What Happens to Other Bulbs When One Series Bulb Breaks?

If one bulb in a simple series circuit breaks, every other bulb in that series circuit goes out. It doesn’t matter whether the remaining bulbs are perfectly healthy or have very low resistance. The broken filament interrupts the only conducting path, so the current through every bulb becomes zero.

The surprising part is where the battery voltage goes. The intact bulbs do not suddenly receive the full supply voltage. Instead, almost the entire supply voltage appears across the gap left by the broken filament.

01Why does one broken bulb turn all the others off?

Imagine two identical bulbs connected in series to a 12 V battery.

Before anything breaks, current travels through bulb A, then bulb B, and back to the battery. Both bulbs light up.

Now suppose the filament inside bulb A snaps.

You might predict that bulb B receives the electricity that previously went through A. After all, B is still connected to the battery.

But that prediction misses one important feature of a series circuit: there is only one path for current to follow.

A broken filament creates a physical gap in that path. With no complete conducting loop, charge cannot continue flowing around the circuit. The current through both bulbs falls to zero.

Think of a road with a single bridge connecting two sides of a river. If the bridge collapses, traffic stops along that route. It doesn’t matter that the road on the other side is still perfectly usable.

The analogy captures the need for a continuous path. It isn’t exact, though: electrons are already present throughout the conducting wires. They don’t need to travel all the way from the battery to the bulb before current can exist.

There is another misconception worth correcting. A bulb uses electrical energy, not electrical current. The current entering a series bulb is the same as the current leaving it. The bulb transfers electrical energy into light and heat, but it doesn’t consume the charge.

02Use conservation laws to explain what happens

You can explain a broken series circuit using two important circuit principles:

• Kirchhoff’s current law, which describes what happens at junctions.
• Kirchhoff’s voltage law, which describes the energy changes around a circuit.

These principles explain the result before you need to calculate an equivalent resistance.

1. Kirchhoff’s current law: current cannot disappear at a junction

Kirchhoff’s current law (KCL) states that the total current entering a junction equals the total current leaving it.

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

In a series circuit, there are no branching paths. The same current must therefore pass through every component.

For two bulbs in series:

\[ I_A=I_B=I \]

Suppose bulb A breaks. The gap prevents current from passing through it, so:

\[ I_A=0 \]

Because the bulbs share one series path, bulb B must also have zero current:

\[ \boxed{I_A=I_B=0} \]

This is why an intact bulb cannot remain illuminated when another bulb in the same series path breaks.

If current continued through B, charge would have to accumulate indefinitely at the break. Instead, a brief redistribution of charge establishes a new steady state in which the current is zero everywhere in the series path.

2. Kirchhoff’s voltage law: where does the battery voltage go?

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

As you travel around a complete circuit, the total potential rises must equal the total potential drops.

For a simple circuit with a battery and two bulbs:

\[ \mathcal{E}=V_A+V_B \]

Here:

• \(\mathcal{E}\) is the battery’s electromotive force (emf), measured in volts.
• \(V_A\) is the potential difference across bulb A.
• \(V_B\) is the potential difference across bulb B.

Now suppose bulb A breaks.

We must include the voltage across the open gap when writing the voltage balance:

\[ \mathcal{E}=V_{\text{gap}}+V_A+V_B \]

For an intact bulb modelled as a resistor, Ohm’s law gives:

\[ V=IR \]

where \(V\) is potential difference in volts, \(I\) is current in amperes, and \(R\) is resistance in ohms.

Since the current through both intact bulbs is now zero:

\[ V_A=0,\qquad V_B=0 \]

Therefore:

\[ \boxed{V_{\text{gap}}=\mathcal{E}} \]

For an ideal 12 V supply, the entire 12 V appears across the open gap.

This is the key distinction: a component can have a voltage across it without carrying current.

The broken filament has a voltage across its gap, even though no current flows through it.

Before and after a bulb breaks

The important electrical changes can be summarised as follows.

These results assume one open filament, negligible leakage, and negligible resistance in the connecting wires.

