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Physics - Current Electricity Question with Solution | TestHub

PhysicsCurrent ElectricityMicroscopicEasy2 minQB

PhysicsEasysingle choice

The number of electrons flowing per second in the filament of a $110 W$ bulb operating at $220 V$ is : $($ Given $e = 1.6 \times 1 0^{- 19} C)$

Options:

Answer:

Solution:

Power $(P) = V . I$ $\Rightarrow 110 = (220) (I) \Rightarrow I = 0.5 A$

Now, $I = \frac{n \cdot e}{t}$ $\Rightarrow 0.5 = (\frac{n}{t}) (1.6 \times 1 0^{- 19}) \Rightarrow \frac{n}{t} = \frac{0.5}{1.6 \times 1 0 ^{- 19}} \Rightarrow \frac{n}{t} = 31.25 \times 1 0^{17}$

Stream:JEESubject:PhysicsTopic:Current ElectricitySubtopic:Microscopic

⏱ 2mℹ️ Source: QB

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