A thin uniform wire AB of length 1 m, an unknown resistance X and a resistance of 12 ohms are connec

Q: Why doesn't the galvanometer have positive and negative terminals?

Unlike a voltmeter or ammeter that measures the exact magnitude and direction of a continuous current, a galvanometer in a meter bridge is used purely as a null detector. We only care about finding the exact point where the needle rests at zero.

Q: Why is the battery connected across the entire wire?

The wire acts as the two lower arms of the Wheatstone bridge. To create a potential difference across the entire bridge network, the main power source must be connected across the extreme ends of the wire.

Q: How do we know which length corresponds to which resistance?

The resistance in the left gap corresponds to the length of the wire from the left end to the jockey. The resistance in the right gap corresponds to the remaining length on the right.

Q: Why does the official answer key say $8\ \Omega$?

The official solution made a substitution error. They incorrectly wrote the ratio as $40/60$ instead of $60/40$. Based on the provided diagram where $X$ is on the left and the balance point is $60\text{ cm}$ from the left, the correct answer is mathematically $18\ \Omega$.

Q: What happens if we swap the battery and the galvanometer?

Interestingly, the null point remains completely unaffected! This is a standard property of a balanced Wheatstone bridge; swapping the source and the detector does not change the balance condition.

Question

A thin uniform wire $AB$ of length $1	ext{ m}$, an unknown resistance $X$ and a resistance of $12\ \Omega$ are connected by thick conducting strips, as shown in the figure. A battery and galvanometer (with a sliding jockey connected to it are also available). Connections are to be made to measure the unknown resistance $X$ using the principle of Wheatstone bridge. Answer the following questions.
(a) Are there positive and negative terminals on the galvanometer?
(b) Copy the figure in your answer book and show the battery and the galvanometer (with jockey) connected at appropriate points.
(c) After appropriate connections are made, it is found that no deflection takes place in the galvanometer when the sliding jockey touches the wire at a distance of $60	ext{ cm}$ from $A$. Obtain the value of the resistance $X$.

Sigma Percentile · Accepted Answer

### The Anatomy of a Meter Bridge

Before diving into the calculations, let's orient ourselves with the physical setup of a meter bridge. A meter bridge is essentially a practical, stretched-out version of the classic Wheatstone bridge. It consists of a uniform wire, typically $1	ext{ meter}$ ($100	ext{ cm}$) long, stretched between two thick metallic strips.

In part (a) of our problem, we are asked if the galvanometer has positive and negative terminals. The answer is a resounding **No**. Unlike a voltmeter or an ammeter, which are designed to measure the exact magnitude and direction of a continuous current, a galvanometer in a meter bridge acts purely as a **null detector**. Its only job is to tell us when the current passing through the middle branch is exactly zero. Because it only needs to detect the presence and direction of current to help us find that zero point, it doesn't require fixed polarity terminals.

### Completing the Circuit

For part (b), we need to complete the circuit by connecting the battery and the galvanometer. Think of the Wheatstone bridge diamond shape. The main power supply (the battery) must drive current through the entire network, so it is connected across the extreme ends of the main wire, which are points $A$ and $D$ in our diagram.

The galvanometer, on the other hand, sits in the central branch. One end is connected to the central thick metallic strip (between the two unknown/known resistances), and the other end is connected to the sliding jockey $J$, which makes contact with the wire.

### The Magic of the Null Point

Now for the calculation in part (c). We are told that the null point is found at a distance of $60	ext{ cm}$ from point $A$. This means the jockey divides the $100	ext{ cm}$ wire into two segments:

$$l_1 = 60	ext{ cm}$$
$$l_2 = 100	ext{ cm} - 60	ext{ cm} = 40	ext{ cm}$$

At the null point, the Wheatstone bridge is perfectly balanced. The core principle of a balanced bridge dictates that the ratio of the resistances in the top gaps must equal the ratio of the resistances of the wire segments directly below them. Since the wire is uniform, the resistance of a segment is directly proportional to its length.

Therefore, we can write the master equation:

$$\frac{R_{	ext{left}}}{R_{	ext{right}}} = \frac{l_1}{l_2}$$

Substituting our known values from the diagram (where $X$ is in the left gap and $12\ \Omega$ is in the right gap):

$$\frac{X}{12} = \frac{60}{40}$$

Simplifying the fraction on the right gives us $\frac{3}{2}$. Multiplying both sides by $12$ yields:

$$X = 12 	imes \frac{3}{2} = 18\ \Omega$$

### A Note on the Official Answer

You might notice that the official answer key for this specific exam year states the answer as $8\ \Omega$. This is a known historical error in the official solutions! The authors mistakenly substituted the lengths in reverse, writing $\frac{X}{12} = \frac{40}{60}$, which leads to $8\ \Omega$. However, based on the provided diagram and the standard laws of physics, **$18\ \Omega$ is the mathematically and physically correct answer**.

Visualized Solution

$P a r t (a) : G a l v an o m e t er T er mina l s$

$P a r t (b) : B a tt er y C o nn ec t i o n$

$P a r t (b) : G a l v an o m e t er C o nn ec t i o n$

$P a r t (c) : N u l l P o in t L e n g t h s$

$W h e a t s t o n e B r i d g e P r in c i pl e$

$F ina l C a l c u l a t i o n$

The Sigma Insight: Electrical Instruments

The Anatomy of a Meter Bridge

Completing the Circuit

The Magic of the Null Point

A Note on the Official Answer

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Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Part(a):GalvanometerTerminals

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Part(b):BatteryConnection

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Part(b):GalvanometerConnection

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Part(c):NullPointLengths

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } WheatstoneBridgePrinciple

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } FinalCalculation