Notice that we haven’t needed to calculate the total resistance of the circuit. Conservation laws establish what happens first. Resistance calculations can then help us find the numerical values.

03How can an experiment demonstrate this?

Simply watching two bulbs go dark tells you what happened, but it doesn’t show where the voltage went.

A controlled investigation can test the explanation directly.

Setting up the circuit

Consider a low-voltage 12.0 V DC supply connected to two bulbs, A and B, in series.

The apparatus is arranged in this order:

1. Connect the positive terminal of the supply to an ammeter.
2. Connect the ammeter to bulb A.
3. Connect bulb A to bulb B.
4. Connect bulb B to a switch, X.
5. Connect the switch back to the negative terminal of the supply.

The ammeter measures the current through the entire series path.

A high-resistance voltmeter can be connected across each bulb and across switch X. The switch provides a controlled way to introduce an open gap, simulating a broken filament.

What should the experiment show?

Before opening the switch, both bulbs have a complete conducting path.

For an illustrative model, assume each bulb has a fixed resistance of 20 Ω.

The total resistance is:

\[ R_{\text{total}}=20+20=40\ \Omega \]

The current is:

\[ I=\frac{V}{R} =\frac{12.0}{40} =0.30\ \text{A} \]

Each bulb has a potential difference of:

\[ V_{\text{bulb}}=IR =0.30\times20 =6.0\ \text{V} \]

Now open switch X.

The predicted steady-state readings are:

These are illustrative readings for the fixed-resistance model, not measurements from a reported experiment.

The results support the open-circuit explanation:

• The supply still provides 12.0 V.
• The current through the circuit is zero.
• Neither intact bulb has a potential difference across it.
• The full supply voltage appears across the open switch.

Opening the switch has reproduced the electrical behaviour of a broken filament.

Why is measuring brightness alone insufficient?

An initial investigation might simply observe that both bulbs turn off when one connection is broken.

However, this observation cannot distinguish between several possibilities:

• The series circuit has developed an open circuit.
• The power supply has failed.
• A connecting wire has become loose.
• Another component has failed.

A better investigation adds the ammeter and voltmeter measurements.

Checking the supply voltage confirms that the battery or power supply is still operating. Measuring across the open gap then shows where the potential difference appears.

There is still one limitation: a voltage reading across the gap establishes that the circuit is open at that location, but it does not prove that a filament, rather than a loose connection, caused the break.

To identify the faulty component, the circuit can be disconnected from the supply and individual components and connections checked for continuity.

The meter connections matter:

• An ammeter must be connected in series.
• A voltmeter must be connected in parallel with the component being measured.

Connecting an ammeter directly across the open gap would create an unintended low-resistance path. It is not the correct way to measure the voltage there.

04What happens if one bulb breaks in a parallel circuit?

Now change the arrangement.

Instead of connecting both bulbs along one path, connect each bulb on its own branch across the same two battery terminals.

What happens when the upper bulb breaks?

The lower bulb stays illuminated.

This is different from a series circuit because the lower bulb has its own complete conducting path. The break in the upper branch does not interrupt that path.

Kirchhoff’s current law describes the arrangement:

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

If bulb A breaks:

\[ I_A=0 \]

Therefore:

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

The current in the surviving branch can continue flowing.

Does the remaining parallel bulb have the same brightness?

Under an important assumption, yes.

If both bulbs are connected directly across an ideal battery with a fixed terminal voltage, the potential difference across the surviving bulb remains unchanged.

For a fixed-resistance bulb:

\[ I_B=\frac{V}{R_B} \]

Since neither \(V\) nor \(R_B\) changes, the current and power of bulb B remain unchanged.

The failed bulb goes out, but the other bulb stays at its original brightness.

However, real batteries are not perfectly ideal. They have internal resistance, which means their terminal voltage can change when the total current changes.

When one parallel branch opens, the total current drawn from the battery decreases. This reduces the voltage lost inside the battery, so the terminal voltage can increase.

The remaining bulb may therefore become brighter.

A common resistor placed before the parallel branches can produce a similar effect.

