Problem 33 - A Complete Resource Book in Physics for JEE Main

Question

An ammeter reads up to \(1 \mathrm{~A}\). Its internal resistance is \(0.81 \Omega\). To increase the range to \(10 \mathrm{~A}\), the value of the required shunt is (A) \(0.03 \Omega\) (B) \(0.3 \Omega\) (C) \(0.9 \Omega\) (D) \(0.09 \Omega\)

Step-by-Step Solution

Answer

The shunt resistance required to increase the range of the ammeter to 10 Amperes is approximately \(0.09 \, \Omega\).

Step 1: Calculate the Ammeter's Maximum Voltage Drop

First, find the maximum voltage drop across the ammeter when it reads its full-scale deflection of 1 Ampere. Use Ohm's law for this, which is \( V = I \times R \), where \( V \) is the voltage, \( I \) is the current, and \( R \) is the resistance. The voltage across the ammeter is \( V = 1A \times 0.81 \, \Omega = 0.81 \, volts \).

Step 2: Calculate the Total Current Required

Second, since the ammeter needs to be able to read up to 10 Amperes, the total current (including the initial 1 Ampere the ammeter can already measure) is 10 Amperes.

Step 3: Determine the Current Across the Shunt Resistor

Third, determine the current that needs to pass through the shunt resistor. This is the total current minus the current the ammeter already handles (so \( I_{shunt} = I_{total} - I_{ammeter} \)). The shunt current is \( I_{shunt} = 10A - 1A = 9A \).

Step 4: Calculate the Shunt Resistance

Finally, determine the shunt resistance using Ohm's law again. Rearrange the equation to solve for the resistance: \( R = \frac{V}{I} \). The shunt resistance is \( R_{shunt} = \frac{0.81 \, volts}{9A} \approx 0.09 \, \Omega \). Therefore, the correct option is (D) \(0.09 \Omega\).