Animated Solution for Physics - Current Electricity: An ideal battery of $4\text{ V}$ and resistance $R$ are connected in series in the primary circuit of a potentiometer of length $1\text{ m}$ and resistance $5\Omega$. The value of $R$ to give a potential difference of $5\text{ mV}$ across $10\text{ cm}$ of potentiometer wire is

The potentiometer is one of the most elegant instruments in classical physics, relying on the beautiful simplicity of potential gradients and balanced circuits. In this problem, we are tasked with finding an unknown resistance $R$ in the primary circuit of a potentiometer.

I know circuits can sometimes look like a tangled mess of wires, but let's take a breath and break this down step by step. Imagine you are walking along the potentiometer wire; the voltage drops steadily with every step you take. This steady drop is the key to unlocking the entire problem!

Let's visualize the setup. We have an ideal battery supplying $4\text{ V}$, connected in series with an unknown resistance $R$ and a potentiometer wire. The wire itself has a total length of $1\text{ m}$ and a resistance of $5\Omega$.

The problem gives us a crucial piece of intel: a $10\text{ cm}$ segment of this wire has a potential difference of $5\text{ mV}$. This small segment is our window into the behavior of the entire wire.

To understand the whole wire, we first need to find the potential gradient ($k$). Think of the potential gradient as the "voltage cost" per meter of wire. It tells us exactly how much potential drops for every unit of length we traverse.

We calculate it by dividing the given potential difference by its corresponding length:

Substituting our values and converting to standard SI units (volts and meters) to avoid any silly mistakes:

Now that we know the "cost" per meter, finding the total voltage across the entire $1\text{ m}$ wire is a breeze. We just multiply the gradient by the total length:

Now, let's zoom out and look at the entire primary circuit. The battery provides a total of $4\text{ V}$. According to Kirchhoff's Voltage Law, this total voltage must be distributed across the components in the series circuit.

In our case, the voltage is shared between the unknown resistor $R$ and the potentiometer wire:

Since we already know the battery voltage and the wire's voltage, we can easily find the voltage across $R$:

We are almost there! Because the resistor $R$ and the potentiometer wire are connected in series, the current flowing through them must be exactly the same. This is a fundamental property of series circuits.

Using Ohm's law ($I=\frac{V}{R}$), we can express this shared current in two ways and equate them:

Let's substitute the values we've worked so hard to find:

Simplifying the left side gives us $0.01$. Now, we just solve for $R$:

And there we have it! By systematically applying the concepts of potential gradient and series circuits, we've found that the unknown resistance $R$ is exactly $395\Omega$.