
Observe the following figure which bulb gets fuzed?

Answer
125.1k+ views
Hint: When the filament of a bulb is broken then the bulb is said to be fused. When it gets fused it does not emit light. When the bulb gets fused, the electric current is not allowed to pass through the filament and the bulb does not glow because the filament of the bulb is broken.
Complete Step by Step Solution:
An electric bulb is a device which produces light when electricity is passed through its terminals. The bulb has two thick contact wires in the centre with a thin wire attached between them. This thin wire is called filament.
One of the thick wires is connected to the metal case at the base of the bulb and the other is connected to the metal tip at the centre of the base. These two form the terminals. When electricity is passed through the terminals of the bulb, the filament gets heated up and produces light.
A bulb is said to be fused if the filament gets broken. Fused bulb doesn’t glow. The two terminals do not directly touch each other to avoid short circuit.
Current through bulb A can be calculated by –
${I_a} = {P_a}/{V_a} = 50/220$
${I_a} = 0.227A$
Current through bulb B can be calculated by –
${I_b} = {P_b}/{V_b} = 100/220$
${I_b} = 0.455A$.
Resistance of bulb A can be calculated by–
${R_a} = {220^2}/50 = 968\Omega $
Resistance of bulb B can be calculated by–
${R_b} = {220^2}/100 = 484\Omega $
Total resistance in series - $R = {R_a} + {R_b} = \left( {968 + 484} \right)\Omega = 1452\Omega $.
Current flowing through circuit is –
$I = V/R$
$ \Rightarrow I = 440/1452 = 0.303A$
Thus, $I > {I_a}$.
Bulb A will fuse because current drawn from the battery will be much greater than that of current in bulb A which results in melting of fuse wire and bulb gets fused.
Note: The tungsten coil (filament) has hot spots. Slowly the tungsten is boiled away and the hot spots boil away faster. Eventually a little vibration breaks the coil or just the turn on current causes the coil to expand as it starts to glow and that motion and stress breaks the coil. That’s why the bulbs often burn out when they are switched on.
If the break is clean, the bulb can be worked again by wiggling it until the broken ends touch and they tack weld themselves together or at least stick. Sometimes the bulb fixes itself when the filament contracts as it cools down and the broken ends are really close. Next time when the bulb is turned on, the 120V or 240V bridges the gap, the tack weld / stick happens.
Complete Step by Step Solution:
An electric bulb is a device which produces light when electricity is passed through its terminals. The bulb has two thick contact wires in the centre with a thin wire attached between them. This thin wire is called filament.
One of the thick wires is connected to the metal case at the base of the bulb and the other is connected to the metal tip at the centre of the base. These two form the terminals. When electricity is passed through the terminals of the bulb, the filament gets heated up and produces light.
A bulb is said to be fused if the filament gets broken. Fused bulb doesn’t glow. The two terminals do not directly touch each other to avoid short circuit.
Current through bulb A can be calculated by –
${I_a} = {P_a}/{V_a} = 50/220$
${I_a} = 0.227A$
Current through bulb B can be calculated by –
${I_b} = {P_b}/{V_b} = 100/220$
${I_b} = 0.455A$.
Resistance of bulb A can be calculated by–
${R_a} = {220^2}/50 = 968\Omega $
Resistance of bulb B can be calculated by–
${R_b} = {220^2}/100 = 484\Omega $
Total resistance in series - $R = {R_a} + {R_b} = \left( {968 + 484} \right)\Omega = 1452\Omega $.
Current flowing through circuit is –
$I = V/R$
$ \Rightarrow I = 440/1452 = 0.303A$
Thus, $I > {I_a}$.
Bulb A will fuse because current drawn from the battery will be much greater than that of current in bulb A which results in melting of fuse wire and bulb gets fused.
Note: The tungsten coil (filament) has hot spots. Slowly the tungsten is boiled away and the hot spots boil away faster. Eventually a little vibration breaks the coil or just the turn on current causes the coil to expand as it starts to glow and that motion and stress breaks the coil. That’s why the bulbs often burn out when they are switched on.
If the break is clean, the bulb can be worked again by wiggling it until the broken ends touch and they tack weld themselves together or at least stick. Sometimes the bulb fixes itself when the filament contracts as it cools down and the broken ends are really close. Next time when the bulb is turned on, the 120V or 240V bridges the gap, the tack weld / stick happens.
Recently Updated Pages
Allylic Substitution Reaction Important Concepts and Tips for JEE

Compressibility Factor Z Important Concepts and Tips for JEE Main

JEE - Alpha, Beta and Gamma Decay Important Concepts and Tips

Fluid Pressure - Important Concepts and Tips for JEE

Differential Equations Chapter For JEE Main Maths

JEE Main 2023 April 15 Shift 1 Question Paper with Answer Key

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility & More

JEE Main Exam Marking Scheme: Detailed Breakdown of Marks and Negative Marking

The formula of the kinetic mass of a photon is Where class 12 physics JEE_Main

JEE Main 2023 January 24 Shift 2 Question Paper with Answer Keys & Solutions

Learn About Angle Of Deviation In Prism: JEE Main Physics 2025

JEE Main 2025: Conversion of Galvanometer Into Ammeter And Voltmeter in Physics

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

JEE Main Login 2045: Step-by-Step Instructions and Details

Dual Nature of Radiation and Matter Class 12 Notes: CBSE Physics Chapter 11

Electric field due to uniformly charged sphere class 12 physics JEE_Main

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Ideal and Non-Ideal Solutions Raoult's Law - JEE
