Question

A $5A$ current is set up in an external circuit by a $6.0V$ storage battery for $6.0\min $. The chemical energy of the battery will be reduced by
$\begin{align}
  & A.1.08\times {{10}^{4}}J \\
 & B.1.08\times {{10}^{-4}}J \\
 & C.1.8\times {{10}^{4}}J \\
 & D.1.8\times {{10}^{-4}}J \\
\end{align}$

Answer 1

Hint: First of all the time given in the minutes should be converted into seconds. This is done by multiplying the value of time in minute with a$60$. Then find out the amount by which the energy is lost. This is done by taking the product of potential supplied by the storage battery, current flowing through the circuit and the time taken for this. These all may help you to solve this question.

Formula used: $E=VIt$
Where $V$ be the potential, $I$ be the current flowing and $t$ be the time taken for the flow.

Complete step by step answer:
Let us look at the values of the terms mentioned in the question. The current flowing through the external circuit is given as,
$I=5A$
The potential supplied by the external circuit is given as,
$V=6V$
And the time taken for this much flow of current is written as,
$t=6\min $
This should be converted to seconds in order to calculate the amount of energy lost.
That is
$t=6\min \times 60=360s$
By this amount of time, the energy reduced can be found by taking the product of potential supplied by the storage battery, current flowing through the circuit and the time taken for this.
Therefore we can write that,
$E=VIt$
Where $V$be the potential, $I$be the current flowing and $t$be the time taken for the flow.
Substituting this values in the equation,
$E=6\times 5\times 360=1.08\times {{10}^{4}}J$

So, the correct answer is “Option A”.

Note: Voltage is defined as the electric potential difference between any two points. It can be also defined as electric tension or in another words electric pressure. This is described as the work required per unit of charge to make a movement in a test charge between the two given points.