Animated Solution for Physics - Current Electricity: In a Wheatstone bridge (see figure), resistances $P$ and $Q$ are approximately equal. When $R=400\Omega$, the bridge is balanced. On interchanging $P$ and $Q$, the value of $R$ for balance is $405\Omega$. The value of $X$ is close to

The Wheatstone bridge is one of the most elegant and fundamental circuits in physics, allowing us to measure unknown resistances with incredible precision. But what happens when the bridge itself has slight imperfections? This problem takes us on a journey to understand how we can cleverly eliminate unknown variables and find the exact resistance we are looking for.

Imagine you are standing in front of a classic Wheatstone bridge. It consists of four resistive arms: $P$, $Q$, $R$, and $X$. A galvanometer bridges the middle, acting as a highly sensitive current detector.

The problem states that the bridge is balanced. This is the magic word! When a Wheatstone bridge is balanced, the potential difference across the galvanometer is exactly zero. This means no current flows through the middle branch, and the circuit behaves as if the galvanometer isn't even there.

Because the bridge is balanced, the ratio of the resistances in the adjacent arms must be perfectly equal. This gives us our master equation:

We are given two distinct scenarios. Let's break them down.

Case 1: The Initial Balance Initially, the bridge is balanced when $R=400\Omega$. Substituting this into our master equation, we get: Let's rearrange this to isolate the ratio of our unknown arms, $P$ and $Q$:

Case 2: The Interchanged Arms Now, the experimenter does something clever. They physically swap the positions of resistors $P$ and $Q$. To achieve balance again, the resistance $R$ must be adjusted to $405\Omega$.

Our master equation adapts to this new reality: Again, let's isolate the ratio of $Q$ to $P$:

We have two equations, but we don't know the individual values of $P$ or $Q$. How do we find $X$?

Look closely at Equation 1 and Equation 2. One has the ratio $P/Q$, and the other has $Q/P$. If we multiply these two equations together, the unknown variables will beautifully annihilate each other!

The left side simply becomes $1$:

Now, it's just a matter of simple algebra. Rearranging for $X^2$, we find:

To find $X$, we take the square root:

Without a calculator, we can estimate this. We know that $\sqrt{400}=20$. So, $X=20\times \sqrt{405}$. Since $20^2=400$, $\sqrt{405}$ is just slightly more than $20$.

The value of $X$ is incredibly close to $402.5\Omega$. By interchanging the arms and using a bit of algebraic elegance, we bypassed the unknown variables entirely!