Question

In $10 s$, a charge of $25C$ leaves a battery and $200J$ of energy are delivered to an outside circuit as a result.
(a) What is the p.d. across the battery?
(b) What current flows from the battery?

Answer 1

In the given question we have to find the potential difference which is the amount of work done when a charge flows through it. Current is defined as the charge per unit time. So, by putting the values of charge and time we will find the current flowing from the battery.

Formula Used:
$V = \dfrac{W}{q}$
where $V$ is the potential difference, $W$ is the work done and $q$ is the charge flowing. Potential difference is also defined as the multiplication of current and resistance.
$V = IR$
$\Rightarrow i = \dfrac{q}{t}$
where $q$ is the charge, $i$ is the current flowing and $t$ is time.

Complete step by step answer:
(a) We know that,
$V = \dfrac{W}{q} - - - - \left( a \right)$ where $V$ is the potential difference, $W$ is the work done and $q$ is the charge flowing.
In the given question the values of $W$, $q$ is given,
$W = 200{\text{ }}J$ and $q = 25{\text{ }}C$
Thus by substituting the values in equation $\left( a \right)$ we get,
$\therefore V = \dfrac{{200}}{{25}} = 8\,volt$

So, the p.d. (potential difference) across the battery is $8{\text{ }}volt$.

(b) We know that,
$q = it$ where $q$ is the charge, $i$ is the current flowing and $t$ is time.
By cross-multiplication we get,
$i = \dfrac{q}{t} - - - - \left( b \right)$
Substituting the values of $q = 25{\text{ }}C$ and $t = 10{\text{ }}s$ in equation $\left( b \right)$ we get,
$\therefore i = \dfrac{{25}}{{10}} = 2.5{\text{ }}A$

So, the current that flows from the battery is $2.5{\text{ }}A$.

Note: We know that potential difference is defined as the amount of work done when unit charge flows through it. The question has emphasized that the energy is given outside the battery which implies that the work is being done by the body. And current is defined as the charge that flows through the conducting medium in unit time.