K. J and K are connected by uninterrupted wire, so they are the same node. However, current can enter or leave through the branches between these locations.

The current in the wire segment between J and K can therefore differ from the current elsewhere in the same node.

Same node means same potential, not necessarily identical current everywhere.

2. Kirchhoff’s voltage law: what happens around a loop?

Now follow a complete circuit loop.

A battery supplies energy to the charges. Resistors transfer electrical energy into other forms, such as thermal energy.

If a charge travels all the way around a closed loop and returns to its starting point, its net change in electrical potential energy must be zero.

Therefore:

\[ \sum \Delta V=0 \]

This is Kirchhoff’s voltage law.

Here, \(\Delta V\) represents the signed change in electric potential, measured in volts (\(\mathrm{V}\)).

For a simple circuit containing a \(12\ \mathrm{V}\) battery and two resistors:

\[ 12-I R_1-I R_2=0 \]

The battery provides a potential rise of \(12\ \mathrm{V}\), while the resistors produce potential drops.

An ideal wire contributes no voltage drop:

\[ V_{\text{wire}}=0\ \mathrm{V} \]

This is another way to understand why every point along an uninterrupted ideal wire belongs to the same node.

3. The correct order for solving a circuit

Before you apply resistance reductions, follow this sequence.

1. Trace every uninterrupted wire and identify which points belong to the same node.
2. Mark the junctions and use Kirchhoff’s current law to account for current entering and leaving.
3. Follow complete loops and use Kirchhoff’s voltage law to account for potential rises and drops.
4. Only now identify which resistors are in series or parallel.
5. Apply equivalent resistance calculations and solve for the unknown quantities.

This order prevents a common mistake: combining resistors based on how they look instead of how they are connected.

Remember these two conditions:

Two resistors drawn beside each other are not necessarily parallel. Two resistors drawn one after another are not necessarily in series.

The connections decide.

03A zero voltmeter reading does not prove two points are the same node

Here is a particularly useful trap.

Suppose you connect a voltmeter between two points in a circuit. It reads \(0.00\ \mathrm{V}\).

Does that prove the points are connected by an uninterrupted wire?

No.

The points might be separate nodes that happen to have equal potentials under the current circuit conditions.

A balanced bridge provides a clear example.

Investigating a balanced bridge

Imagine building a circuit using:

• A \(12.0\ \mathrm{V}\) DC power supply.
• Four \(100\ \Omega\) resistors arranged as two series branches.
• A \(50\ \Omega\) resistor connecting the two branch midpoints.
• A high-impedance voltmeter connected between those midpoints.

Call the left midpoint L and the right midpoint R.

The positive supply terminal connects to the tops of both branches. The bottoms of both branches return to the negative terminal.

The \(50\ \Omega\) resistor connects L to R. The voltmeter is connected in parallel with this bridge resistor.

Before measuring anything, predict the voltmeter reading.

Each branch has a total resistance of:

\[ R_{\text{branch}}=100+100=200\ \Omega \]

The supply voltage is \(12.0\ \mathrm{V}\), so the current in each branch is:

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

\[ I=\frac{12.0}{200} =0.060\ \mathrm{A} \]

The voltage drop across the upper \(100\ \Omega\) resistor in either branch is:

\[ V=IR \]

\[ V=0.060\times100 =6.0\ \mathrm{V} \]

Taking the negative supply terminal as \(0\ \mathrm{V}\), both midpoint nodes are therefore at \(6.0\ \mathrm{V}\).

The potential difference between them is:

\[ V_{LR}=6.0-6.0=0\ \mathrm{V} \]

The current through the \(50\ \Omega\) bridge resistor is also zero:

\[ I_{\text{bridge}}=\frac{0}{50}=0\ \mathrm{A} \]

The predicted voltmeter reading is \(0.00\ \mathrm{V}\), within the instrument’s measurement resolution.

What does this result actually tell us?

It tells us that L and R have equal potentials in this particular circuit configuration.

It does not tell us that L and R are the same node.

There is a \(50\ \Omega\) resistor physically connecting them. That resistor separates the two nodes, even though no current flows through it while the bridge is balanced.

