Understanding Ammeter Range Extension

An ammeter is a crucial instrument for measuring electrical current. The maximum current an ammeter can measure is known as its range. This range is determined by the internal resistance of the meter, often a galvanometer coil ($R_m$), and the maximum current ($I_m$) that the coil can handle without damage.

To measure currents larger than the original range, we need to modify the ammeter. This is typically achieved by connecting a low-resistance resistor, called a shunt resistor ($R_{sh}$), in parallel with the ammeter's internal resistance. The purpose of the shunt resistor is to divert the majority of the current away from the sensitive meter movement, allowing the combination to measure higher total currents.

Given Parameters

Let's identify the values provided in the question:

• The ammeter's original current range ($I_m$): 5 A
• The internal resistance of the ammeter ($R_m$): 0.2 Ω
• The desired new current range ($I$): 25 A

Calculating the Shunt Resistance

When the ammeter is connected to measure a current ($I$) within the new range, the total current splits. A portion ($I_m$) flows through the ammeter's internal resistance ($R_m$), and the remaining current ($I - I_m$) flows through the parallel shunt resistor ($R_{sh}$).

Since the shunt resistor is connected in parallel with the ammeter, the voltage drop across both components must be equal:

Voltage across ammeter = Voltage across shunt resistor

Using Ohm's Law ($V = I \times R$), we can write this as:

$$I_m \times R_m = (I - I_m) \times R_{sh}$$

Now, we plug in the known values to find the required shunt resistance ($R_{sh}$):

$$5 \, \text{A} \times 0.2 \, \Omega = (25 \, \text{A} - 5 \, \text{A}) \times R_{sh}$$

Calculate the left side (voltage across the ammeter):

$$1.0 \, \text{V} = (20 \, \text{A}) \times R_{sh}$$

Now, solve for $R_{sh}$:

$$R_{sh} = \frac{1.0 \, \text{V}}{20 \, \text{A}}$$

$$R_{sh} = 0.05 \, \Omega$$

Determining the Connection Type

The calculation yields a resistance value of $0.05 \, \Omega$. To effectively extend the current measuring capability of the ammeter to 25 A, this calculated resistance must be connected in parallel with the ammeter itself. This parallel connection ensures that only $5$ A passes through the meter when the total current is $25$ A.

Final Answer

Based on the calculation, a resistance of $0.05 \, \Omega$ is needed, and it must be connected in parallel with the meter to achieve the desired 0-25 A range.