What Happens When a Bulb Breaks in Series and Parallel Circuits?

Break one bulb in a series circuit and every bulb goes out. Break one bulb in a parallel circuit and the other bulbs can stay lit. The difference comes down to one question: after the bulb breaks, are there still complete paths for current to flow?

There’s another detail worth understanding. In a parallel circuit, the remaining bulbs usually stay at the same brightness, but they can become brighter if the battery has significant internal resistance. And in a mixed circuit, breaking one bulb can switch off other bulbs in its branch, even when separate branches remain connected.

To predict what happens, start with conservation of charge at junctions and conservation of energy around loops. Resistance calculations come afterwards.

01What happens when a bulb breaks?

The basic outcomes are:

These predictions assume negligible wire resistance and, initially, an ideal voltage supply that maintains a constant terminal voltage.

Before going further, consider a useful prediction.

Two identical bulbs are connected in series to a 12 V battery. One bulb breaks. Does the intact bulb receive the full 12 V, half the voltage, or zero volts?

It receives zero volts in the ideal circuit model. The full 12 V appears across the break instead.

This seems strange until you recognise that voltage across a resistor depends on the current flowing through it. If the current is zero, the voltage drop across an intact resistor is also zero.

02How to test what happens

A simple circuit investigation can distinguish the two arrangements.

Use:

• A low-voltage, current-limited DC power supply.
• Two low-voltage bulbs and their holders.
• Connecting wires.
• Switches to open individual parts of the circuit.
• An ammeter to measure current.
• A voltmeter to measure potential difference.

Use a low-voltage laboratory supply, not mains electricity.

Experiment 1: Two bulbs in series

Connect the positive terminal of the supply to a switch, then an ammeter, followed by bulb A and bulb B. Connect bulb B back to the negative terminal.

There is only one route through the circuit.

Open a switch placed in series with bulb B to simulate a broken filament.

What should happen to the ammeter reading?

It should fall to zero. Both bulbs should go out because there is no longer a complete path around the circuit.

Experiment 2: Two bulbs in parallel

Now connect the positive and negative supply terminals to two separate conducting rails.

Place bulb A in one branch and bulb B in the other. Each branch connects directly between the same two rails.

Give each branch its own switch. Keep an ammeter in the common supply lead, and use a voltmeter across the supply rails.

Open the switch in branch B.

The expected results are different:

• The current in branch B becomes zero.
• Bulb B goes out.
• Bulb A remains illuminated.
• The total current supplied by the battery decreases.
• The voltage across the rails remains constant if the supply is regulated.

The last point is important. The remaining bulb does not automatically receive extra current just because another branch has stopped working.

What the first experiment cannot tell you

Suppose you only measure the total current supplied by the battery.

You would see the current fall in both situations, although for different reasons:

• In series, the total current falls to zero.
• In parallel, the total current falls because one branch no longer carries current.

That measurement alone cannot establish whether the current through the surviving parallel branch has changed.

To improve the investigation, measure the current in each branch separately and monitor the voltage across the supply rails. A regulated supply helps isolate the effect of circuit arrangement from changes in battery voltage.

You can then repeat the parallel investigation with an ordinary battery and compare the results. This reveals the effect of internal resistance, which we’ll examine later.

Opening a switch is electrically equivalent to a broken filament for this investigation: both create an open circuit. It does not reproduce every thermal effect of an actual filament failure, but it tests the electrical principle being investigated.

03The physics: conservation of charge and energy

Two circuit laws explain the results.

1. Conservation of charge at a junction

A junction is a point where three or more conducting paths meet.

Charge cannot continually accumulate at an ordinary circuit junction. Therefore, the total current entering must equal the total current leaving.

This is Kirchhoff’s current law:

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

Current, represented by \(I\), is measured in amperes (A).

For a parallel circuit with two branches:

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

If bulb B breaks, its branch becomes an open circuit:

\[ I_B=0 \]

Therefore:

\[ I_{\text{total}}=I_A \]

Notice what this tells us.

The total current decreases, but Kirchhoff’s current law does not require the current through bulb A to increase.

Its current depends on the voltage across it and its resistance.