This is the difference between:

• Two points being connected by an uninterrupted conductor.
• Two separate points having equal electrical potential.

The first is a property of the circuit’s connections. The second depends on the electrical conditions.

How would you improve the experiment?

The original voltmeter test cannot distinguish a direct wire connection from two separate points that happen to have equal potentials.

A useful revision is to change one resistor in the bridge and repeat the measurement.

If L and R are genuinely the same node, their potential difference must remain zero because an ideal wire forces them to the same potential.

If they are separate nodes connected through a resistor, changing the circuit conditions can produce a non-zero voltage between them.

A direct continuity test can also help. Turn off the supply, isolate the suspected connection from alternative paths, and use an ohmmeter to measure the resistance between the points.

An ideal wire has zero resistance. A \(50\ \Omega\) resistor does not.

In a real investigation, connecting leads and contacts have small resistances, so the measured value will not necessarily be exactly zero.

This is why circuit topology is the most reliable way to identify a node. Voltage measurements provide useful evidence, but they do not replace tracing the connections.

06Questions and solutions

The following questions progress from basic node identification to situations where the distinction between a shared node and equal potential becomes important.

Question 1

An ideal \(12.0\ \mathrm{V}\) supply feeds a \(3.0\ \Omega\) resistor in series with a parallel section.

The parallel section contains a \(4.0\ \Omega\) resistor and an ideal closed switch S, both connected between points X and Y.

Determine:

a) Whether X and Y are the same node.

b) The potential difference across the \(4.0\ \Omega\) resistor.

c) The current through the \(4.0\ \Omega\) resistor and the switch.

d) The total supply current.

Solution 1

X and Y are the same node. The \(4.0\ \Omega\) resistor has zero potential difference and carries zero current, while the switch carries the full \(4.0\ \mathrm{A}\) circuit current.

Part a: Identify the nodes.

The switch is closed and ideal, so it acts as an uninterrupted conducting wire.

It directly connects X and Y.

Therefore:

\[ V_{XY}=0\ \mathrm{V} \]

Part b: Find the voltage across the resistor.

The \(4.0\ \Omega\) resistor is connected between X and Y.

Since both terminals are at the same potential:

\[ \boxed{V_{4\Omega}=0\ \mathrm{V}} \]

Part c: Calculate the currents.

Using Ohm’s law:

\[ I_{4\Omega}=\frac{V}{R} \]

\[ I_{4\Omega}=\frac{0}{4.0} \]

\[ \boxed{I_{4\Omega}=0\ \mathrm{A}} \]

The total circuit resistance is simply the \(3.0\ \Omega\) resistor because the closed switch bypasses the \(4.0\ \Omega\) resistor.

The supply current is:

\[ I_{\text{total}}=\frac{12.0}{3.0} \]

\[ \boxed{I_{\text{total}}=4.0\ \mathrm{A}} \]

All of this current flows through the closed switch.

Part d: Interpret the result.

The resistor carries no current, but the circuit itself is not switched off.

The switch provides a conducting path around the resistor.

For a more detailed treatment of the current and voltage consequences of bypassing a resistor, see How to Solve a Circuit with a Shorted Resistor.

Question 2

Two junctions, J and K, are connected by an uninterrupted ideal wire.

At J:

• \(2.8\ \mathrm{A}\) enters from the left.
• \(0.9\ \mathrm{A}\) enters from below.
• \(1.1\ \mathrm{A}\) leaves upwards.
• An unknown current \(I_{JK}\) flows towards K.

At K:

• \(1.4\ \mathrm{A}\) enters from above.
• \(0.7\ \mathrm{A}\) leaves to the right.
• An unknown current \(I_B\) leaves downwards.

A separate diagonal wire carries \(1.0\ \mathrm{A}\) from the lower-left to the upper-right. It crosses the J-to-K wire using a bridge, without an electrical connection.

Determine:

a) Whether J and K are the same node.

b) The current \(I_{JK}\).

c) The current \(I_B\).

d) Why the diagonal wire’s current must not be included in either junction equation.

Solution 2

J and K are the same node. The current from J to K is \(2.6\ \mathrm{A}\), and the downward current at K is \(3.3\ \mathrm{A}\). The diagonal wire does not contribute to either junction equation.

