Question

Question : How many minutes are required to deliver $3.21 \times {10^6}$ coulombs using a current of 500A used in the commercial production of chlorine ?
A.8.3
B.$5.3 \times {10^4}$
C.6420
D.107

Answer 1

Hint: The electric current is defined as the rate flow of electricity . It is carried by charged particles. So to calculate the electric charge the current flowing in the solution is multiplied by the total time for which it is flowing .

Complete step by step answer:
If a current of I amperes is passed through a solution for t seconds then the quantity of electricity produced in coulombs is given by
$Q = I \times t$
where , I is it current in amperes which is passed through the solution ,
t is the time in seconds
and Q is the quantity of electricity delivered .
Now in the given question we have to find the time required to deliver $3.21 \times {10^6}$ coulombs of electricity when a current of 500A is passed .
So , to calculate time the equation becomes
$t = \dfrac{Q}{I}$
The value of $Q = 3.21 \times {10^6}$ and the value of $I = 500A$
So , on substituting the values in the equation , we get
$t = \dfrac{{3.21 \times {{10}^6}}}{{500}}$ = $6420s$
The value of time which is obtained is in seconds and in the question we have been asked to give the value of time in minutes
Therefore on converting 6420 seconds into minutes we get
$\dfrac{{6420}}{{60}} = 107$
Hence , 107 minutes are required to deliver $3.21 \times {10^6}$ coulombs of electricity when a current of 500A is passed .

So, the correct answer is Option D .

Note:
The value of the quantity of electricity is used in various calculations . It is used to calculate the amount of substance deposited or liberated at any electrode as the quantity of electricity and amount of substance deposited are directly proportional .