Problem 47 - Master Resource Book in JEE Main Physics

Question

A galvanometer of resistance \(100 \Omega\) gives a full scale deflection for a current of \(10^{-5} \mathrm{~A} .\) To convert it into a ammeter capable of measuring upto \(1 \mathrm{~A}\), we should connect a resistance of (a) \(1 \Omega\) in parallel (b) \(10^{-3} \Omega\) in parallel (c) \(10^{5} \Omega\) in series (d) \(100 \Omega\) in series

Step-by-Step Solution

Answer

(b) Connect \(10^{-3} \Omega\) in parallel.

Step 1: Understanding the Problem

We need to convert the galvanometer into an ammeter that can measure up to 1 A. A galvanometer is connected in parallel with a shunt resistor to redirect excess current. The formula for shunting is used to find the resistance needed.

Step 2: Equation for Parallel Resistor

When a resistor is connected in parallel to the galvanometer, let's denote the shunt resistance as \(R_s\). The current through the galvanometer, \(I_g = 10^{-5}\,\text{A}\) and the resistance \(R_g = 100 \,\Omega\). The remaining current \(I - I_g\) must pass through \(R_s\). Using the parallel resistor formula: 1. \(V_g = R_g \times I_g\)2. \(V_s = R_s \times (I - I_g)\)3. Since they are parallel, \(V_g = V_s\)Calculate \(R_s\) using these equations.

Step 3: Calculation of Shunt Resistance

Calculate voltage across the galvanometer:\( V_g = 100 \,\Omega \times 10^{-5} \,\text{A} = 10^{-3} \,\text{V} \).Voltage across the shunt resistor, \( V_s = I_s \times R_s = (1 - 10^{-5}) \,\text{A} \times R_s \).Setting \( V_g = V_s \), we have:\[ 10^{-3} = (1 - 10^{-5}) R_s \]Approximating \((1 - 10^{-5}) \approx 1\), \[ R_s = 10^{-3} \Omega \].

Step 4: Selecting the Correct Option

From our calculation, the appropriate resistance to connect in parallel is \(10^{-3} \Omega\). Check the options: (b) \(10^{-3} \Omega\) is the correct answer among the given choices.

Key Concepts

Shunt Resistor

A shunt resistor is an essential component when converting a galvanometer, a sensitive instrument for detecting and displaying small current changes, into an ammeter that can measure higher currents.
This is done to protect the galvanometer by redirecting most of the current through the shunt, while allowing a very small, measured amount through the galvanometer.
The decision on how much of the current should bypass the galvanometer is crucial.
This is determined by the shunt resistor's value, which ensures that the galvanometer only experiences a fraction of the total current.

• The shunt is always placed in parallel with the galvanometer.
• Its goal is to maintain the galvanometer's deflection for the maximum current it measures.
• Selecting the correct resistor value is vital to avoid damaging the galvanometer.

Using the right shunt resistor effectively allows a galvanometer to serve as an ammeter, thus facilitating the measurement of larger currents in an electrical circuit without impacting the delicate nature of the galvanometer.

Ammeter Design

Designing an ammeter requires a clear understanding of its core components and their roles. Primarily, an ammeter is built by adding a shunt resistor to a galvanometer.
This enables the measurement of higher currents while keeping the galvanometer safe from high current damage. The design process involves:

• Determining the maximum current the ammeter should measure (1 A in this case).
• Choosing a suitable shunt resistor to ensure the galvanometer measures a manageable current.
• Ensuring that the voltage across the galvanometer remains constant regardless of the total current.

By adding the shunt resistor, the current is split between the shunt and the galvanometer. The voltage across both components should be equal due to their parallel configuration. This careful balance ensures accurate and reliable current measurement.

Parallel Circuit Calculation

Understanding the calculations in parallel circuits is important when converting a galvanometer to an ammeter.
In this configuration, both the galvanometer and the shunt resistor share the same voltage.The key steps to the calculation are:

• Calculate the voltage across the galvanometer using: \( V_g = R_g \times I_g \), where \( R_g \) is the resistance, and \( I_g \) is the measured current.
• Calculate the voltage across the shunt resistor using: \( V_s = I_s \times R_s \), where \( I_s \) is the current through the shunt and \( R_s \) is its resistance.
• Set the two voltages equal (\( V_g = V_s \)) since they are in parallel, and solve for the shunt resistance (\( R_s \)).

The final step ensures that the actual current the ammeter can measure is balanced and any excess current is properly calibrated out by the shunt resistor.
This precise calculation makes it possible for the galvanometer's delicate system to be part of a high-current measuring device.