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Chapter 34: Problem 2873

The seave of a galvanometer of resistance \(100 \mathrm{Q}\) contains 25 divisions. It gives a deflection of 1 division on passing a current of $4 \times 10^{-4} \mathrm{~A}$. The resistance in ohm to be added to it. so that it may become a voltmeter of range \(2.5 \mathrm{~V}\) is (A) 250 (B) 300 (C) 150 (D) 100

Short Answer

Expert verified
To convert the given galvanometer into a voltmeter of range 2.5 V, a resistance of \(150 \Omega\) must be added to it. The correct option is (C) 150.

Step by step solution

01

Identify the given values

Here are the given values from the problem statement: Galvanometer resistance (G) = 100 Ω Deflection for a current (I₁) = 1 division Current that causes 1 division deflection (I₁) = 4 × 10⁻⁴ A Desired voltmeter range (V) = 2.5 V
02

Calculate the current for 25 divisions deflection

To calculate the current that causes 25 divisions deflection, we can use the proportionality between the current and the deflection: I₁ / 1 division = I₂ / 25 divisions Solving for I₂ (current for 25 divisions deflection): I₂ = I₁ × 25 divisions I₂ = (4 × 10⁻⁴ A) × 25 divisions I₂ = 10 × 10⁻³ A
03

Calculate the voltage across the galvanometer

Using Ohm's law, we can calculate the voltage across the galvanometer (V_G) when the full-scale deflection of 25 divisions is reached: V_G = I₂ × G V_G = (10 × 10⁻³ A) × (100 Ω) V_G = 1 V
04

Calculate the resistance to be added

Now, to convert the galvanometer into a voltmeter with a range of 2.5 V, we need to find the resistance (R) that needs to be added in series with the galvanometer. We can use the formula: R = (V - V_G) / I₂ R = (2.5 V - 1 V) / (10 × 10⁻³ A) R = 1.5 V / 0.01 A R = 150 Ω Therefore, a resistance of 150 Ω must be added to the galvanometer to convert it into a voltmeter of range 2.5 V. So, the correct option is (C) 150.

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Most popular questions from this chapter

A galvanometer, having a resistance of \(50 \Omega\), gives a full scale deflection for a current of \(0.05 \mathrm{~A}\). The length in meter of a resistance wire of area of cross-section \(2.97 \times 10^{2} \mathrm{~cm}^{2}\) that can be used to convert the galvanometer into an ammeter which can read a maximum of \(5 \mathrm{~A}\) current is (specific resistance of wire $=5 \times 10^{-7} \Omega \mathrm{m}$ ) (A) \(1.5\) (B) 6 (C) 8 (D) 3

A galvanometer of resistance \(200 \Omega\) gives full scale deflection for a current of \(10^{-3} \mathrm{~A}\). To convert it into an ammeter capable of measuring upto \(1 \mathrm{~A}\). What resistance should be connected in parallel with it ? (A) \(2 \times 10^{-1} \Omega\) (B) \(2 \Omega\) (C) \(2 \times 10^{-3} \Omega\) (D) \(2 \times 10^{-6} \Omega\)

A galvanometer of resistance \(200 \Omega\) gives full scale deflection with 15 milli-ampere current. In order to convert it into a \(15 \mathrm{~V}\) range voltmeter. What is the value of resistance connected in series ? (A) \(1000 \Omega\) (B) \(800 \Omega\) (C) \(2500 \Omega\) (D) \(1500 \Omega\)

A galvanometer with resistance \(100 \Omega\) is converted into an ammeter with a resistance of \(0.1\). The galvanometer shows full scale deflection with current of \(100 \mu \mathrm{A}\). Then what will be the minimum current in the circuit for full scale deflection of galvanometer ? (A) \(0.1001 \mathrm{~mA}\) (B) \(100.1 \mathrm{~mA}\) (C) \(1000.1 \mathrm{~mA}\) (D) \(1.001 \mathrm{~mA}\)

What is the relation between figure of merit \((\mathrm{k})\) and current sensitivity \(\left(\mathrm{S}_{1}\right) ?\) (A) \(\mathrm{S}_{1}=\mathrm{k}^{-1}\) (B) \(\mathrm{S}_{1}=(\mathrm{k} / 2)\) (C) \(\mathrm{S}_{1}=\mathrm{kV}\) (D) \(\mathrm{S}_{1}=(\mathrm{k}) \mathrm{I}\)
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