Animated Solution for Physics - Current Electricity: A moving coil galvanometer of resistance $100\Omega$ is used as an ammeter using a resistance $0.1\Omega$. The maximum deflection current in the galvanometer is $100\mu \text{A}$. Find the current in the circuit, so that the ammeter shows maximum deflection.

Imagine you have a highly sensitive moving coil galvanometer. It's a fantastic device for detecting tiny currents, but if you try to measure the current flowing through a standard household circuit with it, the delicate coil will instantly burn out. To solve this, we convert the galvanometer into an ammeter by connecting a very small resistance, known as a shunt ($S$), in parallel with it.

This clever setup acts like a bypass lane on a highway. When the main current $i$ enters the device, it encounters a junction. Because the shunt has a much lower resistance than the galvanometer ($G$), the vast majority of the current takes the path of least resistance through the shunt. Only a tiny, safe fraction of the current ($i_g$) flows through the galvanometer to produce a reading.

Since the galvanometer and the shunt are connected in parallel, the potential difference across both branches must be exactly the same. By applying Ohm's law ($V=IR$) to each branch, we can set their voltage drops equal to each other:

Here, $i_g$ is the current through the galvanometer, and $(i-i_g)$ is the remaining current flowing through the shunt. Our goal is to find the maximum main current $i$ that the ammeter can measure. Let's rearrange the equation to isolate $i$:

This elegant formula tells us exactly how the range of the ammeter scales based on the ratio of the galvanometer's resistance to the shunt's resistance.

Now, let's plug in the specific values given in our problem. We know the galvanometer resistance $G=100\Omega$, the shunt resistance $S=0.1\Omega$, and the maximum safe current for the galvanometer $i_g=100\mu \text{A}=100\times 10^{-6}\text{ A}$.

Substituting these into our master equation:

First, we evaluate the fraction inside the parentheses:

So the equation simplifies to:

To make this number more readable, we can shift the decimal point three places to the left, which converts the units from microamperes to milliamperes:

By adding a tiny $0.1\Omega$ shunt, we successfully increased the measuring range of the galvanometer from a mere $0.1\text{ mA}$ to over $100\text{ mA}$!