A galvanometer of resistance 10 Ω gives full-scale deflection when 1 mA current passes through it. The resistance required to convert it into a voltmeter reading upto 2.5 V is

Question

A galvanometer of resistance 10 Ω gives full-scale deflection when 1 mA current passes through it. The resistance required to convert it into a voltmeter reading upto 2.5 V is (A) 24.9 Ω (B) 249 Ω (C) 2490 Ω (D) 24900 Ω

Tardigrade Admin · Accepted Answer

Here, Ig=1 mA =1×10-3 A, G=10 Ω, V=2.5 V From the figure <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/b488944cad6e288599580f406f9c6c35-.jpeg" /> V=Ig(G+R) or R=(V/Ig)-G Substituting the given values, we get, R=(2.5 V/1×10-3 A)-10 Ω =2500 Ω-10 Ω =2490 Ω

Q. $A$ galvanometer of resistance $10 Ω$ gives full-scale deflection when $1 m A$ current passes through it. The resistance required to convert it into a voltmeter reading upto $2.5 V$ is

$24.9 Ω$

$249 Ω$

$2490 Ω$

$24900 Ω$

Here, $I_{g} = 1 m A$
$= 1 \times 1 0^{- 3} A$ ,
$G = 10 Ω$ ,
$V = 2.5 V$
From the figure

$V = I_{g} (G + R)$ or $R = \frac{V}{I _{g}} - G$
Substituting the given values, we get,
$R = \frac{2.5 V}{1 \times 1 0 ^{- 3} A} - 10 Ω$
$= 2500 Ω - 10 Ω$
$= 2490 Ω$

Q. A galvanometer of resistance 10Ω gives full-scale deflection when 1mA current passes through it. The resistance required to convert it into a voltmeter reading upto 2.5V is

24.9Ω

249Ω

2490Ω

24900Ω