How to Compare Bulb Brightness in Series-Parallel Circuits

Three identical bulbs are connected to a 9 V battery. Bulb A sits on the main wire before the circuit splits into two branches, with bulb B on one branch and bulb C on the other. Which bulb will be brightest?

You might predict equal brightness because the bulbs are identical. But A will be brightest. If each bulb has the same constant resistance, A dissipates four times as much electrical power as B or C.

The trick to comparing bulb brightness in mixed series-parallel circuits is to work out where current splits and how voltage is shared before calculating power. A bulb’s position in the circuit doesn’t determine its brightness. The electrical power it receives does.

01Follow the current before calculating anything

Consider this arrangement:

• An ideal 9 V battery supplies the circuit.
• Bulb A is connected directly after the positive terminal.
• The wire then reaches junction J, where it splits.
• The upper branch passes through bulb B.
• The lower branch passes through bulb C.
• Both branches meet again at junction K, before returning to the battery’s negative terminal.

Bulb A is in series with the parallel combination of B and C.

First, conserve current at the junction

Imagine following the electric charge travelling through A. It reaches J and has two available paths.

Some charge flows through B, while the rest flows through C. At K, the currents recombine.

Here’s the important point: charge isn’t consumed by either bulb.

In steady operation, the current entering a junction must equal the total current leaving it.

Therefore,

\[
I_A=I_B+I_C
\]

where \(I_A\), \(I_B\), and \(I_C\) are the currents through the corresponding bulbs, measured in amperes (\(\mathrm{A}\)).

This is Kirchhoff’s current law, which follows from conservation of electric charge.

Now suppose B and C are identical and have the same resistance.

Both branches connect between J and K, so they experience the same potential difference. Their currents must therefore be equal:

\[
I_B=I_C
\]

Combining the two statements gives

\[
I_A=2I_B=2I_C
\]

Bulb A carries twice as much current as either B or C.

That’s our first useful result. We haven’t needed to calculate equivalent resistance yet.

Next, conserve energy around each loop

A battery transfers energy to the charges moving around the circuit.

As charges pass through the bulbs, electrical energy is transferred into heat and light.

For an ideal battery, the total voltage rise across the battery equals the sum of the voltage drops around any complete circuit loop.

Take conventional current as flowing from the battery’s positive terminal through the bulbs towards its negative terminal. Treat each bulb’s voltage drop as positive in that direction.

There are two loops to consider.

Through A and B:

\[
V_s=V_A+V_B
\]

Through A and C:

\[
V_s=V_A+V_C
\]

Here:

• \(V_s\) is the supply voltage in volts (\(\mathrm{V}\)).
• \(V_A\), \(V_B\), and \(V_C\) are the voltage drops across the corresponding bulbs.

Because both loops contain the same battery and bulb A,

\[
V_B=V_C
\]

This is Kirchhoff’s voltage law, an expression of energy conservation.

Notice what it does not say. It doesn’t say that every bulb receives the full 9 V.

The battery’s voltage is shared between A and the entire parallel section containing B and C.

So why is A brighter?

For a bulb modelled as a resistor, electrical power is

\[
P=IV
\]

where \(P\) is power measured in watts (\(\mathrm{W}\)).

If its resistance \(R\), measured in ohms (\(\Omega\)), is constant, we can also use

\[
P=I^2R
\]

For three identical constant-resistance bulbs, we’ve already established that A carries twice the current through B.

Therefore,

\[
\begin{aligned}
\frac{P_A}{P_B}
&=\frac{I_A^2R}{I_B^2R}\\
&=\frac{(2I_B)^2}{I_B^2}\\
&=4
\end{aligned}
\]

The same applies when comparing A with C.

So the power ranking is

\[
P_A>P_B=P_C
\]

In fact, the power ratio is \(4:1:1\).

This result depends on the bulbs having identical, constant resistances. We’ll deal with real filament bulbs, which don’t quite obey that model, shortly.

02How would you test the brightness prediction?

Suppose you want to investigate this arrangement using a low-voltage DC power supply and three identical filament lamps.

Start with the simplest possible experiment: connect the bulbs as described and compare their brightness by eye.

What should you expect?

The constant-resistance model predicts that A glows more brightly than B and C, while B and C look equally bright.

With suitable matched lamps, that is also the qualitative behaviour we’d expect to observe.

But there’s a problem with the first experimental design.

Looking at the bulbs cannot tell you exactly how much electrical power each bulb receives.

