Animated Solution for Physics - Current Electricity: In an electrical circuit, a battery is connected to pass $20\text{ C}$ of charge through it in a certain given time. The potential difference between two plates of the battery is maintained at $15\text{ V}$. The work done by the battery is ......... $\text{J}$.

Imagine a simple electrical circuit as a bustling city, and the battery as its beating heart. The battery's primary job is to pump life—in the form of electric charge—through the intricate pathways of the circuit. But pumping this charge isn't free; it requires energy. It requires work.

In this problem, we are given a scenario where a battery pushes a total charge of $20\text{ C}$ through a circuit. The battery maintains a steady potential difference of $15\text{ V}$ across its plates. Our mission is to find out exactly how much work the battery had to do to accomplish this feat.

Work Done

To solve this, we need to go back to the very fundamental definition of electric potential difference, commonly known as voltage. What does a voltage of $15\text{ V}$ physically mean?

By definition, the potential difference $V$ between two points is the amount of work done $W$ in moving a unit positive charge $q$ from one point to the other. Mathematically, this is expressed as:

This elegant equation tells us that voltage is essentially energy per unit charge. If a battery has a potential difference of $15\text{ V}$, it means the battery expends $15\text{ Joules}$ of energy for every single Coulomb of charge it pushes through the circuit.

To find the total work done, we simply rearrange our master equation to solve for $W$:

Time

You might have noticed a sneaky phrase in the problem statement: "in a certain given time." Why is this mentioned if we don't have a number for it?

This is a classic distractor designed to test your conceptual clarity. The total work done by a battery depends only on the total amount of charge moved and the potential difference it was moved across. It does not matter if it took one second, one minute, or one hour to move that $20\text{ C}$ of charge; the total energy expended remains exactly the same.

Time would only become relevant if we were asked to calculate the Power—which is the rate at which work is done ($P=\frac{W}{t}$). Since we only need the total work, we can confidently ignore the time aspect.

Now, we are ready to execute the calculation. We substitute our known values into the rearranged formula:

The arithmetic is straightforward. Multiplying $20$ by $15$ gives us $300$. Since we are calculating work, which is a form of energy, our final unit is the Joule ($\text{J}$).

And there we have it! The battery performs exactly $300\text{ Joules}$ of work to push the $20\text{ C}$ charge through the circuit. This energy doesn't just vanish; it is transformed. As the charge flows through the circuit's resistance, this $300\text{ J}$ of electrical energy is dissipated, typically radiating out into the universe as heat or light.