Q.An ammeter of resistance $0\cdot8\ \Omega$ can measure a current up to $1\cdot0$ A. Find the value of shunt resistance required to convert this ammeter to measure a current up to $5\cdot0$ A.

When a galvanometer is converted into an ammeter, a small resistance (shunt) is connected in parallel with it. The purpose is to allow most of the current to bypass the delicate galvanometer coil, so only a small fraction passes through the meter itself. The galvanometer shows full-scale deflection when the current through its coil reaches its maximum rated value, say $I_g$. The shunt carries the remaining current.

Here, the galvanometer resistance is $G = 27\ \Omega$, the shunt resistance is $S = 3\ \Omega$, and the ammeter range is $0$ to $10\ \text{mA}$ — meaning the total current that produces full-scale deflection in the ammeter is $I = 10\ \text{mA}$.

Let’s work through the reasoning step by step.

1. Understand the parallel connection In an ammeter, the galvanometer and shunt are in parallel. So the voltage across both is the same. If $I_g$ is the current through the galvanometer at full deflection, and $I_s$ is the current through the shunt, then:

Also, the total current entering the ammeter is:

2. Express $I_s$ in terms of $I_g$ From the voltage equality:

Substitute into the total current equation:

3. Plug in the given values $G = 27\ \Omega$, $S = 3\ \Omega$, $I = 10\ \text{mA}$:

Therefore:

$I_g = \frac{10}{10} = 1\ \text{mA}$ …

The key idea is that a galvanometer is a sensitive current-measuring device that deflects fully for a small current (its full-scale deflection current, $I_g$). To measure larger currents (as an ammeter does), we need to bypass most of the current around the galvanometer coil, protecting it from burning out. This is done by connecting a shunt — a small resistance — in parallel.

Why parallel? Because a parallel path divides the current. The galvanometer still sees only $I_g$ at full deflection, while the shunt carries the excess current $(I - I_g)$. The shunt resistance $S$ is chosen so that at the desired maximum current $I$, exactly $I_g$ flows through the galvanometer. Since the voltage across parallel branches is equal:

where $G$ is the galvanometer resistance. Solving:

For a large range ($I \gg I_g$), $S$ becomes very small — hence a small resistance in parallel.

Now, let's work through the reasoning step by step:

1. Understand the galvanometer's limitation: A galvanometer is essentially a sensitive moving-coil meter with resistance $G$ (typically 10–100 $\Omega$) and full-scale deflection current $I_g$ (often a few mA). It cannot handle large currents directly — passing a large current through it would permanently damage the coil.

2. Goal of an ammeter: An ammeter must measure a wide range of currents (say 0–1 A or more) while offering very low resistance to the circuit, so it doesn't disturb the current being measured. The galvanometer alone has too high a resistance and too low a current capacity.

3. Why parallel (shunt) works: Connecting a small resistance $S$ in parallel creates a current divider. At full-scale deflection, the total current $I$ entering the ammeter splits: $I_g$ through the galvanometer and $(I - I_g)$ through the shunt. The shunt "steals" the excess current. The parallel combination also reduces the overall ammeter resistance to $\frac{G S}{G+S}$, which is very small — ideal for an ammeter.

4. Derive the shunt value: Using the voltage equality across parallel branches: …

Why a shunt converts a galvanometer into an ammeter

A galvanometer is a sensitive current-measuring device with high resistance $G$ that can only handle a small current $I_g$ before its coil deflects fully. To measure larger currents, we place a low-resistance shunt $S$ in parallel with the galvanometer. Most of the current bypasses the galvanometer through this shunt, while a small fraction flows through the coil to produce the deflection.

The ammeter's effective resistance is the parallel combination of $G$ and $S$, which must be very small so that inserting the ammeter into a circuit doesn't significantly alter the current being measured.

Step-by-step solution

1. Identify what flows through the galvanometer

The problem states that when the ammeter reads its full-scale value $I$, the current through the galvanometer is $0.1\%$ of $I$:

2. Find the current through the shunt

Since the galvanometer and shunt are in parallel, the total current splits between them:

3. Apply the voltage-equality condition

Both the galvanometer and shunt have the same potential difference across them (parallel connection). Using Ohm's law:

Substituting the currents:

Simplifying:

4. Calculate the ammeter's net resistance …