Human vision doesn’t measure watts. Filament temperature affects the amount of visible light produced, and even lamps with the same stated rating may have slightly different properties.

The first design can test the predicted ordering of brightness, but it cannot establish the predicted \(4:1:1\) electrical power ratio.

Improving the experiment: measure voltage and current

Instead of relying only on appearance, add electrical measurements.

The apparatus should be connected in this order:

1. Connect the positive terminal of a suitable low-voltage DC supply to an ammeter.
2. Connect the ammeter to bulb A.
3. Connect A to junction J, where the circuit divides into branches containing B and C.
4. Join the branches again at K, and connect K to the supply’s negative terminal.
5. Connect a voltmeter across A to measure its potential difference. Repeat across B and C.

To measure the current through an individual branch, move an ammeter into that branch in series with its bulb, or use separate branch ammeters.

An ammeter must not be connected directly across a bulb: its low resistance could create a short circuit.

The revised experiment measures the quantities needed to calculate power:

\[
P_A=V_AI_A
\]

\[
P_B=V_BI_B
\]

\[
P_C=V_CI_C
\]

It also lets you test the two conservation laws independently.

At junction J, check whether

\[
I_A=I_B+I_C
\]

Around the loops, check whether

\[
V_s\approx V_A+V_B
\]

and

\[
V_s\approx V_A+V_C
\]

The approximately equal signs allow for measurement uncertainty and small voltage drops in real wires, meters, and the supply.

Readings should be taken after the lamps have reached a reasonably steady temperature, using lamps and equipment rated for the chosen supply.

The investigation described here is a proposed experiment. The numerical results in the next section are predictions from an idealised model, not measurements from an experiment that was performed.

04The misleading rule: higher resistance means brighter

You might have learnt that a higher-resistance bulb dissipates more power.

That’s true when the compared bulbs carry the same current.

But suppose two bulbs experience the same voltage instead.

Using Ohm’s law,

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

For equal voltage, the lower resistance dissipates more power.

Both statements are correct. The condition that stays constant has changed.

The trouble with a mixed circuit is that you cannot assume the same current or voltage applies everywhere.

05A reliable method for any mixed bulb circuit

When a circuit looks complicated, use the same sequence every time.

First: identify the actual paths

Trace the wires, not the positions of the bulbs on the page.

Two bulbs are in parallel only if both ends connect to the same pair of circuit nodes.

Two bulbs carry the same current when there is no branching path between them.

Second: write the junction equations

For every important junction, use conservation of charge.

For example,

\[
I_1=I_2+I_3
\]

This prevents you from treating a branch current as though it were the total supply current.

Third: write the loop equations

Work out how the supply voltage is shared.

For example,

\[
V_s=V_A+V_{BC}
\]

Here \(V_{BC}\) is the voltage across the entire parallel section containing B and C.

This prevents you from giving every bulb the full battery voltage.

Fourth: simplify the resistance

Once the circuit structure is clear, combine genuinely series and parallel resistances.

Use the equivalent resistance to calculate the main current, then work backwards to individual bulbs.

Remember that an equivalent resistance represents an entire section. It isn’t another bulb.

Fifth: calculate each bulb’s power

Use whichever formula matches the information you know:

\[
P=VI,\qquad P=I^2R,\qquad P=\frac{V^2}{R}
\]

Choose the formula after identifying the current and voltage that actually apply to the bulb.

Finally: check conservation

Your current and voltage calculations should satisfy the junction and loop equations.

For an ideal circuit containing only resistive lamps,

\[
P_{\mathrm{source}}
=\sum P_{\mathrm{bulbs}}
\]

The symbol \(\sum\) means to add the powers of all the bulbs.

If you’re unsure about the basic series and parallel relationships, the separate tutorgum guides on series circuits and parallel circuits develop those foundations. The challenge here is recognising which rules apply to each part of the same network.

06Why real filament bulbs are more complicated

So far, we’ve treated bulbs as components with fixed resistance.

That is useful for learning circuit analysis, but a real incandescent filament doesn’t behave like an ideal fixed resistor.

When current flows through a tungsten filament, the filament heats up. Its electrical resistance increases as its temperature rises.

This means resistance can change when the circuit is rearranged.

For example, in our original three-identical-bulb arrangement, A carries more current than B or C.

In steady operation, A becomes hotter and develops a higher operating resistance than the other two.

The simple assumption

\[
R_A=R_B=R_C
\]

is therefore not generally valid for identical incandescent bulbs operating at different temperatures.

What remains valid?

