Question

Calculate the number of electrons passing through a 400 ohm filament of a bulb connected to

12 V supply in 1 minute.​

Answer 1

Given:

Resistance of Bulb, R= 400Ω

Potential across bulb, V= 12V

Time for which current passes through bulb= t

∴ t= 1 min= 60 s

To Find:

Number number of electrons passing through filament of given bulb in 1 minute or 60 seconds

Solution:

We know that,

• According to Ohm's Law, for a conductor

$\pink{\boxed{\bf{V=IR}}}$

• Relation between current and charge is given by

$\purple{\boxed{\bf{I=\frac{Q}{t}}}}$

• According to Quantisation of Charge,

$\orange{\boxed{\bf{Q=ne}}}$

where,

V is voltage across conductor

I is current passing through conductor

R is the resistance of conductor

Q is charge flowing through conductor

t is time

n is number of electrons

e is charge of 1 electron i.e. 1.6×10⁻¹⁹C

Now,

Let the current flowing through bulb be 'I'

So, on using ohm's law, we get

$\longrightarrow\rm{V=IR}$

$\longrightarrow\rm{12=I(400)}$

$\longrightarrow\rm{I=\dfrac{\cancel{12}}{\cancel{400}}}$

$\longrightarrow\rm{I=\dfrac{3}{100}A}$

Now,

Let the charge flowing through conductor be 'Q'

So,

$\longrightarrow\rm{I=\dfrac{Q}{t}}$

$\longrightarrow\rm{\dfrac{3}{100}=\dfrac{Q}{60}}$

$\longrightarrow\rm{\dfrac{Q}{60}=\dfrac{3}{100}}$

$\longrightarrow\rm{Q=\dfrac{3\times60}{100}}$

$\longrightarrow\rm{Q=\dfrac{\cancel{180}}{\cancel{100}}}$

$\longrightarrow\rm{Q=\dfrac{9}{5}C}$

Now,

Let there be 'n' number of electrons passing through the filament of bulb in 60 s

So,

$\longrightarrow\rm{Q=ne}$

$\longrightarrow\rm{\dfrac{9}{5}=n(1.6\times10^{-19})}$

$\longrightarrow\rm{n(1.6\times10^{-19})=\dfrac{9}{5}}$

$\longrightarrow\rm{n=\dfrac{9}{5\times1.6\times10^{-19}}}$

$\longrightarrow\rm{n=\dfrac{9}{8\times10^{-19}}}$

$\longrightarrow\rm{n=\dfrac{1.125}{10^{-19}}}$

$\longrightarrow\rm\green{n=1.125\times10^{19}}$

Hence, there are 1.125×10¹⁹ electrons passing through the filament of the bulb in 1 min.

Answer 2

GiveN :

• Resistance of Bulb (R) = 400 ohms
• Potential Difference (V) = 12 V
• Time (t) = 1 min = 60 sec

To FinD :

• Number of electrons passing through the circuit.

SolutioN :

Use Ohm's Law to calculate the current flowing through circuit :

$\underbrace{\sf{Current \: passing \: through \: circuit}}$

$\implies \sf{V \: = \: IR} \\ \\ \implies \sf{I \: = \: \dfrac{V}{R}} \\ \\ \implies \sf{I \: = \: \dfrac{12}{400}} \\ \\ \implies \sf{I \: = \: \dfrac{3}{100}} \: \: \: \: \: \: ...(1)$

_____________________________

Now, we have to find charge flowing through circuit :

$\underbrace{\sf{Charge \: in \: the \: circuit}}$

$\implies \sf{I \: = \: \dfrac{Q}{t}}$

Put Value of I from (1)

$\implies \sf{Q \: = \: I \: \times \: t} \\ \\ \implies \sf{Q \: = \: \dfrac{3}{100} \: \times \: 60} \\ \\ \implies \sf{Q \: = \: \dfrac{18}{10}} \\ \\ \implies \sf{Q \: = \: \dfrac{9}{5}} \\ \\ \implies \sf{Q \: = \: \dfrac{9}{5}} \: \: \: \: \: \: ...(2)$

____________________________

Now, we have to calculate number of electrons :

$\underbrace{\sf{Number \: of \: electrons}}$

$\implies \sf{Q \: = \: ne}$

Put value of Q from (2)

$\implies \sf{ \dfrac{9}{5} \: = \: n \: \times \: 1.6 \: \times \: 10^{-19}} \\ \\ \implies \sf{n \: = \: \dfrac{9}{8 \: \times \: 10^{-19}}} \\ \\ \implies \sf{n \: = \: 1.125 \: \times \: 10^{19}}$