The important distinction is that parallel branches share the same terminal voltage, not necessarily the same current.

For more practice with this distinction, see How to Compare Bulb Brightness in Series-Parallel Circuits.

05Worked examples

06Questions and solutions

Question 1

Two identical bulbs, A and B, each have a resistance of 6.0 Ω. They are connected in series to an ideal 12.0 V battery.

Bulb A breaks.

Calculate:

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

(b) The current through bulb B after the break.

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

Solution 1

Bulb B goes out. After A breaks, the current through both bulbs is zero, the voltage across B is zero, and the full 12.0 V appears across the broken filament.

Step 1: Calculate the initial total resistance.

The bulbs are in series:

\[ R_{\text{total}}=6.0+6.0 =12.0\ \Omega \]

Step 2: Calculate the initial current.

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

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

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

Step 3: Find the current after A breaks.

The open filament interrupts the only current path.

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

Step 4: Find the voltage across bulb B.

\[ V_B=IR_B \]

\[ V_B=0\times6.0 \]

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

Step 5: Find the voltage across the gap.

Applying Kirchhoff’s voltage law:

\[ V_{\text{gap}}+V_A+V_B=12.0 \]

Since both intact resistive components have zero voltage drop:

\[ \boxed{V_{\text{gap}}=12.0\ \text{V}} \]

The fact that bulb B is still physically intact doesn’t matter. It has no current and therefore produces no light.

Question 2

An ideal 18.0 V battery is connected to three bulbs in series:

• Bulb A: 3.0 Ω
• Bulb B: 6.0 Ω
• Bulb C: 9.0 Ω

Bulb B breaks.

(a) Calculate the initial current through the circuit.

(b) Calculate the initial potential difference across each bulb.

(c) Determine the current and potential difference across bulbs A and C after B breaks.

(d) A student argues that bulb C must receive the entire 18.0 V because it has the greatest resistance. Explain why this prediction is incorrect.

Solution 2

After B breaks, bulbs A and C both have zero current and zero potential difference across them. The entire 18.0 V appears across the broken filament in B.

Step 1: Calculate the initial total resistance.

\[ R_{\text{total}}=3.0+6.0+9.0 \]

\[ R_{\text{total}}=18.0\ \Omega \]

Step 2: Calculate the initial current.

\[ I=\frac{V}{R_{\text{total}}} \]

\[ I=\frac{18.0}{18.0} \]

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

Step 3: Calculate the voltage across each bulb.

For bulb A:

\[ V_A=IR_A=1.0\times3.0 \]

\[ V_A=3.0\ \text{V} \]

For bulb B:

\[ V_B=1.0\times6.0 =6.0\ \text{V} \]

For bulb C:

\[ V_C=1.0\times9.0 =9.0\ \text{V} \]

Check the result using KVL:

\[ 3.0+6.0+9.0=18.0\ \text{V} \]

Step 4: Determine the current after B breaks.

There is now an open circuit.

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

This applies to both A and C because they remain in the same interrupted series path.

Step 5: Calculate the voltage across A and C.

For A:

\[ V_A=0\times3.0 =\boxed{0\ \text{V}} \]

For C:

\[ V_C=0\times9.0 =\boxed{0\ \text{V}} \]

The gap across B therefore has:

\[ V_{\text{gap}}=18.0\ \text{V} \]

Step 6: Evaluate the student’s prediction.

The student’s reasoning is tempting because bulb C has the greatest resistance. Before the break, it does have the largest voltage drop.

However, the relationship \(V=IR\) depends on the current flowing through the component.

After B breaks, the current is zero. Therefore, even though C has the largest resistance, its voltage drop is zero.

The mistake is applying the original voltage-divider reasoning after the circuit conditions have changed.

Question 3

An ideal 12.0 V battery is connected to a 2.0 Ω resistor in series with two parallel bulbs, each with a resistance of 6.0 Ω.

One bulb breaks.

Assume the bulbs behave as fixed resistors.

(a) Calculate the total current before the break.

(b) Calculate the current and electrical power dissipated by each bulb before the break.