Conservation of charge still requires

\[
I_A=I_B+I_C
\]

Conservation of energy still requires

\[
V_s=V_A+V_B=V_A+V_C
\]

And electrical power is still calculated using

\[
P=VI
\]

The change is that we cannot assign a single fixed resistance to each bulb and assume it stays unchanged as current changes.

For real bulbs, measuring current and voltage is often more reliable than predicting power from a cold resistance measurement.

There’s another limitation: electrical power and perceived brightness are not identical quantities.

An incandescent bulb converts much of its electrical input into thermal radiation outside the visible range. Its visible-light output changes with filament temperature.

For matched bulbs, comparing electrical powers is a useful way to predict brightness ordering. But a power ratio of \(4:1\) does not necessarily mean the eye will perceive one real bulb as exactly four times brighter.

Different lamp technologies, efficiencies, and ratings make this distinction even more important.

07Questions and solutions

For Questions 1-3 and 5, assume ideal DC voltage sources, negligible connecting-wire resistance, and bulbs with the stated constant resistances. Compare brightness using electrical power, assuming the light outputs are comparable for a given power.

Question 4 explicitly concerns real tungsten-filament bulbs.

Question 1

Three identical bulbs, A, B, and C, each have a constant resistance of \(6.0\ \Omega\).

A and B are connected in series along the upper branch of a circuit. C is alone on the lower branch. The two branches are connected in parallel across an ideal 12 V battery.

Calculate:

• The current through each bulb.
• The electrical power dissipated by each bulb.
• The brightness ranking.
• The total current supplied by the battery.

Solution 1

Bulb C is brightest, while A and B are equally dim. Their electrical powers are \(P_C=24\ \mathrm{W}\) and \(P_A=P_B=6.0\ \mathrm{W}\). The battery supplies \(3.0\ \mathrm{A}\).

The key is that A and B share the upper branch’s voltage, while C receives the full supply voltage.

Step 1: Identify the current and voltage relationships.

The upper branch contains A and B in series, so

\[
I_A=I_B
\]

Both complete branches have 12 V across them.

Therefore,

\[
V_A+V_B=12\ \mathrm{V}
\]

and

\[
V_C=12\ \mathrm{V}
\]

Step 2: Calculate the branch currents.

The upper branch resistance is

\[
R_{AB}=6.0+6.0=12\ \Omega
\]

Thus,

\[
\begin{aligned}
I_A=I_B&=\frac{12\ \mathrm{V}}{12\ \Omega}\\
&=1.0\ \mathrm{A}
\end{aligned}
\]

For C,

\[
\begin{aligned}
I_C&=\frac{12\ \mathrm{V}}{6.0\ \Omega}\\
&=2.0\ \mathrm{A}
\end{aligned}
\]

Step 3: Calculate power.

Using \(P=I^2R\),

\[
\begin{aligned}
P_A&=(1.0)^2(6.0)=6.0\ \mathrm{W}\\
P_B&=(1.0)^2(6.0)=6.0\ \mathrm{W}\\
P_C&=(2.0)^2(6.0)=24\ \mathrm{W}
\end{aligned}
\]

C dissipates four times as much power as either A or B.

Step 4: Calculate the source current.

At the junction, the branch currents add:

\[
\begin{aligned}
I_{\mathrm{total}}&=I_A+I_C\\
&=1.0+2.0\\
&=3.0\ \mathrm{A}
\end{aligned}
\]

Checking the total power,

\[
\begin{aligned}
P_{\mathrm{source}}&=VI\\
&=(12)(3.0)\\
&=36\ \mathrm{W}
\end{aligned}
\]

The bulb powers also total \(6.0+6.0+24=36\ \mathrm{W}\).

The common mistake is to assume every bulb in a parallel circuit receives the full battery voltage. The upper and lower branches each receive 12 V, but A and B must share their branch’s 12 V because they’re in series.

Question 2

Three identical bulbs, A, B, and C, each have a constant resistance of \(4.0\ \Omega\), and are connected to an ideal 12 V battery.

Initially, A is in series with the parallel combination of B and C.

The circuit is then rewired so that all three bulbs are connected in series, with the same battery.

(a) Calculate the power dissipated by each bulb before and after the change. Which bulbs brighten, dim, or remain unchanged?

(b) Calculate the change in total power delivered by the battery. Explain why a bulb can retain exactly the same brightness despite being moved from a parallel arrangement into a series arrangement.

(c) Now suppose C has a resistance of \(8.0\ \Omega\), while A and B remain \(4.0\ \Omega\). For this altered set of bulbs, calculate B’s power before and after the same rewiring. Does the unchanged-brightness result survive?

