A galvanometer with full-scale deflection of 10 mA is to be converted to a voltmeter with full-scale deflection of 5v.  if a series resistance of 498Ω is used for the conversion, the resistance of the galvanometer is ?

Nwese

5 years ago

Myschool the answer is 2ohms

paiseols

4 years ago

R=(V/Ig)-r
we are looking for resistance in galvanometer,r
r=(v/Ig)-R
convert 10ma to A=10/1000=0.01A
r=(5/0.01)-498Ω
r=500-498
r=2Ω

OladosuMayowa304

4 months ago

You can actually guess the answer using normal emf equation

Solzzz

5 years ago

It has a series resistance of 498Ω according to jamb 2010, there the answer is 2Ω 😌

zamix

7 years ago

please explain where you got 490

Tribunal244

1 year ago

When converting galvanometer to voltmeter:
V= Ig x (Rg + Rs)
Now plug in the values:
5 = 0.01 x (Rg + 498)
Divide both sides by 0.01:
500 = Rg + 498
Now subtract 498 from both sides:
Rg = 500 – 498
Final Answer: 2

igwewise

9 years ago

The question here @49 is abnormal.