Part a: Identify the node.

J and K are connected by an uninterrupted ideal wire.

Therefore, they have the same potential and belong to one node.

However, current can enter or leave this node at different junctions.

Part b: Find the current from J to K.

Apply Kirchhoff’s current law at J.

The total incoming current is:

\[ I_{\text{in}}=2.8+0.9 \]

\[ I_{\text{in}}=3.7\ \mathrm{A} \]

The total outgoing current is:

\[ I_{\text{out}}=1.1+I_{JK} \]

Equating the two:

\[ 3.7=1.1+I_{JK} \]

\[ \boxed{I_{JK}=2.6\ \mathrm{A}} \]

The current flows from J towards K.

Part c: Find the downward current at K.

Apply Kirchhoff’s current law again, this time at K.

The incoming currents are \(2.6\ \mathrm{A}\) from J and \(1.4\ \mathrm{A}\) from above.

The outgoing currents are \(0.7\ \mathrm{A}\) to the right and \(I_B\) downwards.

Therefore:

\[ 2.6+1.4=0.7+I_B \]

\[ 4.0=0.7+I_B \]

\[ \boxed{I_B=3.3\ \mathrm{A}} \]

Part d: Explain the crossing.

The diagonal conductor crosses the J-to-K wire without a junction.

Its current does not enter or leave the shared node.

It has its own continuous conducting path, separate from the J-to-K connection.

A common mistake is to include every current shown near a junction, even when the conductors do not actually connect.

Question 3

An ideal \(12.0\ \mathrm{V}\) supply connects two parallel branches.

The left branch contains a \(2.0\ \Omega\) resistor followed by a \(4.0\ \Omega\) resistor.

The right branch contains a \(3.0\ \Omega\) resistor followed by another \(3.0\ \Omega\) resistor.

The midpoint conductors L and R cross on the diagram at a bridge with no junction dot. They are not electrically connected.

A high-impedance voltmeter is connected with its red lead at L and its black lead at R.

Determine:

a) The current in each branch.

b) The potential at L relative to the negative supply terminal.

c) The potential at R relative to the negative supply terminal.

d) The voltmeter reading.

e) Whether L and R are the same node, explaining why equal total branch resistances do not make their midpoint potentials equal.

Solution 3

L and R are separate nodes, and the voltmeter reads \(+2.0\ \mathrm{V}\), with L at the higher potential.

Even though both branches have the same total resistance and carry equal currents, the voltage drops across their upper resistors are different.

Part a: Calculate the branch currents.

For the left branch:

\[ R_L=2.0+4.0=6.0\ \Omega \]

\[ I_L=\frac{12.0}{6.0} \]

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

For the right branch:

\[ R_R=3.0+3.0=6.0\ \Omega \]

\[ I_R=\frac{12.0}{6.0} \]

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

The two branches carry equal currents.

Part b: Calculate the potential at L.

Take the negative supply terminal as \(0\ \mathrm{V}\).

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

\[ V=IR \]

\[ V=2.0\times2.0 =4.0\ \mathrm{V} \]

Starting at the positive terminal at \(12.0\ \mathrm{V}\), the potential falls by \(4.0\ \mathrm{V}\).

Therefore:

\[ V_L=12.0-4.0 \]

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

Part c: Calculate the potential at R.

The voltage drop across the upper \(3.0\ \Omega\) resistor is:

\[ V=2.0\times3.0 \]

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

Therefore:

\[ V_R=12.0-6.0 \]

\[ \boxed{V_R=6.0\ \mathrm{V}} \]

Part d: Find the voltmeter reading.

The red lead is at L and the black lead is at R.

Therefore:

\[ V_{LR}=V_L-V_R \]

\[ V_{LR}=8.0-6.0 \]

\[ \boxed{V_{LR}=+2.0\ \mathrm{V}} \]

Part e: Explain the node distinction.

L and R are not the same node because the crossing has no electrical connection.

The equal total branch resistances only establish that the branch currents are equal.

They do not establish that the voltage drops across individual resistors are equal.

The upper resistors produce different voltage drops:

• Left branch: \(4.0\ \mathrm{V}\).
• Right branch: \(6.0\ \mathrm{V}\).