Solution 2

With three identical bulbs, A dims from 16 W to 4.0 W, while B and C remain at 4.0 W each. Total battery power halves from 24 W to 12 W. The unchanged brightness is a special consequence of equal resistances; it disappears when C becomes \(8.0\ \Omega\).

Part (a): Compare the two arrangements.

Initially, the B-C equivalent resistance is

\[
R_{BC}=\frac{4.0\times4.0}{4.0+4.0}
=2.0\ \Omega
\]

The total resistance is

\[
R_{\mathrm{eq}}=4.0+2.0=6.0\ \Omega
\]

Therefore,

\[
\begin{aligned}
I_A&=\frac{12\ \mathrm{V}}{6.0\ \Omega}\\
&=2.0\ \mathrm{A}
\end{aligned}
\]

The parallel branches are identical, so the current divides equally:

\[
I_B=I_C=1.0\ \mathrm{A}
\]

Using \(P=I^2R\),

\[
\begin{aligned}
P_A&=(2.0)^2(4.0)=16\ \mathrm{W}\\
P_B&=(1.0)^2(4.0)=4.0\ \mathrm{W}\\
P_C&=(1.0)^2(4.0)=4.0\ \mathrm{W}
\end{aligned}
\]

After rewiring, all three bulbs are in series.

The new resistance is

\[
R_{\mathrm{eq}}=4.0+4.0+4.0
=12\ \Omega
\]

The same current now passes through every bulb:

\[
I=\frac{12\ \mathrm{V}}{12\ \Omega}
=1.0\ \mathrm{A}
\]

Each dissipates

\[
P=(1.0)^2(4.0)=4.0\ \mathrm{W}
\]

So A dims, while B and C remain unchanged.

Part (b): Compare source power and explain the surprise.

Initially,

\[
P_{\mathrm{source}}=(12)(2.0)
=24\ \mathrm{W}
\]

After rewiring,

\[
P_{\mathrm{source}}=(12)(1.0)
=12\ \mathrm{W}
\]

The total source power has halved.

The tempting prediction is that B and C must change brightness because they’re no longer in parallel.

But brightness depends on the actual current and voltage through each bulb, not whether its connection has been given a different name.

Initially, the total current through A was \(2.0\ \mathrm{A}\), split equally into two \(1.0\ \mathrm{A}\) branches.

After rewiring, the entire circuit carries \(1.0\ \mathrm{A}\).

B and C therefore carry exactly the same current as before.

For equal fixed resistances, this means exactly the same power.

Part (c): Change C to \(8.0\ \Omega\).

Initially, B and C are still in parallel:

\[
\begin{aligned}
R_{BC}
&=\frac{4.0\times8.0}{4.0+8.0}\\
&=\frac83\ \Omega
\end{aligned}
\]

The total resistance is

\[
R_{\mathrm{eq}}=4.0+\frac83
=\frac{20}{3}\ \Omega
\]

Therefore,

\[
\begin{aligned}
I_A&=\frac{12}{20/3}\\
&=1.8\ \mathrm{A}
\end{aligned}
\]

The parallel section receives

\[
\begin{aligned}
V_{BC}&=I_AR_{BC}\\
&=(1.8)\left(\frac83\right)\\
&=4.8\ \mathrm{V}
\end{aligned}
\]

So B carries

\[
I_B=\frac{4.8}{4.0}
=1.2\ \mathrm{A}
\]

Its initial power is

\[
P_B=(1.2)^2(4.0)
=5.76\ \mathrm{W}
\]

After rewiring all three bulbs in series,

\[
R_{\mathrm{eq}}=4.0+4.0+8.0
=16\ \Omega
\]

The new current is

\[
I=\frac{12}{16}
=0.75\ \mathrm{A}
\]

B now dissipates

\[
P_B=(0.75)^2(4.0)
=2.25\ \mathrm{W}
\]

B dims from \(5.76\ \mathrm{W}\) to \(2.25\ \mathrm{W}\).

The earlier equality depended on the branch currents and the new series current happening to match. Once the resistances change, that matching condition disappears.

Question 3

Four identical bulbs, A, B, C, and D, each have a constant resistance of \(3.0\ \Omega\).

An ideal 12 V battery is connected to bulb A, followed by a junction with two branches:

• The upper branch contains bulb B alone.
• The lower branch contains bulbs C and D in series.

The branches rejoin before returning to the battery.

While operating, bulb C suddenly goes dark.

