Question

A moving coil galvanometer has 150 equal divisions, Its current sensitivity is 10 divisions per milli-ampere and voltage sensitivity is 2 divisions per milli-volt. In order that each divisions reads \(1 \mathrm{~V}\), What will be the resistance in ohms needed to be connected in series with the coil? (A) \(10^{3}\) (B) 99995 (C) 9995 (D) \(10^{5}\)

Short answer

The resistance needed to be connected in series with the coil for each division to read 1V is 9995 ohms (Option C).

Step-by-step solution

Understand the provided information

We are given: - Total number of divisions of the galvanometer: 150 - Current sensitivity: 10 divisions per mA (milliampere) - Voltage sensitivity: 2 divisions per mV (millivolt) - Our goal is to have each division represent 1V, so we need to find the resistance.

Calculate the current and voltage for one division

We can calculate the current value for one division using the current sensitivity: 1 division = \(\frac{1}{10}\) mA = 0.1 mA Similarly, we can calculate the voltage value for one division using the voltage sensitivity: 1 division = \(\frac{1}{2}\) mV = 0.5 mV

Convert values to standard units

To make our calculations easier, let's convert mA and mV to A (Ampere) and V (Volt) respectively: Current for 1 division = 0.1 mA = \(0.1 \times 10^{-3}\) A Voltage for 1 division = 0.5 mV = \(0.5 \times 10^{-3}\) V

Use Ohm's Law to find the original resistance

Using Ohm's Law: \(V = IR\) Where V = voltage, I = current, and R = resistance. Let's find the original resistance of the coil: \(\frac{0.5 \times 10^{-3}}{0.1 \times 10^{-3}} = R\) R = 5 ohms

Determine the voltage per division with the desired 1V reading

Since we want each division to read 1V, the total voltage for 150 divisions must equal 150V, because: 1 division = 1V

Calculate the new current for the desired 1V per division

The current for 1 division remains the same, 0.1 mA or \(0.1 \times 10^{-3}\) A. Therefore, the total current for 150 divisions is: Total current = 0.1 mA \(\times\) 150 = 15 mA = \(15 \times 10^{-3}\) A

Use Ohm's Law to find the new resistance with the desired 1V per division

With the new total voltage and total current, we can find the desired resistance using Ohm's law: \(V = IR\) \(150 = (15 \times 10^{-3})R\) R = \(150 \div (15 \times 10^{-3})\) R ≈ 10000 ohms

Calculate the resistance to be connected in series

We've found the original resistance (5 ohms) and the desired resistance (10000 ohms). To find the resistance needed to be connected in series, subtract the original resistance from the desired resistance: 10000 ohms - 5 ohms = 9995 ohms So, the resistance needed to be connected in series is 9995 ohms (Option C).