Why Series Circuits Go Dark When One Bulb Fails but Parallel Ones Don’t

A string of decorative lights goes completely dark because one tiny bulb has failed. In the next room, the tube light keeps glowing even though the fan has stopped.

This everyday puzzle is one of the most reliably tested ideas in Current Electricity. It appears in the Karnataka PU board syllabus, in KCET Physics, in NEET and in JEE. A student who understands why it happens can solve a whole family of circuit questions in seconds, without memorising anything extra.

Instead of a lecture, this guide works through three small cases, each ending with a quick question of your own. Then it gives you the board-exam answer in four lines and shows where the idea returns in Class 10, Class 12 and the entrance exams.

Case One: The Festival Lights That All Died Together

What happened

It is the night before the festival. You plug in a long string of small bulbs, and the whole string stays dark. You check bulb after bulb until you find one with a broken filament. Replace that single bulb, and every other bulb lights up again.

The rule behind it

Electric current flows only through a complete, closed loop from one terminal of the battery to the other. In a series circuit, the bulbs are joined end to end, so the charge has only one route. A broken filament opens that route at one point, and with the loop open, the current falls to zero everywhere.

Here is the part students often miss. Every other bulb now carries no current, so the voltage across it is V = IR = 0. The entire battery voltage appears across the gap. That is what an examiner is hinting at when asking for “the potential difference across the faulty bulb.”

The same story in numbers

Three identical 12 Ω bulbs are connected to an ideal 12 V battery.

Series bulbs are dim even when they work, because each gets only a third of the voltage, and power varies as V²/R. Each series bulb therefore produces about one ninth of the power it would produce on the full 12 V.

Case Two: The Tube Light That Outlived the Fan

In the next room, the ceiling fan stops because its capacitor fails. The tube light and the television carry on as if nothing had happened.

The appliances in a house are in parallel. Each one sits on its own branch, connected directly across the same two supply lines, so each branch has the full supply voltage and its own complete loop to the source. When the fan’s branch opens, the other branches still have a complete path, the same voltage and the same current as before.

One thing does change: the total current drawn from the supply falls, because one branch has gone. In the numbers above, the main-line current dropped from 3 A to 2 A.

A small twist: the battery’s own resistance

A real battery has a little internal resistance r, so its terminal voltage is V = E − Ir. When one parallel branch opens, the total current falls, Ir becomes smaller, and V rises slightly. The surviving bulbs therefore glow marginally brighter, a favourite statement-type question in NEET and JEE.

Why home wiring works this way

Parallel wiring gives every appliance the full mains voltage and lets each one be switched on or off independently. See how domestic electric circuits are put together. A fuse, by contrast, is placed in series on purpose, so that when too much current flows it melts, opens the circuit and protects everything else. That is explained in how fuses and MCBs protect your home.

Case Three: Which Bulb Is Brighter?

Two bulbs are rated 60 W and 100 W at 220 V. They are connected first in series and then in parallel across a 220 V supply. Which glows brighter each time?

The trick is to pick the power formula by what the two bulbs share. In series they share the current, so use P = I²R. In parallel they share the voltage, so use P = V²/R.

Step 1: find the resistances using R = V²/P. The 60 W bulb has 220²/60 ≈ 807 Ω, and the 100 W bulb has 220²/100 = 484 Ω.

Step 2: series. The current is I = 220/(807 + 484) ≈ 0.17 A. Then P = I²R gives about 23.4 W for the 60 W bulb and about 14.1 W for the 100 W bulb. The 60 W bulb is brighter because its resistance is higher.

Step 3: parallel. Each bulb gets the full 220 V, so each delivers its rated power, and the 100 W bulb is brighter.

When the bulb is shorted instead of broken

A broken filament is an open circuit. A short is the opposite: the bulb is bypassed, and the rest of the circuit gets more voltage. Short one of the three 12 Ω bulbs in the series string from Case One, and the other two share 12 V, getting 6 V each instead of 4 V. Power per bulb rises from 16/12 ≈ 1.33 W to 36/12 = 3 W, a 2.25-fold increase. “Fails” and “shorted” give opposite answers, so read the question twice.

Where the Formulas Come From

Series. The same current I flows through every resistor, and the voltages add, so V = V₁ + V₂ + V₃. Then IR_eq = IR₁ + IR₂ + IR₃, which gives R_eq = R₁ + R₂ + R₃.

Parallel. The same voltage V acts across every branch, and the currents add, so I = I₁ + I₂ + I₃. Then V/R_eq = V/R₁ + V/R₂ + V/R₃, which gives 1/R_eq = 1/R₁ + 1/R₂ + 1/R₃.

Each formula comes from one simple idea: current is shared in series, and voltage is shared in parallel. The full details of how resistors combine are worth a second read.

Where This Shows Up Next

In Class 10, series and parallel circuits sit in the chapter on Electricity (Chapter 11 in the CBSE book and Chapter 12 in the Karnataka SSLC textbook). In Class 12 and II PUC, the same reasoning grows into Current Electricity, with Kirchhoff’s rules, bridge circuits and measuring instruments. Questions in KCET, NEET UG and JEE Main all rest on it, from quick MCQs on brightness and meter readings to multi-resistor networks, so it repays the time. Syllabus portions change from year to year, so confirm the details with your board and the latest exam syllabus.

To build speed, work through these steps for solving resistance numericals and then see how to solve circuit problems fast in the board exam. Students who carry board and entrance preparation together can also read about balancing board exams and competitive exam preparation in Science PU.

Frequently Asked Questions

Q1. Why does a series circuit stop working when one bulb fails?

A series circuit has only one path for the current. When a filament breaks, that path is opened, the current falls to zero everywhere, and every bulb goes dark.

Q2. Why do parallel bulbs keep glowing when one fails?

Each bulb sits on its own branch directly across the supply. A broken branch carries no current, but the other branches still form complete loops and keep the same voltage, so they glow as before.

Q3. Why are homes wired in parallel instead of series?

Parallel wiring gives every appliance the full supply voltage and lets each one be switched on or off independently. In series, appliances would share the voltage and all stop together if one failed.

Q4. Which bulb glows brighter in series, and which in parallel?

In series, the same current flows through both, so the bulb with the higher resistance (the lower wattage rating) glows brighter, since P = I²R. In parallel, the voltage is the same, so the bulb with the lower resistance (the higher wattage rating) glows brighter, since P = V²/R.

Q5. What happens to the other bulbs when one parallel bulb is removed?

With an ideal battery, nothing changes for them: each remaining bulb still has the full voltage across it. With a real battery that has internal resistance, the total current falls, so the terminal voltage rises slightly and the remaining bulbs glow marginally brighter.

Paths Decide Everything

Once you stop memorising and start asking “how many paths does the current have?”, the dark string of lights stops being a puzzle. It becomes a rule you can apply to any circuit that a board paper, KCET, NEET or JEE can throw at you.

If you would like help turning this kind of reasoning into exam-hall speed, our team can walk you through how the board-synchronized programmes at Deeksha’s PU colleges.