The following changes are observed:

• A becomes brighter.
• B becomes dimmer.
• D becomes brighter.

There are two proposed faults:

1. The filament of C has broken, leaving an open circuit.
2. A conducting wire has created an ideal short circuit directly across C, bypassing C but not D.

Identify the fault.

Calculate the power dissipated by each bulb after the fault, and explain why the other proposed fault cannot account for the observations.

Solution 3

The fault is a short circuit directly across C. After the fault, A dissipates approximately \(21.3\ \mathrm{W}\), B and D each dissipate approximately \(5.33\ \mathrm{W}\), and C dissipates \(0\ \mathrm{W}\).

This explains why A and D brighten while B dims.

An open filament in C would instead make D go dark and reduce A’s power.

Step 1: Work out the original current distribution.

Before the fault, C and D are in series:

\[
R_{CD}=3.0+3.0=6.0\ \Omega
\]

The upper branch containing B has resistance \(3.0\ \Omega\).

The parallel combination is

\[
\begin{aligned}
R_{BCD}
&=\frac{3.0\times6.0}{3.0+6.0}\\
&=2.0\ \Omega
\end{aligned}
\]

Including A,

\[
R_{\mathrm{eq}}=3.0+2.0
=5.0\ \Omega
\]

Therefore,

\[
I_A=\frac{12}{5.0}
=2.4\ \mathrm{A}
\]

The voltage across the parallel section is

\[
V_{BCD}=(2.4)(2.0)
=4.8\ \mathrm{V}
\]

So the original branch currents are

\[
\begin{aligned}
I_B&=\frac{4.8}{3.0}=1.6\ \mathrm{A}\\
I_C=I_D&=\frac{4.8}{6.0}=0.80\ \mathrm{A}
\end{aligned}
\]

The junction check is

\[
1.6+0.80=2.4\ \mathrm{A}
\]

Original powers are therefore

\[
\begin{aligned}
P_A&=(2.4)^2(3.0)=17.28\ \mathrm{W}\\
P_B&=(1.6)^2(3.0)=7.68\ \mathrm{W}\\
P_C=P_D&=(0.80)^2(3.0)\\
&=1.92\ \mathrm{W}
\end{aligned}
\]

We can now compare these values with the faulty circuit.

Step 2: Analyse a short across C.

An ideal wire across C provides a zero-resistance bypass.

The potential difference across C becomes zero, so C dissipates no power.

However, D remains connected in the lower branch.

The circuit now contains A in series with the parallel combination of B and D.

Because B and D each have \(3.0\ \Omega\),

\[
R_{BD}=\frac{3.0\times3.0}{3.0+3.0}
=1.5\ \Omega
\]

The new total resistance is

\[
R_{\mathrm{eq}}=3.0+1.5
=4.5\ \Omega
\]

The current through A becomes

\[
\begin{aligned}
I_A&=\frac{12\ \mathrm{V}}{4.5\ \Omega}\\
&=\frac83\ \mathrm{A}\\
&\approx2.67\ \mathrm{A}
\end{aligned}
\]

A now carries more current than before, so it becomes brighter.

The voltage across the B-D parallel section is

\[
\begin{aligned}
V_{BD}&=I_AR_{BD}\\
&=\left(\frac83\right)(1.5)\\
&=4.0\ \mathrm{V}
\end{aligned}
\]

Therefore,

\[
I_B=I_D=\frac{4.0}{3.0}
=\frac43\ \mathrm{A}
\]

Step 3: Calculate the new powers.

For A,

\[
\begin{aligned}
P_A&=I_A^2R_A\\
&=\left(\frac83\right)^2(3.0)\\
&\approx21.3\ \mathrm{W}
\end{aligned}
\]

For B and D,

\[
\begin{aligned}
P_B=P_D&=\left(\frac43\right)^2(3.0)\\
&\approx5.33\ \mathrm{W}
\end{aligned}
\]

For the bypassed bulb C,

\[
P_C=0\ \mathrm{W}
\]

Comparing with the original powers:

Every change matches the observations.

Step 4: Rule out the open-circuit fault.

If C’s filament breaks, the entire lower branch stops conducting because C and D are in series.

D would go dark.

Only A and B would remain in the complete conducting path.

The total resistance would be

\[
R_{\mathrm{eq}}=3.0+3.0
=6.0\ \Omega
\]

The current would become

\[
I=\frac{12}{6.0}
=2.0\ \mathrm{A}
\]

A would dissipate

\[
P_A=(2.0)^2(3.0)
=12\ \mathrm{W}
\]

That is less than its original \(17.28\ \mathrm{W}\), so A would dim rather than brighten.