2. Conservation of energy around a loop

A loop is a complete path around a circuit.

As charge moves through a circuit, the energy supplied by the battery is transferred to circuit components. The total energy gained and lost around a complete loop must balance.

This is Kirchhoff’s voltage law:

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

For a simple battery and a series of resistors:

\[ \mathcal E=V_1+V_2+\cdots \]

Here:

• \(\mathcal E\) is the electromotive force (emf) of the supply, measured in volts (V).
• \(V_1,V_2,\ldots\) are the potential differences across individual components, also measured in volts.

For a resistor, Ohm’s law gives:

\[ V=IR \]

where \(R\) is resistance, measured in ohms (\(\Omega\)).

The important idea is that a broken bulb changes which complete loops exist.

Why every bulb goes out in series

A series circuit has only one path for current.

Every bulb lies along that same path.

If one filament breaks, the path is interrupted. There is no complete loop through the battery and the bulbs.

Consequently:

\[ I=0 \]

For every intact bulb:

\[ V=IR=0 \]

There is no current through any bulb, so none transfers electrical energy into light.

The full supply voltage appears across the open gap in the ideal model.

This also explains a common misconception: an intact bulb does not automatically receive the battery’s full voltage just because it is the only working bulb left.

The circuit still needs a complete loop.

Why the other bulbs stay on in parallel

A parallel circuit has multiple branches.

Each branch provides a separate path between the same two junctions.

If bulb B breaks, its branch is interrupted, but the branch containing bulb A still forms a complete loop with the battery.

For an ideal voltage supply:

\[ V_A=V_{\text{supply}} \]

If bulb A has resistance \(R_A\), its current is:

\[ I_A=\frac{V_{\text{supply}}}{R_A} \]

Opening another branch does not change either quantity.

The remaining bulb therefore keeps the same current and, in the fixed-resistance model, the same electrical power.

Why equivalent resistance comes afterwards

You can also calculate what happens using equivalent resistance.

For resistors in series:

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

For resistors in parallel:

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

When a parallel branch opens, the equivalent resistance of the remaining network increases.

However, equivalent resistance alone does not tell you what happens to the current in an individual surviving branch.

First identify the complete loops using Kirchhoff’s laws. Then use resistance calculations to find the currents.

For a more detailed explanation of these foundations, see Series Circuits: Current, Voltage, Resistance and Power and Parallel Circuits: Voltage, Current, Resistance and Power.

06An important exception: What happens with a real battery?

So far, we’ve assumed an ideal voltage supply.

An ideal supply maintains a constant terminal voltage regardless of the current drawn from it.

Real batteries have internal resistance.

A simple battery model consists of an ideal emf, \(\mathcal E\), in series with an internal resistance, \(r\).

The terminal voltage is:

\[ V_{\text{terminal}}=\mathcal E-I_{\text{total}}r \]

When more current flows, more voltage is lost across the battery’s internal resistance.

When a parallel branch breaks, the total current decreases. The voltage lost inside the battery also decreases.

This means the terminal voltage can increase.

The remaining bulb now receives a slightly larger potential difference, so its current and power can increase.

The distinction is:

This is not a contradiction of Kirchhoff’s laws. The laws still apply. The supply voltage across the external circuit has changed.

We’ll examine the numerical effect in Question 4.

07What if the circuit contains both series and parallel bulbs?

Not every circuit fits neatly into one category.

Consider a circuit with two parallel branches:

• Branch 1 contains bulbs A and B in series.
• Branch 2 contains bulb C on its own.

If bulb B breaks, what happens to A and C?

Bulb A goes out because it shares a branch with B. Bulb C stays on because its separate branch still forms a complete loop.

The fact that the overall circuit contains parallel branches does not mean every bulb operates independently.

There is another complication.

Sometimes a resistor or bulb is placed in series with the entire parallel network, before the current reaches the junction.

If one branch breaks, the total current changes. That changes the voltage drop across the shared series component, which can change the voltage across every surviving branch.

This is why you should always trace the actual circuit rather than simply label it a parallel circuit.

