Animated Solution for Physics - Magnetic Effects of Current: The resistance of a galvanometer is $50\Omega$ and the maximum current which can be passed through it is $0.002\text{A}$. What resistance must be connected to it in order to convert it into an ammeter of range $0-0.5\text{A}$?

To measure large currents, we cannot simply place a delicate galvanometer directly into the circuit. The delicate coil inside would instantly burn out under the heavy electrical load. Instead, we must engineer a clever bypass route for the excess current. We achieve this by connecting a very small resistance, known as a shunt, in parallel with the galvanometer.

When the total current $I$ enters this parallel combination, it splits into two distinct paths. A small, safe fraction of the current, denoted as $I_g$, flows through the galvanometer. The vast majority of the current, which is $I-I_g$, takes the path of least resistance through the shunt $S$.

Because the galvanometer and the shunt are connected in parallel, the fundamental laws of circuits dictate that the potential difference across both branches must be exactly the same.

Using Ohm's law ($V=IR$), the voltage across the galvanometer is its current times its resistance ($I_{g}G$). Similarly, the voltage across the shunt is the remaining current times the shunt resistance ($(I-I_g)S$). Equating these two potential drops gives us our master equation:

Let's bring in the specific values provided in the problem. The maximum safe current for the galvanometer is $I_g=0.002\text{A}$, its internal resistance is $G=50\Omega$, and we want to design an ammeter capable of measuring a total current of $I=0.5\text{A}$. Substituting these values into our master equation yields:

Now, we carefully compute the values on both sides. On the left, $0.002\times 50$ simplifies cleanly to $0.1$. On the right, the current flowing through the shunt is $0.5-0.002=0.498\text{A}$. This updates our equation to:

To isolate the required shunt resistance $S$, we divide:

While this fraction might look slightly messy, notice that the denominator $0.498$ is extremely close to $0.5$. In competitive exams, making such safe approximations is a crucial time-saving technique. Approximating the denominator gives:

Thus, a shunt resistance of $0.2\Omega$ is required to successfully convert this galvanometer into the desired ammeter.