B would dissipate \(12\ \mathrm{W}\), becoming brighter, while D would be off.

Those predictions contradict the observations.

The diagnostic lesson is that a dark bulb doesn’t, by itself, tell you whether the component is open or shorted.

You need to examine what happens to the other branches. Their brightness changes reveal how the fault altered the current paths and voltage distribution.

Question 4

Three identical tungsten-filament bulbs are connected in the familiar arrangement: A in series with parallel bulbs B and C.

An ideal 10.0 V supply powers the circuit.

After the filaments reach steady operating temperatures, instruments give these readings:

• Current through A: \(0.40\ \mathrm{A}\).
• Voltage across A: \(7.5\ \mathrm{V}\).
• Voltage across B: \(2.5\ \mathrm{V}\).

Assume B and C have identical electrical characteristics and operate under identical conditions.

(a) Calculate the current, voltage, and electrical power of each bulb.

(b) Determine the operating resistance of A and B.

(c) A student claims that because the bulbs are identical, A must always dissipate exactly four times the power of B. Evaluate this claim using the measurements.

(d) Can the measurements establish that A appears exactly six times as bright as B? Explain.

Solution 4

A dissipates \(3.0\ \mathrm{W}\), while B and C each dissipate \(0.50\ \mathrm{W}\). The electrical power ratio is \(6:1:1\), not \(4:1:1\). The measurements show that A has a higher operating resistance than B and C.

They do not establish an exact ratio of perceived brightness.

Part (a): Apply the junction and loop constraints.

Because B and C have identical electrical characteristics and share the same voltage,

\[
I_B=I_C
\]

At the junction,

\[
I_A=I_B+I_C
\]

Therefore,

\[
I_B=I_C=\frac{0.40}{2}
=0.20\ \mathrm{A}
\]

Both parallel bulbs share the same voltage:

\[
V_B=V_C=2.5\ \mathrm{V}
\]

The loop check is

\[
7.5+2.5=10.0\ \mathrm{V}
\]

Now calculate electrical power using the measured current and voltage.

\[
\begin{aligned}
P_A&=(7.5)(0.40)\\
&=3.0\ \mathrm{W}\\[4pt]
P_B&=(2.5)(0.20)\\
&=0.50\ \mathrm{W}\\[4pt]
P_C&=(2.5)(0.20)\\
&=0.50\ \mathrm{W}
\end{aligned}
\]

These results are consistent with the total electrical power supplied:

\[
P_{\mathrm{source}}=(10.0)(0.40)
=4.0\ \mathrm{W}
\]

And

\[
3.0+0.50+0.50=4.0\ \mathrm{W}
\]

Part (b): Calculate operating resistance.

Resistance at a particular operating point is given by the measured voltage divided by current:

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

For A,

\[
\begin{aligned}
R_A&=\frac{7.5\ \mathrm{V}}{0.40\ \mathrm{A}}\\
&=18.75\ \Omega
\end{aligned}
\]

For B,

\[
\begin{aligned}
R_B&=\frac{2.5\ \mathrm{V}}{0.20\ \mathrm{A}}\\
&=12.5\ \Omega
\end{aligned}
\]

C has the same operating resistance as B.

Therefore,

\[
\frac{R_A}{R_B}=\frac{18.75}{12.5}
=1.5
\]

A has 1.5 times the operating resistance of B.

Part (c): Explain the failed fixed-resistance prediction.

The \(4:1\) result comes from assuming equal resistances and using

\[
P=I^2R
\]

The current through A really is twice the current through B:

\[
I_A=2I_B
\]

But the resistances are not equal.

Using the measured operating resistances,

\[
\begin{aligned}
\frac{P_A}{P_B}
&=\frac{I_A^2R_A}{I_B^2R_B}\\
&=4\frac{R_A}{R_B}\\
&=4(1.5)\\
&=6
\end{aligned}
\]

The ratio is \(6:1\).

A hotter tungsten filament has higher resistance, so identical bulbs operating at different temperatures do not necessarily have identical resistances.

The conservation equations remain correct. It is the constant-resistance assumption that fails.

Part (d): Distinguish electrical power from brightness.

The measurements establish that A dissipates six times as much electrical power as B.

They do not show that A appears exactly six times as bright.

The proportion of electrical power converted into visible light changes with filament temperature, and perceived brightness depends on the light emitted and the viewing conditions.