A useful exam method is:

1. Mark the broken filament as an open circuit.
2. Trace every possible path from the positive to the negative supply terminal.
3. Identify which complete loops remain.
4. Apply Kirchhoff’s current law at the junctions.
5. Apply Kirchhoff’s voltage law around the remaining loops.
6. Calculate individual currents and power only after establishing the circuit behaviour.

Now apply that reasoning to some less obvious situations.

08Questions and solutions

Question 1

Three identical bulbs, P, Q, and R, each have a resistance of 3.0 Ω. They are connected in series to an ideal 9.0 V battery.

Bulb Q breaks.

Calculate:

(a) The current through the circuit before the break.

(b) The current after the break.

(c) The potential difference across bulb P and the open gap after the break.

Explain why the other two bulbs also go out.

Solution 1

All three bulbs go out. The current becomes zero throughout the circuit, the voltage across bulb P becomes zero, and the full 9.0 V appears across the open gap.

(a) Current before the break

All three bulbs carry the same current because they are in series.

Using Kirchhoff’s voltage law and Ohm’s law:

\[ \mathcal E=I(R_P+R_Q+R_R) \]

Substitute:

\[ 9.0=I(3.0+3.0+3.0) \]

\[ I=\frac{9.0}{9.0} \]

\[ \boxed{I=1.00\text{ A}} \]

(b) Current after the break

Bulb Q creates an open circuit.

There is no complete loop, so:

\[ \boxed{I=0\text{ A}} \]

(c) Potential differences

For bulb P:

\[ V_P=IR_P \]

\[ V_P=0\times3.0 \]

\[ \boxed{V_P=0\text{ V}} \]

Using Kirchhoff’s voltage law:

\[ 9.0=V_P+V_{\text{gap}} \]

Therefore:

\[ \boxed{V_{\text{gap}}=9.0\text{ V}} \]

The battery still provides an emf, but the open circuit prevents current from flowing through any of the bulbs.

Question 2

Three bulbs are connected in parallel across an ideal 12.0 V supply.

Their resistances are:

• Bulb A: 4.0 Ω
• Bulb B: 6.0 Ω
• Bulb C: 12.0 Ω

Bulb C breaks.

Calculate:

(a) The current through each bulb before the break.

(b) The total current supplied by the battery before the break.

(c) The total current after the break.

(d) The power dissipated by bulbs A and B after the break.

Explain why the surviving bulbs do not become brighter.

Solution 2

Bulbs A and B remain illuminated at their original brightness. The total current decreases from 6.00 A to 5.00 A because the current through branch C becomes zero.

(a) Current through each bulb before the break

Each branch has the same potential difference as the supply:

\[ V_A=V_B=V_C=12.0\text{ V} \]

Using Ohm’s law:

For A:

\[ I_A=\frac{12.0}{4.0} =\boxed{3.00\text{ A}} \]

For B:

\[ I_B=\frac{12.0}{6.0} =\boxed{2.00\text{ A}} \]

For C:

\[ I_C=\frac{12.0}{12.0} =\boxed{1.00\text{ A}} \]

(b) Total current before the break

Apply Kirchhoff’s current law:

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

\[ I_{\text{total}}=3.00+2.00+1.00 \]

\[ \boxed{I_{\text{total}}=6.00\text{ A}} \]

(c) Total current after the break

The current through C becomes zero:

\[ I_C=0 \]

The ideal supply maintains 12.0 V across A and B.

Therefore:

\[ I_{\text{total}}=3.00+2.00+0 \]

\[ \boxed{I_{\text{total}}=5.00\text{ A}} \]

(d) Power dissipated by A and B

Use:

\[ P=\frac{V^2}{R} \]

For A:

\[ P_A=\frac{12.0^2}{4.0} \]

\[ \boxed{P_A=36.0\text{ W}} \]

For B:

\[ P_B=\frac{12.0^2}{6.0} \]

\[ \boxed{P_B=24.0\text{ W}} \]

Both power values are unchanged.

The tempting prediction is that A and B should become brighter because the battery has fewer branches to supply. However, the voltage across each surviving branch remains 12.0 V.

The current through each surviving bulb is therefore unchanged.

Question 3

An ideal 12.0 V supply connects two parallel branches.

