Animated Solution for Physics - Current Electricity: Two identical moving coil galvanometer have $10\Omega$ resistance and full scale deflection at $2\mu \text{A}$ current. One of them is converted into a voltmeter of $100\text{ mV}$ full scale reading and the other into an Ammeter of $1\text{ mA}$ full scale current using appropriate resistors. These are then used to measure the voltage and current in the Ohm's law experiment with $R=1000\Omega$ resistor by using an ideal cell. Which of the following statement(s) is/are correct ?

When Instruments Interfere

Imagine you are a detective trying to measure the speed of a car, but the very act of placing your radar gun on the road slows the car down. This is exactly what happens in electrical circuits! Real-world instruments are never perfectly "invisible." A voltmeter draws a tiny bit of current, and an ammeter introduces a tiny bit of resistance. This beautiful JEE Advanced problem forces us to confront this reality by building our own instruments from scratch and analyzing their "observer effect" on an Ohm's law experiment.

We start with a humble moving coil galvanometer. It has a resistance of $R_g=10\Omega$ and achieves full-scale deflection at a microscopic current of $I_g=2\mu \text{A}$. Our first task is to convert this into a voltmeter capable of reading up to $V=100\text{ mV}$.

To do this, we must connect a large resistance $R_V$ in series with the galvanometer. The total voltage across this combination is given by Ohm's law:

Substituting our known values, we get:

Solving for the total resistance $(10+R_V)$, we find it to be exactly $50000\Omega$, or $50\text{ k}\Omega$. This means the series resistance we added is $49990\Omega$. Since the total resistance of our custom voltmeter is $50\text{ k}\Omega$, option (B) is incorrect.

Next, we take an identical galvanometer and convert it into an ammeter that can measure up to $I=1\text{ mA}$. To achieve this, we provide a "bypass lane" for the excess current by connecting a very small resistance, called a shunt ($S$), in parallel.

The voltage across the galvanometer must equal the voltage across the shunt:

Plugging in the numbers:

Because $2\mu \text{A}$ is incredibly small compared to $1\text{ mA}$, we can safely approximate the current flowing through the shunt $(I-I_g)$ as just $10^{-3}\text{ A}$. This gives:

The equivalent resistance of the ammeter is the parallel combination of $10\Omega$ and $0.02\Omega$, which is overwhelmingly dominated by the smaller value. Thus, the ammeter's resistance is approximately $0.02\Omega$. This makes option (C) perfectly correct!

Now for the grand finale. We connect our custom ammeter and voltmeter to measure a $R=1000\Omega$ resistor. In a perfect world, the measured resistance $R_{\text{measured}}=\frac{V_{\text{reading}}}{I_{\text{reading}}}$ would be exactly $1000\Omega$. But our instruments are not perfect.

The voltmeter is connected in parallel with the $1000\Omega$ resistor. Therefore, the voltmeter isn't just measuring the voltage across the resistor; it's measuring the voltage across the parallel combination of the resistor and itself! Meanwhile, the ammeter measures the total current entering this parallel setup.

Consequently, the ratio $\frac{V_{\text{reading}}}{I_{\text{reading}}}$ gives us the equivalent resistance of this parallel combination:

This value falls squarely in the range $978\Omega <R<982\Omega$, making option (A) correct.

Finally, what if we replace the ideal cell with a real cell having an internal resistance of $r=5\Omega$?

It's tempting to think this will mess up our measurement. However, while the internal resistance will lower the total current $I_{\text{total}}$ and the terminal voltage $V_{\text{reading}}$, it lowers them proportionally. The ratio of the voltmeter reading to the ammeter reading depends strictly on the passive components in that part of the circuit—namely, the $1000\Omega$ resistor and the $50\text{ k}\Omega$ voltmeter.

Therefore, the measured resistance remains stubbornly fixed at $980.39\Omega$. It will certainly not exceed $1000\Omega$, rendering option (D) incorrect. The beauty of Ohm's law is that the geometry of the passive circuit dictates the measurement, regardless of the power source driving it.