Consequently, the midpoint potentials differ by \(2.0\ \mathrm{V}\).

Question 4

An ideal \(12.0\ \mathrm{V}\) supply feeds two parallel branches.

The first branch contains a \(3.0\ \Omega\) resistor in series with a section between points A and B. This section contains a \(6.0\ \Omega\) resistor in parallel with an ideal switch S.

The second branch contains a \(4.0\ \Omega\) resistor directly across the supply.

Consider two situations:

• Situation 1: Switch S is closed.
• Situation 2: Switch S is open.

For each situation, determine:

a) Whether A and B are the same node.

b) The current through the \(6.0\ \Omega\) resistor.

c) The current through the switch.

d) The potential difference across A and B.

e) The total supply current.

Explain why the current in the \(4.0\ \Omega\) branch changes or remains constant.

Solution 4

When S is closed, A and B are the same node, the \(6.0\ \Omega\) resistor carries zero current, and the total supply current is \(7.0\ \mathrm{A}\).

When S is open, A and B are separate nodes, the \(6.0\ \Omega\) resistor carries \(1.33\ \mathrm{A}\), and the total supply current falls to \(4.33\ \mathrm{A}\).

Situation 1: Switch closed

Step 1: Identify the nodes.

The closed ideal switch directly connects A and B.

Therefore:

\[ V_{AB}=0\ \mathrm{V} \]

The \(6.0\ \Omega\) resistor is bypassed.

Step 2: Calculate the first branch current.

Only the \(3.0\ \Omega\) resistor limits the current in this branch.

\[ I_1=\frac{12.0}{3.0} \]

\[ I_1=4.0\ \mathrm{A} \]

All \(4.0\ \mathrm{A}\) flows through the closed switch.

The \(6.0\ \Omega\) resistor carries:

\[ I_{6\Omega}=\frac{0}{6.0} \]

\[ \boxed{I_{6\Omega}=0\ \mathrm{A}} \]

The switch carries:

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

Step 3: Calculate the second branch current.

The \(4.0\ \Omega\) resistor is connected directly across the ideal \(12.0\ \mathrm{V}\) supply.

\[ I_2=\frac{12.0}{4.0} \]

\[ I_2=3.0\ \mathrm{A} \]

Step 4: Calculate the total supply current.

Using Kirchhoff’s current law:

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

\[ I_{\text{total}}=4.0+3.0 \]

\[ \boxed{I_{\text{total}}=7.0\ \mathrm{A}} \]

Situation 2: Switch open

Step 5: Identify the nodes again.

The open switch no longer provides a conducting path between A and B.

The \(6.0\ \Omega\) resistor is now the only connection between these points in that section.

Therefore, A and B are separate nodes.

Step 6: Calculate the first branch current.

The \(3.0\ \Omega\) and \(6.0\ \Omega\) resistors are now in series.

\[ R_1=3.0+6.0 \]

\[ R_1=9.0\ \Omega \]

The current is:

\[ I_1=\frac{12.0}{9.0} \]

\[ \boxed{I_1=1.33\ \mathrm{A}} \]

The switch is open, so:

\[ \boxed{I_S=0\ \mathrm{A}} \]

The entire first-branch current passes through the \(6.0\ \Omega\) resistor:

\[ \boxed{I_{6\Omega}=1.33\ \mathrm{A}} \]

Step 7: Find the potential difference between A and B.

The voltage across the \(6.0\ \Omega\) resistor is:

\[ V_{AB}=IR \]

\[ V_{AB}=\frac{12.0}{9.0}\times6.0 \]

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

The remaining \(4.0\ \mathrm{V}\) is across the \(3.0\ \Omega\) resistor.

The loop voltage drops add to the \(12.0\ \mathrm{V}\) supply.

Step 8: Find the total supply current.

The second branch still carries:

\[ I_2=\frac{12.0}{4.0}=3.0\ \mathrm{A} \]

Therefore:

\[ I_{\text{total}}=\frac{12.0}{9.0}+3.0 \]

\[ \boxed{I_{\text{total}}=4.33\ \mathrm{A}} \]

Why does the second branch current stay constant?

