Problem 83 - Physics

Question

The coil of a galvanometer has a resistance of \(20.0 \Omega\), and its meter deflects full scale when a current of \(6.20 \mathrm{~mA}\) passes through it. To make the galvanometer into an ammeter, a \(24.8-\mathrm{m} \Omega\) shunt resistor is added to it. What is the maximum current that this ammeter can read?

Step-by-Step Solution

Answer

The maximum current the ammeter can read is approximately 5.002 A.

Step 1: Understand the concept

The galvanometer acts as a detection unit for current, but only measures a small range of current (its full-scale deflection current). To convert it into an ammeter capable of measuring larger currents, a shunt resistor (of low resistance) is placed parallel to it. This allows additional current to bypass the galvanometer coil, therefore, extending the range of current measurement.

Step 2: Calculate the total current

The total current flowing in the circuit is divided between the galvanometer coil and the shunt resistor. Since the galvanometer coil deflects full scale for a current of 6.20 mA, this current will pass through it, and the remaining current will flow through the shunt resistor. The total current follows the relation \( I = I_G + I_S \), where \( I_G = 6.20 \, \mathrm{mA} \) is the current through the galvanometer and \( I_S \) is the current through the shunt resistor.

Step 3: Apply current division rule

Apply the current division rule for parallel circuits. This states that the current through each branch is inversely proportional to its resistance and given by:\[ I_S = I \times \frac{R_G}{R_G + R_S} \]Here, \( R_G = 20.0 \, \Omega \) and \( R_S = 24.8 \, \mathrm{m} \Omega = 0.0248 \, \Omega \). Use the current condition in the galvanometer for full scale, \( I_G = 6.20 \, \mathrm{mA} \). Substituting this into the equation:\[ I = I_G + (I_G \times \frac{R_G}{R_S}) \]

Step 4: Solve for maximum current

Substitute given values:\[ I = 6.20 \, \mathrm{mA} + \left(6.20\, \mathrm{mA}\times \frac{20.0\, \Omega}{0.0248\, \Omega} \right) \]Solve this to obtain the total current \(I\): \[ I = 6.20 \, \mathrm{mA} + 6.20 \, \mathrm{mA} \times 806.45 \approx 5002.18 \, \mathrm{mA} \]Convert this to Amperes: \[ I \approx 5.002 \, \mathrm{A} \].

Step 5: Final Step: Write down the final answer

The maximum current this ammeter can read is approximately \(5.002 \, \mathrm{A}\).

Key Concepts

Shunt Resistor

A shunt resistor is a small resistance component used in electrical circuits, primarily to extend the range of instruments like galvanometers. By connecting a shunt resistor in parallel with a galvanometer, it allows a large proportion of the current to "shunt" or bypass the sensitive galvanometer coil. This is necessary because galvanometers are designed to measure only small currents, often limited to milliamperes.

• The resistance value of a shunt resistor is typically very low. In our exercise, it totals only 24.8 milli-ohms, a fraction compared to the galvanometer's 20 ohms.
• By maintaining this low resistance, more current is directed through the shunt compared to the galvanometer. This division of current is key in achieving accurate measurements at higher current scales.

The main function of the shunt is to "keep safe" the galvanometer while reading large currents. Without it, exceeding the galvanometer's capacity will damage the system. Thus, a well-chosen shunt resistor is critical for safety and precision.

Current Division Rule

The current division rule helps us understand how current divides across branches in a parallel circuit. It's especially useful when converting a galvanometer into an ammeter using a shunt resistor. This rule states that the current through a particular branch is inversely proportional to the resistance of that branch. In simpler words, less resistance leads to more current flowing through that path, which is why low-resistance shunt resistors are used.
Here's the formula:

• \[ I_S = I \times \frac{R_G}{R_G + R_S} \]

Where:

• \( I_S \) is the current through the shunt resistor.
• \( I \) is the total current entering the parallel combination.
• \( R_G \) and \( R_S \) are the resistances of the galvanometer and the shunt respectively.

This calculation is deterministic for converting and upgrading the measurement capacity of a galvanometer by exploiting the distribution of current based on resistance.

Parallel Circuit Calculation

In any parallel circuit, voltage across all branches is identical, but the current differs based on resistance. Calculating the total current flowing through a parallel circuit involves understanding that each branch "competes" based on its resistance. For the galvanometer and shunt resistor, we say the total current (\( I \)) is the sum of the individual branch currents:

• \[ I = I_G + I_S \]
• \(I_G\) is the current through the galvanometer.
• \(I_S\) is the current through the shunt resistor, calculated using the current division rule.

One applies Ohm's Law and the formulae involving resistance to realize such calculations.
Understanding these principles helps in deriving the maximum current measurable, as seen in the example, by summing the individual currents through each component. Each component's resistance determines its current, thus safeguarding the galvanometer from excess current while enhancing utility.