How many electrons flow per sec. through the filament of a 220 V and 110 W electric bulb :-

3.125 \times 10^{18}

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3.125 \times 10^{17}

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312.5 \times 10^{18}

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31.25 \times 10^{18}

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Solution

The correct option is A $3.125 \times 10^{18}$

If $n$ be the number of electrons , then
number of electron per sec $= n e =$ current (I).
now , $P = V I, I = P / V$
thus, $n = I / e = \frac{P}{e V} = \frac{110}{1.6 \times 10^{- 19} \times 220} = 3.125 \times 10^{18}$

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How many electrons flow per sec. through the filament of a 220 V and 110 W electric bulb :-

The correct option is A 3.125×1018 If n be the number of electrons , thennumber of electron per sec =ne= current (I).now , P=VI,I=P/Vthus, n=I/e=PeV=1101.6×10−19×220=3.125×1018

The correct option is A $3.125 \times 10^{18}$

If $n$ be the number of electrons , then
number of electron per sec $= n e =$ current (I).
now , $P = V I, I = P / V$
thus, $n = I / e = \frac{P}{e V} = \frac{110}{1.6 \times 10^{- 19} \times 220} = 3.125 \times 10^{18}$