The second branch remains connected directly across an ideal supply that maintains \(12.0\ \mathrm{V}\).

Its resistance also remains \(4.0\ \Omega\).

Therefore, its current remains \(3.0\ \mathrm{A}\), regardless of whether the switch in the other branch is open or closed.

Question 5

An ideal \(12.0\ \mathrm{V}\) supply connects two parallel branches.

The left branch contains two \(100\ \Omega\) resistors in series, with midpoint L.

The right branch also initially contains two \(100\ \Omega\) resistors in series, with midpoint R.

A \(50\ \Omega\) resistor connects L and R.

A high-impedance voltmeter is connected across L and R.

All wires are ideal, the supply voltage remains constant, and all resistors are ohmic.

Initial configuration

a) Calculate the potential at L and R relative to the negative supply terminal.

b) Calculate the current through the \(50\ \Omega\) bridge resistor.

c) A student observes that the voltmeter reads zero and concludes that L and R are the same node. Explain why this conclusion is incorrect.

Modified configuration

The lower-right \(100\ \Omega\) resistor is replaced with a \(200\ \Omega\) resistor. The other three \(100\ \Omega\) resistors and the \(50\ \Omega\) bridge resistor remain unchanged.

d) Determine the new potentials at L and R.

e) Calculate the magnitude and direction of the current through the \(50\ \Omega\) bridge resistor.

f) Check that Kirchhoff’s current law is satisfied at both L and R.

g) Explain why treating the two midpoints as one node, or calculating the two voltage dividers independently, produces an incorrect result.

Solution 5

L and R are not the same node. Initially, they happen to have equal potentials of \(6.0\ \mathrm{V}\). After the lower-right resistor is changed, their potentials become \(6.6\ \mathrm{V}\) and \(7.2\ \mathrm{V}\), producing a \(12\ \mathrm{mA}\) current through the bridge resistor from R to L.

This question tests the difference between a connection and an electrical condition. A zero voltage reading does not automatically mean that two points are one node.

Parts a and b: Solve the initial configuration.

Initially, both branches contain two \(100\ \Omega\) resistors in series.

Each branch therefore has:

\[ R_{\text{branch}}=100+100 \]

\[ R_{\text{branch}}=200\ \Omega \]

The current in either branch is:

\[ I=\frac{12.0}{200} \]

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

The voltage drop across each upper resistor is:

\[ V=IR \]

\[ V=0.060\times100 \]

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

Taking the negative supply terminal as \(0\ \mathrm{V}\):

\[ \boxed{V_L=6.0\ \mathrm{V}} \]

\[ \boxed{V_R=6.0\ \mathrm{V}} \]

The voltage across the bridge resistor is:

\[ V_{LR}=6.0-6.0=0\ \mathrm{V} \]

Therefore:

\[ I_{\text{bridge}}=\frac{0}{50} \]

\[ \boxed{I_{\text{bridge}}=0\ \mathrm{A}} \]

Part c: Explain the misconception.

The student is correct that L and R have equal potentials.

However, the \(50\ \Omega\) resistor remains between them.

They are separate nodes because there is no uninterrupted ideal wire connecting them.

The zero voltmeter reading is caused by the balanced resistor arrangement, not by L and R being the same node.

The distinction becomes obvious when the circuit changes.

Parts d and e: Solve the modified configuration.

The lower-right resistor is now \(200\ \Omega\).

The left branch still has two \(100\ \Omega\) resistors, but the right branch now has a \(100\ \Omega\) upper resistor and a \(200\ \Omega\) lower resistor.

The bridge resistor connects the two midpoint nodes, so the branches can no longer be solved as independent voltage dividers.

We must apply Kirchhoff’s current law at L and R.

Take the negative supply terminal as \(0\ \mathrm{V}\), and let the unknown midpoint potentials be \(V_L\) and \(V_R\).

At L, the current entering from the upper-left resistor and the bridge must equal the current leaving through the lower-left resistor:

\[ \frac{12-V_L}{100} + \frac{V_R-V_L}{50} = \frac{V_L}{100} \]

At R, the current entering from the upper-right resistor must equal the current leaving through the lower-right resistor and the bridge:

\[ \frac{12-V_R}{100} = \frac{V_R}{200} + \frac{V_R-V_L}{50} \]

These are two simultaneous equations.