• The first branch contains bulb A (4.0 Ω) and bulb B (8.0 Ω) in series.
• The second branch contains bulb C (6.0 Ω) on its own.

Bulb B breaks.

Calculate:

(a) The current through each bulb before the break.

(b) The total current supplied by the battery before the break.

(c) Which bulbs remain illuminated after the break.

(d) The power dissipated by bulb C before and after the break.

(e) The potential difference across bulb A after the break.

Solution 3

Bulbs A and B both go out because they share the same series branch. Bulb C remains illuminated and maintains its original power of 24.0 W.

(a) Current through each bulb before the break

Start with the first branch.

Bulbs A and B are in series, so their resistances add:

\[ R_{AB}=4.0+8.0=12.0\ \Omega \]

The potential difference across this branch is 12.0 V.

Using Ohm’s law:

\[ I_{AB}=\frac{12.0}{12.0} \]

\[ \boxed{I_A=I_B=1.00\text{ A}} \]

Bulb C is connected directly across the supply:

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

\[ \boxed{I_C=2.00\text{ A}} \]

(b) Total current before the break

Apply Kirchhoff’s current law at the junction:

\[ I_{\text{total}}=I_{AB}+I_C \]

\[ I_{\text{total}}=1.00+2.00 \]

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

(c) Which bulbs remain illuminated?

When B breaks, the entire upper branch becomes an open circuit.

Since A and B are in series, the current through both becomes zero.

Therefore:

• A goes out.
• B goes out.
• C stays illuminated.

The parallel arrangement preserves the lower branch, not every individual bulb.

(d) Power dissipated by C

Before the break:

\[ P_C=VI_C \]

\[ P_C=12.0\times2.00 \]

\[ \boxed{P_C=24.0\text{ W}} \]

After the break, C is still connected across the ideal 12.0 V supply.

Its current remains 2.00 A.

Therefore:

\[ P_C=12.0\times2.00 \]

\[ \boxed{P_C=24.0\text{ W}} \]

Its brightness remains unchanged in the fixed-resistance model.

(e) Potential difference across A after the break

Since the current through A is zero:

\[ V_A=IR_A \]

\[ V_A=0\times4.0 \]

\[ \boxed{V_A=0\text{ V}} \]

The voltage across the open gap at B is the full 12.0 V.

The important distinction is between a parallel network and an individual parallel branch. A and B belong to the same branch, so they depend on each other to maintain a complete current path.

Question 4

A battery has an emf of 12.0 V and an internal resistance of 2.0 Ω.

Two identical bulbs, A and B, each have a constant resistance of 8.0 Ω. They are connected in parallel across the battery.

Bulb B breaks.

Calculate:

(a) The current through bulb A before the break.

(b) The terminal voltage of the battery before the break.

(c) The current through bulb A after the break.

(d) The power dissipated by A before and after the break.

(e) The percentage increase in A’s electrical power.

Solution 4

Bulb A becomes brighter in the fixed-resistance model. Its current increases from 1.00 A to 1.20 A, and its power increases from 8.00 W to 11.52 W.

The reason is that opening branch B reduces the current through the battery’s internal resistance, increasing the terminal voltage.

(a) Current through A before the break

The two bulbs have identical resistances and are connected in parallel.

Therefore:

\[ I_A=I_B \]

Using Kirchhoff’s current law:

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

The battery’s emf must supply the voltage drop across its internal resistance and the external parallel branches.

Using Kirchhoff’s voltage law:

\[ \mathcal E=I_{\text{total}}r+I_AR \]

Substitute the known values:

\[ 12.0=2.0(I_A+I_B)+8.0I_A \]

Since \(I_A=I_B\):

\[ 12.0=2.0(2I_A)+8.0I_A \]

\[ 12.0=12.0I_A \]

Therefore:

\[ \boxed{I_A=1.00\text{ A}} \]

(b) Terminal voltage before the break

The terminal voltage is the potential difference across the external circuit.

\[ V_{\text{terminal}}=\mathcal E-I_{\text{total}}r \]

The total current is:

\[ I_{\text{total}}=1.00+1.00=2.00\text{ A} \]

Therefore:

\[ V_{\text{terminal}}=12.0-(2.00\times2.0) \]

\[ \boxed{V_{\text{terminal}}=8.00\text{ V}} \]

(c) Current through A after the break

Bulb B now carries zero current.

