Animated Solution for Physics - Current Electricity: A potentiometer wire $PQ$ of 1 m length is connected to a standard cell $E_1$. Another cell $E_2$ of emf 1.02 V is connected with a resistance $r$ and switch $S$ (as shown in figure). With switch $S$ open, the null position is obtained at a distance of 49 cm from $Q$. The potential gradient in the potentiometer wire is

Finding the Potential Gradient

Welcome to a classic exploration of the potentiometer! This device is an elegant piece of classical physics, relying on the simple principle of balancing potential drops. Let's decode the setup and find the potential gradient of the wire.

Our circuit is divided into two main parts. The primary circuit contains the standard cell $E_1$ which drives a steady current through the 1-meter long potentiometer wire $PQ$. This steady current establishes a uniform potential drop across the wire, known as the potential gradient ($k$).

The secondary circuit contains the cell we are testing, $E_2$, which has an electromotive force (emf) of $1.02\text{ V}$. Notice that it is connected in series with a parallel combination of a resistor $r$ and a switch $S$, followed by a galvanometer $G$.

Here lies a common trap! The problem states that the switch $S$ is open, meaning the current must pass through the resistor $r$. You might wonder, won't this resistor cause a voltage drop and mess up our readings?

The beauty of the null point method is that at the exact balance point, the galvanometer shows zero deflection. This means absolutely zero current flows through the secondary circuit. According to Ohm's law ($V=IR$), if the current $I$ is zero, the voltage drop across the resistor $r$ is exactly zero. The resistor $r$ acts merely as a protective device to prevent large currents from damaging the sensitive galvanometer when the jockey is far from the null point. It has absolutely no effect on the final balancing length!

The problem states that the null position is obtained at a distance of $49\text{ cm}$ from point $Q$. However, to apply our potentiometer formula, we must measure the balancing length from the zero-potential reference point, which is point $P$ (where the positive terminals of both cells are connected).

Since the total length of the wire is $1\text{ m}$ (or $100\text{ cm}$), the balancing length $l$ measured from $P$ is:

The core principle of the potentiometer states that at the null point, the potential drop across the balancing length of the wire perfectly matches the emf of the cell in the secondary circuit. Mathematically, this is expressed as:

We need to find the potential gradient $k$. Rearranging the formula gives us:

Now, we simply substitute the known values into our master equation:

Notice how the numbers are perfectly orchestrated. $1.02$ is exactly twice $0.51$. Performing the division yields:

This is our potential gradient. Looking at the given options, we can confidently conclude that option (a) is the correct answer.

As a thought experiment, consider what would happen if the switch $S$ were closed. The resistor $r$ would be short-circuited. Would the null point change? No! Since the current is already zero at the null point, removing $r$ doesn't alter the balance condition. It simply increases the sensitivity of the galvanometer, making it easier to pinpoint the exact null location.