For these identical bulbs, A is expected to appear brighter, and B and C should appear equally bright. But the exact perceived-brightness ratio cannot be obtained from the electrical measurements alone.

Question 5

An ideal 14 V battery supplies three resistive bulbs in the following arrangement:

• Bulb A is in series with a parallel combination of B and C.
• Bulb A has a fixed resistance of \(3.0\ \Omega\).
• Bulb C has a fixed resistance of \(9.0\ \Omega\).
• Bulb B has an adjustable positive resistance \(R_B\).

Assume every selected resistance remains constant during operation, and that the bulbs’ light output can be ranked by their dissipated electrical power.

A designer wants B to be brighter than both A and C.

(a) Is there any positive value of \(R_B\) that makes B at least as bright as A? Prove your answer without testing individual resistance values. Find the smallest possible ratio \(P_A/P_B\).

(b) The designer replaces A with a \(2.0\ \Omega\) bulb, leaving C at \(9.0\ \Omega\). Find the complete range of \(R_B\) values for which B is strictly brighter than both other bulbs.

(c) Verify your result for \(R_B=6.0\ \Omega\) by calculating the power of all three bulbs.

(d) Explain why simply making \(R_B\) extremely small does not guarantee that B becomes the brightest bulb.

Solution 5

With A at \(3.0\ \Omega\) and C at \(9.0\ \Omega\), B can never equal or exceed A’s power. The smallest possible ratio is \(P_A/P_B=4/3\).

After A is replaced with \(2.0\ \Omega\), B becomes strictly the brightest bulb precisely when

\[
4.5\ \Omega<R_B<9.0\ \Omega
\]

At \(R_B=6.0\ \Omega\), B dissipates \(13.5\ \mathrm{W}\), A dissipates \(12.5\ \mathrm{W}\), and C dissipates \(9.0\ \mathrm{W}\).

The surprising constraint is that increasing B’s current also increases the current through A.

Part (a): Show why B cannot match A.

Start with conservation of current.

At the junction,

\[
I_A=I_B+I_C
\]

B and C have the same voltage across them.

Therefore,

\[
I_B=\frac{V_{BC}}{R_B}
\]

and

\[
I_C=\frac{V_{BC}}{R_C}
\]

Dividing these relationships gives

\[
\frac{I_C}{I_B}=\frac{R_B}{R_C}
\]

Define the positive, dimensionless ratio

\[
x=\frac{R_B}{R_C}
\]

Therefore,

\[
I_C=xI_B
\]

and conservation at the junction becomes

\[
I_A=I_B(1+x)
\]

Now compare the powers of A and B.

\[
\begin{aligned}
\frac{P_A}{P_B}
&=\frac{I_A^2R_A}{I_B^2R_B}\\
&=\frac{R_A}{R_B}(1+x)^2
\end{aligned}
\]

Because \(R_B=xR_C\),

\[
\frac{P_A}{P_B}
=\frac{R_A}{R_C}\frac{(1+x)^2}{x}
\]

For the original values, \(R_A=3.0\ \Omega\) and \(R_C=9.0\ \Omega\):

\[
\begin{aligned}
\frac{P_A}{P_B}
&=\frac13\frac{(1+x)^2}{x}\\
&=\frac43+\frac{(x-1)^2}{3x}
\end{aligned}
\]

Here’s the decisive constraint.

For every positive value of \(x\),

\[
(x-1)^2\geq0
\]

and

\[
3x>0
\]

Therefore,

\[
\frac{(x-1)^2}{3x}\geq0
\]

So

\[
\frac{P_A}{P_B}\geq\frac43
\]

Equality occurs when \(x=1\), meaning

\[
R_B=R_C=9.0\ \Omega
\]

Even at its best possible setting relative to A, B has only three-quarters of A’s power.

Therefore, no positive value of \(R_B\) can make B equally bright or brighter than A.

Notice that the supply voltage cancelled from the ratio. Increasing the ideal battery voltage cannot remove this particular limitation while the resistances remain fixed.

Part (b): Replace A with \(2.0\ \Omega\).