(c) Calculate the total current and power dissipated by the surviving bulb after the break.

(d) Explain why the surviving parallel bulb becomes brighter, even though it is still connected in parallel.

Solution 3

The surviving bulb becomes brighter. Its power increases from 8.64 W to 13.5 W because the voltage across the parallel section increases when the current through the common resistor decreases.

Step 1: Calculate the equivalent resistance of the parallel bulbs before the break.

For two resistors in parallel:

\[ \frac{1}{R_{\text{parallel}}} =\frac{1}{6.0}+\frac{1}{6.0} \]

Therefore:

\[ R_{\text{parallel}}=3.0\ \Omega \]

Step 2: Calculate the total circuit resistance.

The 2.0 Ω resistor is in series with the parallel section.

\[ R_{\text{total}}=2.0+3.0 \]

\[ R_{\text{total}}=5.0\ \Omega \]

Step 3: Calculate the total current.

\[ I_{\text{total}}=\frac{12.0}{5.0} \]

\[ \boxed{I_{\text{total}}=2.4\ \text{A}} \]

Step 4: Calculate the voltage across the parallel section.

The voltage across the common resistor is:

\[ V_{\text{common}}=IR \]

\[ V_{\text{common}}=2.4\times2.0 =4.8\ \text{V} \]

Therefore, the voltage across the parallel bulbs is:

\[ V_{\text{parallel}}=12.0-4.8 \]

\[ V_{\text{parallel}}=7.2\ \text{V} \]

Step 5: Calculate the current through each bulb.

Both bulbs have the same potential difference:

\[ I_{\text{bulb}}=\frac{7.2}{6.0} \]

\[ I_{\text{bulb}}=1.2\ \text{A} \]

Step 6: Calculate the initial power of each bulb.

\[ P=I^2R \]

\[ P=1.2^2\times6.0 \]

\[ \boxed{P=8.64\ \text{W}} \]

Step 7: Recalculate the circuit after one bulb breaks.

Only one parallel branch remains complete.

The external equivalent resistance becomes:

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

The total resistance is now:

\[ R_{\text{total}}=6.0+2.0 \]

\[ R_{\text{total}}=8.0\ \Omega \]

The new total current is:

\[ I_{\text{total}}=\frac{12.0}{8.0} \]

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

Step 8: Calculate the new voltage across the surviving bulb.

The common resistor now has a voltage drop of:

\[ V_{\text{common}}=1.5\times2.0 =3.0\ \text{V} \]

Therefore:

\[ V_{\text{bulb}}=12.0-3.0 \]

\[ V_{\text{bulb}}=9.0\ \text{V} \]

The surviving bulb’s current is:

\[ I_{\text{bulb}}=\frac{9.0}{6.0} =1.5\ \text{A} \]

Step 9: Calculate the new power.

\[ P=I^2R \]

\[ P=1.5^2\times6.0 \]

\[ \boxed{P=13.5\ \text{W}} \]

The surviving bulb’s power increases by:

\[ \frac{13.5-8.64}{8.64}\times100 =56.25\% \]

The student who predicts unchanged brightness is assuming that the voltage across the parallel branches stays constant.

But the common resistor prevents that. When one branch opens, the total current decreases, reducing the voltage drop across the common resistor. More voltage is then available to the surviving bulb.

Question 4

A student designs an investigation to test what happens when a bulb breaks in a series circuit.

The apparatus consists of:

• A 12.0 V low-voltage DC supply.
• Two bulbs, A and B, each modelled as a 20 Ω resistor.
• An ammeter connected in series.
• A removable link, X, that can introduce an open circuit.
• Voltmeters to measure the supply voltage, the voltage across each bulb, and the voltage across X.

The following illustrative readings are recorded:

(a) Explain how these results support the open-circuit model.

(b) Explain why observing that both bulbs go dark would not, by itself, establish the cause of the failure.

(c) The student claims that these readings prove bulb A’s filament has broken. Evaluate this conclusion.

(d) Describe one improvement to the investigation that would help identify the actual faulty component.

