Understanding Ammeter Range Extension

An ammeter is a device designed to measure electrical current. Each ammeter has a specific maximum current it can measure, known as its range. This limit is often determined by the internal resistance of the ammeter's coil and its power handling capacity.

To measure currents that exceed the ammeter's original range, a technique involving a parallel resistor, called a shunt resistor, is employed. This shunt resistor has a very low resistance value. By connecting it in parallel with the ammeter, it diverts the majority of the current away from the ammeter's internal coil, allowing only a fraction of the total current to pass through the meter itself. This protects the ammeter and enables it to measure much larger currents accurately.

Ammeter Specifications

Let's list the information provided in the question:

• The ammeter's original current range ($I_g$): 10 A
• The ammeter's internal resistance ($R_g$): 0.1 $\Omega$
• The desired new current range ($I$): 30 A

Calculating the Shunt Resistance

Our goal is to find the value of the shunt resistance ($R_s$) that needs to be connected in parallel with the ammeter to extend its range from 10 A to 30 A.

When the total current ($I$) is 30 A, the current flowing through the ammeter's internal coil ($I_g$) will still be its maximum rated current, which is 10 A. This is because the shunt resistor diverts the excess current.

The current that must flow through the shunt resistor ($I_s$) is the difference between the total desired current and the current the ammeter can handle:

$$I_s = I - I_g$$

Plugging in the values:

$$I_s = 30 \text{ A} - 10 \text{ A} = 20 \text{ A}$$

Since the shunt resistor ($R_s$) is connected in parallel with the ammeter (internal resistance $R_g$), the voltage drop across both components must be the same. We can express this using Ohm's Law ($V = I \times R$):

$$V_s = V_g$$

$$I_s \times R_s = I_g \times R_g$$

Now, we can rearrange the formula to solve for the shunt resistance ($R_s$):

$$R_s = \frac{I_g \times R_g}{I_s}$$

Let's substitute the known values into the equation:

$$R_s = \frac{10 \text{ A} \times 0.1 \Omega}{20 \text{ A}}$$

First, calculate the numerator ($I_g \times R_g$):

$$10 \text{ A} \times 0.1 \Omega = 1 \text{ V}$$

This 1 V represents the voltage drop across the ammeter when it's measuring its maximum current (10 A).

Now, calculate $R_s$:

$$R_s = \frac{1 \text{ V}}{20 \text{ A}}$$

$$R_s = 0.05 \Omega$$

Therefore, a resistance of 0.05 $\Omega$ needs to be added in shunt (parallel) with the ammeter to increase its range to 30 A.

Conclusion

The calculation shows that adding a 0.05 $\Omega$ resistor in shunt with the ammeter is required to achieve the desired range extension. This matches the option specifying 0.05 $\Omega$ in shunt with the meter.