Simplifying the first:

\[ 12-V_L+2V_R-2V_L=V_L \]

\[ 4V_L-2V_R=12 \]

Simplifying the second:

\[ 24-2V_R=V_R+4V_R-4V_L \]

\[ 7V_R-4V_L=24 \]

We now have:

\[ 4V_L-2V_R=12 \]

\[ 7V_R-4V_L=24 \]

From the first equation:

\[ V_R=2V_L-6 \]

Substitute into the second:

\[ 7(2V_L-6)-4V_L=24 \]

\[ 14V_L-42-4V_L=24 \]

\[ 10V_L=66 \]

Therefore:

\[ \boxed{V_L=6.6\ \mathrm{V}} \]

Substituting back:

\[ V_R=2(6.6)-6 \]

\[ \boxed{V_R=7.2\ \mathrm{V}} \]

The voltage difference across the bridge is:

\[ V_R-V_L=7.2-6.6 \]

\[ V_R-V_L=0.6\ \mathrm{V} \]

The bridge current is:

\[ I_{\text{bridge}}=\frac{0.6}{50} \]

\[ \boxed{I_{\text{bridge}}=0.012\ \mathrm{A}} \]

Therefore, the bridge current is \(12\ \mathrm{mA}\), flowing from R towards L.

Part f: Check Kirchhoff’s current law.

At L, the current entering from the upper-left resistor is:

\[ I_{\text{upper-left}} =\frac{12.0-6.6}{100} \]

\[ I_{\text{upper-left}}=0.054\ \mathrm{A} =54\ \mathrm{mA} \]

The bridge supplies another \(12\ \mathrm{mA}\) into L.

The current leaving through the lower-left resistor is:

\[ I_{\text{lower-left}} =\frac{6.6}{100} \]

\[ I_{\text{lower-left}}=66\ \mathrm{mA} \]

Therefore:

\[ 54+12=66\ \mathrm{mA} \]

Kirchhoff’s current law is satisfied at L.

At R, the current entering from the upper-right resistor is:

\[ I_{\text{upper-right}} =\frac{12.0-7.2}{100} \]

\[ I_{\text{upper-right}}=48\ \mathrm{mA} \]

The current leaving through the lower-right resistor is:

\[ I_{\text{lower-right}} =\frac{7.2}{200} \]

\[ I_{\text{lower-right}}=36\ \mathrm{mA} \]

The remaining \(12\ \mathrm{mA}\) flows through the bridge towards L.

Therefore:

\[ 48=36+12 \]

Kirchhoff’s current law is also satisfied at R.

The total current supplied is:

\[ I_{\text{total}}=54+48 \]

\[ I_{\text{total}}=102\ \mathrm{mA} \]

The returning current is:

\[ 66+36=102\ \mathrm{mA} \]

The current balances across the complete circuit.

Part g: Explain why the tempting methods fail.

There are two tempting approaches.

The first is to merge L and R into one node because the initial voltmeter reading was zero.

That would force \(V_L=V_R\) under every subsequent condition. But the bridge contains a \(50\ \Omega\) resistor, not an ideal wire. When the resistor arrangement changes, a potential difference can develop across it.

The second tempting approach is to calculate each modified branch as an independent voltage divider.

Doing that would give:

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

and

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

Those values ignore the current flowing through the \(50\ \Omega\) bridge resistor.

The bridge connects the two voltage-divider networks and changes the currents in both. The midpoint potentials must therefore be found together using Kirchhoff’s current law.

The correct values are \(6.6\ \mathrm{V}\) and \(7.2\ \mathrm{V}\).

The lesson is that equal potential at one particular operating condition does not establish a permanent electrical connection.

07A final node-identification check

Before simplifying any unfamiliar circuit, ask yourself these questions:

The most important distinction is simple:

A wire connection determines whether two points are the same node. Their potential difference tells you about the electrical conditions of the circuit.

Once you can separate those ideas, complicated-looking circuits become much easier to redraw and analyse. The next useful step is learning how to use the identified nodes to calculate branch currents and compare voltage drops in mixed series-parallel circuits.