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A current of 1 A is drawn by a filament of an electric bulb. The number of electrons passing through a cross-section of the filament in 8 mins would be equal to ________. (Given, charge of electron (e) is equal to 1.6×10−19 C)
3×1020
3×1021
3×1019
3×1022
3×1021
To calculate the number of electrons (n) flowing through a conductor, we use the quantization of charge Q=n×e and the relation Q=I×t. Given current I=1A and time t=8minutes=8×60=480seconds, the total charge Q=1×480=480C.
To calculate the number of electrons (n) flowing through a conductor, we use the quantization of charge Q=n×e and the relation Q=I×t. Given current I=1A and time t=8minutes=8×60=480seconds, the total charge Q=1×480=480C.
Think of electrons as water drops flowing through a pipe; if you know the flow rate (current) and how long the tap is open (time), you can calculate the total number of drops that have passed through a specific point.
Q=I×t — Total charge passing through cross-section
n=eQ — Total number of electrons
The number of electrons n is determined by dividing the total charge Q by the elementary charge of an electron e=1.6×10−19C. Substituting the values: n=1.6×10−19480=300×1019=3×1021.
1 Ampere is defined as 1 Coulomb per second (1A=1C/s).
Time must be converted to SI units (seconds) before calculation (8min=480s).
The elementary charge e is approximately 1.602×10−19C.
Circuit design and analysis
Determining charge carrier density in conductors
Elementary charge e=1.6×10−19C
Option B represents the correct order of magnitude calculation: 480/1.6=300.
B is correct — Using n=eI×t, we get n=1.6×10−191×480=3×1021 electrons.
Always ensure time is in seconds when dealing with current formulas; examiners often provide minutes to test unit conversion skills.
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