Understanding Ammeter Range Extension with a Shunt

An ammeter is an instrument used to measure electric current. A typical meter movement, like the 60 microampere meter described, is designed to measure relatively small currents. To measure larger currents, a parallel resistor, called a shunt resistor, is connected across the meter terminals. This shunt resistor diverts the majority of the current, allowing only a small, proportional amount to pass through the sensitive meter movement.

In this problem, we have a meter with a full-scale deflection current of 60 microamperes (\(\text{I}_{\text{m}}\)) and an internal resistance of 100 ohms (\(\text{R}_{\text{m}}\)). We want to use this meter to measure a total current (\(\text{I}\)) of 100 milliamperes.

Calculating Current Distribution in Ammeter Shunt Circuit

When the shunt resistor is connected in parallel with the meter, the total current \(\text{I}\) entering the parallel combination splits into two paths: one through the meter (\(\text{I}_{\text{m}}\)) and one through the shunt resistor (\(\text{I}_{\text{sh}}\)). The total current is the sum of these two currents:

\(\text{I} = \text{I}_{\text{m}} + \text{I}_{\text{sh}}\)

Our goal is to find the value of the current through the shunt (\(\text{I}_{\text{sh}}\)). We can rearrange the equation above to solve for \(\text{I}_{\text{sh}}\):

\(\text{I}_{\text{sh}} = \text{I} - \text{I}_{\text{m}}\)

Step-by-Step Calculation

First, let's make sure all current values are in the same units.

• The total current to be measured is \(\text{I} = 100\ \text{mA}\).
• The full-scale current of the meter is \(\text{I}_{\text{m}} = 60\ \mu\text{A}\).

We need to convert 60 \(\mu\text{A}\) to milliamperes (mA).

\(1\ \text{mA} = 1000\ \mu\text{A}\)

So, \(1\ \mu\text{A} = \frac{1}{1000}\ \text{mA} = 0.001\ \text{mA}\)

Therefore, \(\text{I}_{\text{m}} = 60\ \mu\text{A} = 60 \times 0.001\ \text{mA} = 0.06\ \text{mA}\).

Now we can calculate the current through the shunt (\(\text{I}_{\text{sh}}\)):

\(\text{I}_{\text{sh}} = \text{I} - \text{I}_{\text{m}}\)

\(\text{I}_{\text{sh}} = 100\ \text{mA} - 0.06\ \text{mA}\)

\(\text{I}_{\text{sh}} = 99.94\ \text{mA}\)

The calculated current through the shunt resistor is 99.94 mA. Looking at the options provided, 99.9 mA is the closest value.

Understanding Shunt Resistance (Optional but helpful)

Although the question only asks for the shunt current, it's useful to understand how the shunt resistance (\(\text{R}_{\text{sh}}\)) is determined. Since the meter and the shunt are in parallel, the voltage across them is the same.

\(\text{V}_{\text{m}} = \text{V}_{\text{sh}}\)

Using Ohm's Law (\(\text{V} = \text{I} \times \text{R}\)):

\(\text{I}_{\text{m}} \text{R}_{\text{m}} = \text{I}_{\text{sh}} \text{R}_{\text{sh}}\)

We know \(\text{I}_{\text{m}} = 0.06\ \text{mA}\), \(\text{R}_{\text{m}} = 100\ \Omega\), and we just calculated \(\text{I}_{\text{sh}} = 99.94\ \text{mA}\). We can calculate \(\text{R}_{\text{sh}}\):

\(\text{R}_{\text{sh}} = \frac{\text{I}_{\text{m}} \text{R}_{\text{m}}}{\text{I}_{\text{sh}}}\)

Let's use amperes (A) for consistency in the calculation, or ensure units cancel correctly. Using mA and Ohms:

\(\text{R}_{\text{sh}} = \frac{0.06\ \text{mA} \times 100\ \Omega}{99.94\ \text{mA}}\)

\(\text{R}_{\text{sh}} = \frac{6\ \text{mA}\ \Omega}{99.94\ \text{mA}}\)

\(\text{R}_{\text{sh}} \approx 0.060036\ \Omega\)

This shows that a very small shunt resistance is needed to divert most of the current.

Summary of Calculation

To find the current through the shunt (\(\text{I}_{\text{sh}}\)), subtract the meter's full-scale current (\(\text{I}_{\text{m}}\)) from the total current to be measured (\(\text{I}\)).

• Total current (\(\text{I}\)) = 100 mA
• Meter current (\(\text{I}_{\text{m}}\)) = 60 µA = 0.06 mA
• Shunt current (\(\text{I}_{\text{sh}}\)) = \(\text{I} - \text{I}_{\text{m}} = 100\ \text{mA} - 0.06\ \text{mA} = 99.94\ \text{mA}\)

The closest option is 99.9 mA.

Revision Table: Ammeter Shunt Concepts

Additional Information: Ammeter Design

Designing an ammeter for a specific range involves selecting the appropriate shunt resistor. The formula \(\text{R}_{\text{sh}} = \frac{\text{I}_{\text{m}} \text{R}_{\text{m}}}{\text{I} - \text{I}_{\text{m}}}\) is used. A higher range requires a smaller shunt resistance. Conversely, to extend a voltmeter's range, a multiplier resistor is added in series. Understanding these techniques is fundamental in electrical measurements and circuit design. The accuracy of the ammeter depends on the precision of both the meter movement and the shunt resistor. Temperature changes can affect resistance values, which might require using special alloys or temperature compensation circuits for high-precision ammeters.