Solution 4

The readings demonstrate an open circuit at X, but they do not prove that bulb A’s filament is the faulty component. The current and voltage measurements support the circuit model, while the location and physical cause of the break require further investigation.

Step 1: Check the readings when X is closed.

The two bulbs have a combined resistance of:

\[ R_{\text{total}}=20+20 =40\ \Omega \]

Using Ohm’s law:

\[ I=\frac{12.0}{40} \]

\[ I=0.30\ \text{A} \]

The voltage across each bulb is:

\[ V=IR \]

\[ V=0.30\times20 =6.0\ \text{V} \]

These values agree with the initial readings.

Step 2: Explain the readings when X is opened.

Opening X interrupts the only conducting path.

Therefore:

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

The potential difference across each intact bulb is:

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

The full supply voltage appears across X:

\[ V_X=12.0\ \text{V} \]

These results agree with both Kirchhoff’s current law and Kirchhoff’s voltage law.

Step 3: Evaluate the observation-only investigation.

Seeing both bulbs turn off is consistent with an open circuit, but it doesn’t establish that an open circuit is the actual cause.

A failed power supply or a loose connection could also leave both bulbs dark.

The additional supply-voltage measurement rules out a collapsed supply in this particular investigation. The ammeter confirms that the steady-state current is zero, while the voltmeter shows that the potential difference is across X.

Step 4: Evaluate the claim about bulb A.

The student’s conclusion is too specific.

The experiment deliberately opens X. It demonstrates the behaviour of a series circuit when a conducting path is interrupted.

It does not establish that bulb A’s filament has broken. A break at any point in the same series path could produce similar readings.

Step 5: Improve the investigation.

To identify the faulty component, disconnect the supply and check the continuity of individual bulbs and connecting leads.

Alternatively, place the removable link at the position of the suspected filament so that opening it directly models that component’s failure.

The key improvement is to distinguish between demonstrating the electrical behaviour of an open circuit and identifying the physical location of a fault.

Question 5

A battery has an emf of 18.0 V and an internal resistance of 1.0 Ω.

It is connected to two parallel branches:

• Branch 1 contains bulbs A and B in series.
• Branch 2 contains bulbs C and D in series.

Each bulb has a resistance of 4.0 Ω. Treat all intact bulbs as fixed resistors, and assume ideal connecting wires and no leakage between branches.

The filament in bulb A breaks.

(a) Identify which bulbs remain illuminated immediately after the circuit reaches steady state.

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

(c) Calculate the terminal voltage of the battery and the power dissipated by each bulb before the break.

(d) Calculate the new total current and terminal voltage after A breaks.

(e) Calculate the new power dissipated by bulbs C and D.

(f) Explain why the surviving parallel branch becomes brighter, and determine what would happen to its brightness if the battery had zero internal resistance.

Solution 5

Bulbs A and B both go out, while bulbs C and D remain illuminated and become brighter. The key is that the battery has internal resistance, so removing one parallel branch increases the terminal voltage available to the surviving branch.

Step 1: Analyse the two branches before the break.

Each branch contains two 4.0 Ω bulbs in series.

Therefore:

\[ R_{\text{branch}}=4.0+4.0 \]

\[ R_{\text{branch}}=8.0\ \Omega \]

The two branches are in parallel:

\[ \frac{1}{R_{\text{load}}} =\frac{1}{8.0}+\frac{1}{8.0} \]

\[ R_{\text{load}}=4.0\ \Omega \]

Step 2: Include the battery’s internal resistance.

The internal resistance is in series with the external load.

\[ R_{\text{total}}=4.0+1.0 \]

\[ R_{\text{total}}=5.0\ \Omega \]

The total battery current is:

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

\[ I_{\text{total}}=\frac{18.0}{5.0} \]

\[ \boxed{I_{\text{total}}=3.6\ \text{A}} \]

Step 3: Calculate the initial terminal voltage.

