Question

4.8 *10power20 electron flows through a circuit in 20 hours Calculate the magnitude of the current flowing in the circuit (Charge on an electron = 1.6 * 10 power -19
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Answer 1

Current flows through the circuit = 0.2666 m A

Explanation:

• Current flows through any circuit is defined as the rate of change of total charge flowing through the circuit and is given as

                  $Current = \frac{total charge}{time taken}$

• Current is a scalar quantity although it have direction that is moving from positive terminal to negative terminal also known as direction of conventional current

Given, number of electrons = 4.8 * 10²⁰

charge on one electron = 1.6 * 10⁻¹⁹ C

total time taken by the charges to flow = 20 h = 20 *60*60 sec

=> Total charge in the circuit is given by quantisation of charge

Total charge = Number of charges × charge on one electron

=> q= n e

=> q =  4.8 * 10²⁰ × 1.6 * 10⁻¹⁹

=> q = 4.8 × 16 C

=> Current = $\frac{charge}{time}$

=> I = $\frac{4.8 * 16}{20 * 60 * 60}$

=> I = 0.2666 * 10⁻³ A

or I = 0.2666 m A

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Answer 2

Answer:

4.8 *10power20 electron flows through a circuit in 20 hours, the magnitude of the current flowing in the circuit is, $1.066mA$

Explanation:

• From the question,

       Number of electrons $n =4.8\times 10^{20}$

       Time, $t = 20h = 20\times 60\times 60 = 72000s$

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

• Let I be the current flowing through an electrical circuit t be the time of flow and Q be the total charge.  

        We have the expression for the total charge as,

        $Q = It$    

        That is,

        $I = \frac{Q}{t}$       ...(1)

        Then we have another expression for Q as,

         $Q = ne$     ...(2)

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

        We can use these equations for solving the given question.

        Then,

        Substituting the given values into equation (2)

        $Q = 4.8\times 10^{20} \times 1.6\times10 ^{20}=76.8C$

        Now, Using equation(2),

        $I = \frac{76.8}{76000} = 0.001066 A$

        $I= 1.066\ mA$