Problem 179 - A Complete Resource Book in Physics for JEE Main

Question

A galvanometer having a coil resistance of \(100 \Omega\) gives a full scale deflection, when a current of \(1 \mathrm{~mA}\) is passed through it. The value of the resistance, which can convert this galvanometer into ammeter giving a full scale deflection for a current of \(10 \mathrm{~A}\), is (A) \(2 \Omega\) (B) \(0.1 \Omega\) (C) \(3 \Omega\) (D) \(0.01 \Omega\)

Step-by-Step Solution

Answer

The value of the shunt resistance that can convert this galvanometer into an ammeter giving a full scale deflection for a current of \(10 \mathrm{~A}\) is approximately \(0.01 \Omega\), which corresponds to option (D).

Step 1: Write down the known values

We are given the following information: - Coil resistance of the galvanometer, \(R_g = 100 \Omega\) - Full scale deflection current in the galvanometer, \(I_g = 1 \mathrm{~mA} = 0.001 \mathrm{~A}\) - Desired full scale deflection current in the ammeter, \(I_A = 10 \mathrm{~A}\)

Step 2: Write down the formula for total current

The total current in the circuit is the sum of the currents in the galvanometer and the shunt resistor: \[I_A = I_g + I_s\] Where \(I_s\) is the current in the shunt resistor.

Step 3: Calculate the current in the shunt resistor

Using the total current formula, we can calculate the current in the shunt resistor: \[I_s = I_A - I_g = 10 \mathrm{~A} - 0.001 \mathrm{~A} = 9.999 \mathrm{~A}\]

Step 4: Write down the formula for the shunt resistance

Since the galvanometer and shunt resistance are in parallel, the voltage across them is the same. Therefore, we can use Ohm's law to find the shunt resistance: \[V = I_g \cdot R_g = I_s \cdot R_s\] Where \(V\) is the voltage across the galvanometer and shunt resistor, and \(R_s\) is the shunt resistance.

Step 5: Calculate the shunt resistance

Rearranging the formula for the shunt resistance, we get: \[R_s = \frac{I_g \cdot R_g}{I_s}\] Now, plug in the given values and calculate \(R_s\): \[R_s = \frac{0.001 \mathrm{~A} \cdot 100 \Omega}{9.999 \mathrm{~A}} = 0.01001 \Omega \approx 0.01 \Omega\] The value of the shunt resistance that can convert this galvanometer into an ammeter giving a full scale deflection for a current of \(10 \mathrm{~A}\) is approximately \(0.01 \Omega\), which corresponds to option (D).