The current relationship remains

\[
I_A=I_B(1+x)
\]

But now,

\[
\frac{R_A}{R_C}=\frac29
\]

So

\[
\frac{P_A}{P_B}
=\frac29\frac{(1+x)^2}{x}
\]

For B to be brighter than A,

\[
P_B>P_A
\]

which is equivalent to

\[
\frac{P_A}{P_B}<1
\]

Substituting,

\[
\frac29\frac{(1+x)^2}{x}<1
\]

Because \(x>0\), we can multiply through without reversing the inequality:

\[
2(1+x)^2<9x
\]

Expanding and simplifying,

\[
2x^2-5x+2<0
\]

Factorising,

\[
(2x-1)(x-2)<0
\]

This product is negative between its two roots:

\[
\frac12<x<2
\]

Since

\[
x=\frac{R_B}{9.0\ \Omega}
\]

the condition for B to outshine A is

\[
4.5\ \Omega<R_B<18\ \Omega
\]

But B must also outshine C.

B and C share the same voltage. Therefore,

\[
P_B=\frac{V_{BC}^2}{R_B}
\]

and

\[
P_C=\frac{V_{BC}^2}{R_C}
\]

For B to dissipate more power than C, its resistance must be smaller:

\[
R_B<R_C=9.0\ \Omega
\]

Combining both conditions gives

\[
4.5\ \Omega<R_B<9.0\ \Omega
\]

This is the complete range in which B is strictly brighter than both A and C.

At either boundary, B ties with another bulb, so the inequalities must be strict.

Part (c): Verify the result at \(R_B=6.0\ \Omega\).

Now the resistances are

\[
R_A=2.0\ \Omega,\quad
R_B=6.0\ \Omega,\quad
R_C=9.0\ \Omega
\]

The parallel resistance is

\[
\begin{aligned}
R_{BC}
&=\frac{6.0\times9.0}{6.0+9.0}\\
&=3.6\ \Omega
\end{aligned}
\]

Total resistance:

\[
R_{\mathrm{eq}}=2.0+3.6
=5.6\ \Omega
\]

Main current:

\[
\begin{aligned}
I_A&=\frac{14\ \mathrm{V}}{5.6\ \Omega}\\
&=2.5\ \mathrm{A}
\end{aligned}
\]

The voltage across A is

\[
V_A=(2.5)(2.0)=5.0\ \mathrm{V}
\]

Therefore, the parallel section receives

\[
V_B=V_C=14-5.0=9.0\ \mathrm{V}
\]

The branch currents are

\[
\begin{aligned}
I_B&=\frac{9.0}{6.0}=1.5\ \mathrm{A}\\
I_C&=\frac{9.0}{9.0}=1.0\ \mathrm{A}
\end{aligned}
\]

And the junction check works:

\[
1.5+1.0=2.5\ \mathrm{A}
\]

Finally, the individual powers are

\[
\begin{aligned}
P_A&=(5.0)(2.5)=12.5\ \mathrm{W}\\
P_B&=(9.0)(1.5)=13.5\ \mathrm{W}\\
P_C&=(9.0)(1.0)=9.0\ \mathrm{W}
\end{aligned}
\]

Therefore,

\[
P_B>P_A>P_C
\]

B is the brightest, exactly as the derived resistance range predicts.

Part (d): Why doesn’t an extremely small B resistance solve the problem?

The tempting argument is that B is in parallel with C, so reducing B’s resistance increases B’s current.

That’s true when comparing B with C at the same voltage.

But A is upstream of both branches.

Every increase in B’s current also contributes to the current through A.

In fact,

\[
I_A=I_B+I_C
\]

And A dissipates power proportional to the square of that current.

As \(R_B\) approaches zero, B behaves increasingly like a conducting bypass.

Its voltage drop and dissipated power approach zero, while the remaining source voltage appears mainly across A.

So B becomes less powerful, not more.

At the opposite extreme, when \(R_B\) becomes extremely large, B carries almost no current and again dissipates very little power.

Neither extreme makes B the brightest.

The winning resistance must balance the current B receives against the voltage remaining across the parallel section.

That is why the best setting is not necessarily the smallest available resistance.

07What the brightness comparison tells you next

The most reliable way to compare bulb brightness is to separate three ideas:

• Junctions determine how current is shared.
• Complete loops determine how voltage is shared.
• The current and voltage across each bulb determine its electrical power.

The resistance calculations come after you’ve established those relationships.

For ideal identical bulbs, this gives exact power comparisons. For real incandescent bulbs, the current and voltage laws still work, but the resistance changes as the filaments heat.

That leads to the next useful experimental question: if a bulb’s resistance changes with current, how can we predict its operating point in a mixed circuit?

The answer is to measure or model its current-voltage characteristic, then combine that behaviour with the same junction and loop constraints we’ve used here. Instead of assuming a fixed resistance, you find the currents and voltages that satisfy both the bulb’s behaviour and the circuit’s conservation laws.