Therefore:

\[ I_{\text{total}}=I_A \]

Apply Kirchhoff’s voltage law:

\[ \mathcal E=I_Ar+I_AR_A \]

Substitute:

\[ 12.0=2.0I_A+8.0I_A \]

\[ 12.0=10.0I_A \]

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

The terminal voltage is now:

\[ V_{\text{terminal}}=12.0-(1.20\times2.0) \]

\[ \boxed{V_{\text{terminal}}=9.60\text{ V}} \]

(d) Power dissipated by A

Use:

\[ P=I^2R \]

Before the break:

\[ P_{\text{before}}=1.00^2\times8.0 \]

\[ \boxed{P_{\text{before}}=8.00\text{ W}} \]

After the break:

\[ P_{\text{after}}=1.20^2\times8.0 \]

\[ \boxed{P_{\text{after}}=11.52\text{ W}} \]

(e) Percentage increase in power

The percentage increase is:

\[ \frac{P_{\text{after}}-P_{\text{before}}} {P_{\text{before}}}\times100 \]

Substitute:

\[ =\frac{11.52-8.00}{8.00}\times100 \]

\[ \boxed{44.0\%} \]

The remaining bulb becomes brighter because the battery’s internal voltage drop decreases.

This is why the statement that a surviving parallel bulb always maintains exactly the same brightness needs an assumption: the supply must maintain a constant terminal voltage.

Question 5

This final problem tests a subtle distinction between two circuits that contain exactly the same components.

Both arrangements use:

• An ideal 12.0 V supply.
• Three bulbs, A, B, and C, each modelled as a constant 4.0 Ω resistor.
• One additional resistor, \(R_0=2.0\ \Omega\).

In both circuits, bulb B breaks.

However, the position of the 2.0 Ω resistor is different.

Arrangement I: Common series resistor

The 2.0 Ω resistor is connected between the positive supply terminal and a junction. After that resistor, the circuit splits into two parallel branches:

• One branch contains bulb A alone.
• The other branch contains bulbs B and C in series.

Arrangement II: Resistor in the A branch

The supply connects directly to two parallel branches:

• One branch contains the 2.0 Ω resistor in series with bulb A.
• The other branch contains bulbs B and C in series.

Determine:

(a) The current through A before the break in each arrangement.

(b) The power dissipated by A before the break in each arrangement.

(c) The current through A after B breaks in each arrangement.

(d) Whether A becomes brighter, dimmer, or stays at the same brightness in each arrangement.

Explain why moving the same resistor changes the result.

Solution 5

Bulb A becomes brighter in Arrangement I but maintains the same brightness in Arrangement II.

Although the components are identical, their connections are different. In Arrangement I, the resistor is shared by both branches, so removing one branch changes the voltage available to the other. In Arrangement II, the resistor is confined to A’s own branch, so the current through A remains unchanged.

This is the key distinction.

Arrangement I: Common series resistor

Step 1: Calculate the voltage across the parallel section before the break.

The resistance of the B-C branch is:

\[ R_{BC}=4.0+4.0=8.0\ \Omega \]

Let the voltage across the parallel section be \(V_p\).

The current through A is:

\[ I_A=\frac{V_p}{4.0} \]

The current through the B-C branch is:

\[ I_{BC}=\frac{V_p}{8.0} \]

Using Kirchhoff’s current law:

\[ I_{\text{total}}=I_A+I_{BC} \]

Substitute:

\[ I_{\text{total}} =\frac{V_p}{4.0}+\frac{V_p}{8.0} \]

\[ I_{\text{total}}=\frac{3V_p}{8} \]

Step 2: Apply Kirchhoff’s voltage law around the complete circuit.

