A galvanometer has a resistance of 5Ω. By using a shunt wire of resistance 0.05Ω, the galvanometer could be converted to an ammeter capable of reading 2Amp. What is the current through the galvanometer?

Olawale4

9 years ago

current in galvanometer, Ig=?

current in shunt, Is= 2A

Resistance in galvanometer,Rg= 5ohms

Resistance in shunt, Rs= 0.05ohms

using the 4mula,

IsRs=IgRg

2x0.05=5xIg

0.1=5Ig

Ig=0.1/5

Ig=0.02A

Ig=20mA

Answer=c

Salvatore_xoxo

9 years ago

I = 2A, