Question

a current of 40mA flows through a circuit for 2 minutes and 30 seconds, calculate the no. of electrons flowing in the circuit

Answer 1

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I = 40 × 10^-6 A
t = 30 s
q = I t
= 40×10^-6×30
= 12×10^-4 C
Let number of electrons be n.
Charge on 1 electron = 1.6×10^-19 C
n×1.6×10^-19 = 1.2×10^-3
n = 1.2/1.6 × 10^16
= 0.75×10^16
= 75×10^14

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Answer 2

Answer:

A current of 40mA flows through a circuit for 2 minutes and 30 seconds, the no. of electrons flowing in the circuit will be $3.07\times 10^{20}$ electrons

Explanation:

• In the question, it is given that,

      Time,  $2 min 30 s=( 2\times 60 s)+30s= 1230s$

      Current,  $I = 40mA = 40\times 10^{-3} A$  

      Charge of 1 electron,  $e = 1.6\times 10^{-19}C$

• Consider a conductor of current flowing I, Q be the total charge of electrons and t be the time.

       Then, we have the expression for current as,

       $I=\frac{Q}{t}$  

      That is,

       $Q = It$            ...(1)

       Here Q is given by the expression,

       $Q = ne$  

       Or

       $n= \frac{Q}{e}$             ...(2)

       Where e is the charge of one electron and n is the number of electrons.

• Substitute the values given in the question into equation (1)

      $Q = (40\times 10^{-3}\times 1230 )= 49.2C$

      Then equation(2) becomes,

      $n= \frac{49.2}{1.6\times 10^-19}= 3.07\times 10^{20}$ electrons