Animated Solution for Physics - Magnetic Effects of Current: A moving coil galvanometer, having a resistance $G$, produces full scale deflection when a current $I_g$ flows through it. This galvanometer can be converted into (i) an ammeter of range 0 to $I_0(I_0>I_g)$ by connecting a shunt resistance $R_A$ to it and (ii) into a voltmeter of range 0 to $V(V=G{I}_0)$ by connecting a series resistance $R_V$ to it. Then,

The Tale of Ammeters and Voltmeters

Imagine you have a delicate moving coil galvanometer. It's a highly sensitive instrument, designed to measure tiny currents. But what if you need to measure a massive current, or a large voltage? You can't just hook it up directly; it would burn out instantly! This is where the magic of circuit design comes in. By adding simple resistors in clever ways, we can transform this fragile device into a robust ammeter or a high-range voltmeter.

To convert a galvanometer into an ammeter, we need to ensure that most of the large current bypasses the sensitive galvanometer coil. We achieve this by connecting a very small resistance, called a shunt resistance ($R_A$), in parallel with the galvanometer.

Let the total current we want to measure be $I_0$, and the maximum safe current for the galvanometer be $I_g$. When $I_0$ enters the parallel combination, it splits. A small portion $I_g$ goes through the galvanometer (resistance $G$), and the rest, $(I_0-I_g)$, flows through the shunt $R_A$.

Because they are in parallel, the potential difference across both branches must be identical. Therefore, we can write:

Rearranging this to solve for the shunt resistance $R_A$, we get:

Now, let's turn our galvanometer into a voltmeter. A voltmeter must have a very high resistance so that it doesn't draw significant current from the circuit it's measuring. We do this by connecting a large resistance ($R_V$) in series with the galvanometer.

In a series circuit, the same current flows through all components. For full-scale deflection, this current is $I_g$. The total voltage $V$ across the combination is the sum of the voltage drops:

The problem gives us a specific condition for this voltmeter: its maximum range $V$ is equal to $I_{0}G$. Substituting this into our equation:

Now, let's isolate $R_V$:

Product and Ratio

We now have the expressions for both the shunt resistance $R_A$ and the series resistance $R_V$. The question asks us to find their product and their ratio.

First, let's find the product $R_A{R}_V$:

Notice how beautifully the terms $(I_0-I_g)$ and $I_g$ cancel out! We are left with a surprisingly simple result:

Next, let's find the ratio $\frac{R_A}{R_V}$:

When dividing fractions, we multiply by the reciprocal. The $G$ terms cancel out, and we get:

Comparing our derived expressions with the given options, we can confidently conclude that option (b) is the correct answer. This problem beautifully illustrates how fundamental circuit laws—Ohm's Law and Kirchhoff's Laws—can be used to design practical measuring instruments.