The supply voltage is shared between the common resistor and the parallel section:

\[ 12.0=2.0I_{\text{total}}+V_p \]

Substitute the expression for total current:

\[ 12.0=2.0\left(\frac{3V_p}{8}\right)+V_p \]

\[ 12.0=\frac{3V_p}{4}+V_p \]

\[ 12.0=\frac{7V_p}{4} \]

Therefore:

\[ \boxed{V_p=\frac{48}{7}=6.86\text{ V}} \]

Step 3: Calculate the current through A before the break.

\[ I_A=\frac{V_p}{R_A} \]

\[ I_A=\frac{48/7}{4.0} \]

\[ \boxed{I_A=\frac{12}{7}=1.71\text{ A}} \]

Step 4: Calculate the power dissipated by A before the break.

Using:

\[ P_A=I_A^2R_A \]

\[ P_A=\left(\frac{12}{7}\right)^2\times4.0 \]

\[ \boxed{P_A=\frac{576}{49}=11.76\text{ W}} \]

Step 5: Calculate the current after B breaks.

The B-C branch is now open.

Consequently:

\[ I_{BC}=0 \]

Only the common resistor and bulb A remain in the complete current path.

They are now in series.

Applying Kirchhoff’s voltage law:

\[ 12.0=I_A(2.0+4.0) \]

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

\[ \boxed{I_A=2.00\text{ A}} \]

Step 6: Calculate A’s new power.

\[ P_A=I_A^2R_A \]

\[ P_A=2.00^2\times4.0 \]

\[ \boxed{P_A=16.0\text{ W}} \]

The power has increased from 11.76 W to 16.0 W.

The percentage increase is:

\[ \frac{16.0-11.76}{11.76}\times100 \]

Using the unrounded initial value:

\[ \boxed{36.1\%} \]

Therefore, bulb A becomes brighter in Arrangement I.

The physical reason is that the total current through the common resistor decreases when the B-C branch opens. The voltage drop across that resistor decreases, leaving a larger potential difference across A.

The voltage across the parallel section rises from approximately 6.86 V to 8.00 V.

Arrangement II: Resistor in the A branch

Step 7: Calculate the current through A before the break.

In this arrangement, the 2.0 Ω resistor and bulb A form one complete branch.

That branch is connected directly across the ideal 12.0 V supply.

Applying Kirchhoff’s voltage law:

\[ 12.0=I_A(2.0+4.0) \]

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

\[ \boxed{I_A=2.00\text{ A}} \]

Step 8: Calculate the power dissipated by A.

\[ P_A=I_A^2R_A \]

\[ P_A=2.00^2\times4.0 \]

\[ \boxed{P_A=16.0\text{ W}} \]

Step 9: Determine what happens after B breaks.

The B-C branch becomes an open circuit.

However, the branch containing the 2.0 Ω resistor and A remains connected directly across the ideal supply.

Its potential difference is still 12.0 V.

Therefore:

\[ 12.0=I_A(2.0+4.0) \]

\[ \boxed{I_A=2.00\text{ A}} \]

The power remains:

\[ \boxed{P_A=16.0\text{ W}} \]

Bulb A maintains the same brightness.

Why are the results different?

The tempting prediction is that A should behave identically in both arrangements because the same components are present and A is connected in parallel with the B-C branch.

But being in parallel only guarantees that components share the same potential difference between their connecting nodes. It does not guarantee that those nodes are directly connected to the ideal supply terminals.

In Arrangement I, the parallel section sits after a common resistor. Its voltage depends on the total current through that resistor. When branch B-C opens, that current decreases, the common resistor’s voltage drop decreases, and A receives a larger voltage.

In Arrangement II, the resistor is only in A’s branch. The entire A branch remains connected directly across the fixed 12.0 V supply. Opening the other branch cannot change its voltage or current.

The same components can therefore produce different outcomes simply because their connections are different.

This calculation assumes an ideal supply, negligible wire resistance, and fixed bulb resistances. Actual filament bulbs have temperature-dependent resistance, so the calculated power changes are an idealised comparison rather than an exact prediction of visible brightness.

The key decision whenever a bulb breaks is to identify which complete loops survive. Once you know that, use Kirchhoff’s current law to track current at junctions and Kirchhoff’s voltage law to determine the potential differences around those loops.

The next step in analysing more complicated circuits is comparing the power of bulbs when their branch voltages or currents change. That’s where the same reasoning becomes useful for comparing bulb brightness in series-parallel circuits.