Animated Solution for Physics - Magnetic Effects of Current: For full scale deflection of total 50 divisions, 50 mV voltage is required in galvanometer. The resistance of galvanometer if its current sensitivity is 2 div/mA will be

The Moving Coil Galvanometer is a fascinating and highly sensitive piece of equipment that forms the heart of many electrical measuring instruments, such as ammeters and voltmeters. In this problem, we are tasked with finding the internal resistance of a galvanometer given its full-scale deflection parameters. To truly appreciate the solution, we must take a thrilling journey into the physics of how a galvanometer actually works, exploring the interplay between electricity, magnetism, and mechanics. Let's dive deep into the physics and math behind this!

Before we crunch the numbers, let's visualize the physical reality of a Moving Coil Galvanometer. Imagine a lightweight rectangular coil of wire, consisting of many turns, suspended within the strong magnetic field of a permanent horseshoe magnet. This coil is attached to a delicate hairspring and a pointer that sweeps across a calibrated scale.

When an electric current flows through this coil, the magnetic field exerts a force on the moving charges. Because the current flows in opposite directions on either side of the coil, these forces create a magnetic torque that tries to twist the coil.

The equation for this magnetic torque is:

Where: - $N$ is the number of turns in the coil. - $I$ is the current flowing through it. - $A$ is the area of the coil. - $B$ is the strength of the magnetic field. - $\theta$ is the angle between the magnetic field and the area vector of the coil. (In a radial magnetic field, $\sin \theta$ is always $1$).

As the coil twists, the hairspring winds up, creating a restoring torque that opposes the magnetic torque. This restoring torque is proportional to the angle of twist, $\alpha$:

Where $k$ is the torsional constant of the spring. The coil will stop twisting and come to rest when these two torques perfectly balance each other:

From this beautiful equilibrium, we can see that the angle of deflection $\alpha$ is directly proportional to the current $I$:

This linear relationship is why the scale on a galvanometer is evenly spaced!

Now, let's talk about Current Sensitivity. It is defined as the deflection produced per unit of current. Mathematically, it is $\frac{\alpha }{I}$. In practical terms, as given in our problem, it tells us how many divisions the pointer will move for every milliampere of current flowing through it.

The problem states that the current sensitivity is $2\text{ div/mA}$. This means a tiny current of just $1\text{ mA}$ is enough to push the pointer across $2\text{ divisions}$ on the scale.

We are also told that the scale has a total of $50\text{ divisions}$. To make the pointer swing all the way to the 50th division—a state we call the full-scale deflection—we need to apply a specific voltage, which is given as $50\text{ mV}$.

To find the resistance of the galvanometer coil, we first need to know the exact amount of current that causes this full-scale deflection. We call this the full-scale deflection current, denoted by $I_g$.

Think about it logically: if $1\text{ mA}$ of current causes a deflection of $2\text{ divisions}$, how much current is needed to cause a deflection of $50\text{ divisions}$?

We can set up a simple ratio or use the formula:

Let's substitute the values we have into our equation:

Calculating this gives us the raw current required:

So, a current of $25\text{ mA}$ flowing through the galvanometer will push the pointer exactly to the end of the scale. This is a critical threshold; passing a current larger than $I_g$ could damage the delicate coil or the spring!

Finding the Resistance

Now we have two crucial pieces of information for the full-scale deflection state: 1. The voltage across the galvanometer: $V=50\text{ mV}$ 2. The current through the galvanometer: $I_g=25\text{ mA}$

The galvanometer coil is essentially a long piece of thin wire, which means it acts as a resistor. Therefore, it must obey Ohm's Law, which states that the voltage across a resistor is directly proportional to the current flowing through it.

The fundamental formula is:

Where $R_g$ is the internal resistance of the galvanometer. Rearranging this formula to solve for $R_g$, we get:

Let's plug in our values and see what happens:

Notice how beautifully the units align in this problem. Both the voltage and the current are given in "milli" units ($10^{-3}$). When we divide them, the "milli" prefixes cancel each other out perfectly, saving us from messy scientific notation!

The $10^{-3}$ terms vanish, leaving us with a simple fraction:

Executing this final atomic compute, we arrive at our answer:

The internal resistance of the galvanometer is $2\Omega$.

This problem beautifully illustrates how the mechanical properties of a galvanometer (like its scale divisions and sensitivity) are intimately connected to its electrical properties (like voltage, current, and resistance).

As you continue your physics journey, consider how this galvanometer could be modified. If you wanted to use this exact galvanometer to measure a much larger voltage, say $10\text{ V}$, what would you need to do? You would need to connect a high-value resistor in series with it to limit the current to $I_g$. Conversely, to measure a large current, you would connect a low-value shunt resistor in parallel. Understanding these core relationships is the key to mastering the design and function of all electrical measuring instruments. Keep practicing, and always look for the physical meaning behind the formulas!