Animated Solution for Physics - Magnetic Effects of Current: A moving coil galvanometer allows a full scale current of $10^{-4}\text{ A}$. A series resistance of $2\text{ M}\Omega$ is required to convert the above galvanometer into a voltmeter of range $0-5\text{ V}$. Therefore, the value of shunt resistance required to convert the above galvanometer into an ammeter of range $0.10\text{ mA}$ is

Welcome to an intriguing problem from the JEE Main 2019 exam! This question takes us into the fascinating world of electrical measuring instruments, specifically the moving coil galvanometer. We will explore how this sensitive device can be transformed into a voltmeter and an ammeter. However, as we dive into the mathematics, we will uncover a surprising twist that makes this problem quite unique. Let's embark on this journey!

A moving coil galvanometer is a highly sensitive instrument used to detect small currents. But its utility doesn't stop there. By cleverly adding resistors, we can convert it into a voltmeter to measure potential difference or an ammeter to measure larger currents.

To convert a galvanometer into a voltmeter, we connect a very high resistance in series with it. Why? Because a voltmeter is connected in parallel across the component whose voltage we want to measure. If its resistance were low, it would draw a significant amount of current, altering the very circuit it's trying to measure! The high series resistance ensures it draws negligible current.

Conversely, to convert it into an ammeter, we connect a very small resistance, known as a shunt, in parallel. This provides a low-resistance bypass path for the majority of the current, protecting the delicate galvanometer coil while allowing us to measure large currents.

Let's look at the data provided for the voltmeter conversion. We are given the full-scale deflection current of the galvanometer, $I_g=10^{-4}\text{ A}$. The series resistance added is $R_S=2\text{ M}\Omega =2\times 10^6\Omega$, and the maximum voltage it can measure is $V_{\max }=5\text{ V}$.

The fundamental equation governing this conversion is derived from Ohm's Law. The total voltage $V_{\max }$ is equal to the current $I_g$ multiplied by the total resistance of the circuit, which is the sum of the galvanometer's internal resistance $R_G$ and the series resistance $R_S$.

Now, let's substitute our known values into this equation to find the internal resistance of the galvanometer, $R_G$.

To isolate $R_G$, we first divide both sides by $10^{-4}$:

Now, we solve for $R_G$:

Wait a minute! We have arrived at a negative value for the resistance. In the realm of classical physics and standard electrical components, resistance is a measure of opposition to current flow and is inherently a positive quantity. A negative resistance of $-1.95\text{ M}\Omega$ is physically impossible for a simple galvanometer coil.

What does this mean? It implies that the numerical values provided in the problem statement are inconsistent. The examiner likely made an error in setting up the parameters. Because the foundational data is flawed, we cannot proceed to calculate the shunt resistance for the ammeter conversion.

In a competitive exam like JEE, encountering such a question can be confusing. However, if you are confident in your concepts and calculations, trust your math! Such questions are typically identified as erroneous, and bonus marks are awarded to all candidates. It's a great reminder that while physics describes reality, the numbers we plug into our equations must also reflect that reality!