
Observe the following figure which bulb gets fuzed?

Answer
146.7k+ 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.
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