The terminal voltage is the emf minus the voltage lost across the internal resistance:

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

\[ V_{\text{terminal}}=18.0-(3.6\times1.0) \]

\[ \boxed{V_{\text{terminal}}=14.4\ \text{V}} \]

This is the voltage across the two external parallel branches.

Step 4: Calculate the current through each branch.

The branches have equal resistance, so they carry equal currents.

\[ I_{\text{branch}}=\frac{14.4}{8.0} \]

\[ \boxed{I_{\text{branch}}=1.8\ \text{A}} \]

The total current is therefore:

\[ I_{\text{total}}=1.8+1.8 =3.6\ \text{A} \]

This confirms Kirchhoff’s current law at the junction.

Since A and B are in series, both carry 1.8 A. The same applies to C and D.

Step 5: Calculate the initial power of each bulb.

For each 4.0 Ω bulb:

\[ P=I^2R \]

\[ P=1.8^2\times4.0 \]

\[ \boxed{P=12.96\ \text{W}} \]

Before the break, all four bulbs dissipate 12.96 W each in this model.

Step 6: Analyse the circuit after bulb A breaks.

The break interrupts branch 1.

Bulb B is still intact, but it is in series with the broken bulb A. It therefore also carries zero current.

Neither A nor B is illuminated.

Branch 2, containing C and D, still has a complete conducting path.

The external load now consists of only one branch:

\[ R_{\text{load}}=8.0\ \Omega \]

Including internal resistance:

\[ R_{\text{total}}=8.0+1.0 \]

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

Step 7: Calculate the new total current.

\[ I_{\text{total}}=\frac{18.0}{9.0} \]

\[ \boxed{I_{\text{total}}=2.0\ \text{A}} \]

Notice that the total current has decreased from 3.6 A to 2.0 A.

Step 8: Calculate the new terminal voltage.

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

\[ V_{\text{terminal}}=18.0-(2.0\times1.0) \]

\[ \boxed{V_{\text{terminal}}=16.0\ \text{V}} \]

The terminal voltage has increased from 14.4 V to 16.0 V.

This happens because the reduced total current produces a smaller voltage drop across the battery’s internal resistance.

Step 9: Calculate the current through bulbs C and D.

The surviving branch has a resistance of 8.0 Ω.

\[ I_{\text{branch}}=\frac{16.0}{8.0} \]

\[ \boxed{I_{\text{branch}}=2.0\ \text{A}} \]

Since C and D are in series, both carry this current.

Step 10: Calculate their new power.

For each surviving bulb:

\[ P=I^2R \]

\[ P=2.0^2\times4.0 \]

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

Their power has increased from 12.96 W to 16.0 W.

The increase is:

\[ \frac{16.0-12.96}{12.96}\times100 \approx23.5\% \]

Therefore, C and D become brighter in the fixed-resistance model.

Step 11: Consider what changes if the battery has zero internal resistance.

This is the important final check.

If \(r=0\), the terminal voltage is always equal to the emf:

\[ V_{\text{terminal}}=18.0\ \text{V} \]

Before the break, each branch carries:

\[ I_{\text{branch}}=\frac{18.0}{8.0} =2.25\ \text{A} \]

After A breaks, branch 1 opens, but branch 2 still has the same 18.0 V across it.

Its current therefore remains:

\[ I_{\text{branch}}=\frac{18.0}{8.0} =2.25\ \text{A} \]

The power of C and D remains:

\[ P=2.25^2\times4.0 \]

\[ P=20.25\ \text{W} \]

Thus, with an ideal battery, C and D would remain at exactly the same brightness.

The tempting prediction that parallel bulbs are always unaffected is only correct when their supply voltage stays fixed.

In this question, the battery’s emf stays fixed, but its terminal voltage changes because of internal resistance. That shared resistance couples the two otherwise separate branches.

This is why the most reliable approach to a bulb-failure problem is to trace the conducting paths first, apply Kirchhoff’s laws second, and only then calculate resistance, current, voltage, and power.

Once you can handle these failures, the next useful step is comparing brightness in more complicated mixed circuits. The same reasoning extends naturally to finding branch currents in a three-branch parallel circuit.