!. A current of 1 A is drawn by a filament of an electric bulb. Number of electrons passing through a
cross section of the filament in 16 seconds would be roughly
2. What is the maximum resistance which can be made using five resistors each of 1/5 Ω?
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Explanation:
1) I = Q/T
Also Q = ne
Therefore I = ne/t
e = 1.6 * [tex]10^{-19}[/tex]
I = 1A
t = 16 seconds
N = (I * t)/e
= (1 * 16)/1.6 * [tex]10^{-19}[/tex]
= 10 * [tex]10^{19}[/tex]
= [tex]10^{20}[/tex]
Option A is your answer
2) The net resistance is maximum when all the resistors are connected in series.
Rs = R1 + R2 + R3 +R4 + R5
= 1/5 + 1/5 + 1/5 + 1/5 + 1/5
= 5/5
= 1 Ω
Option D is your answer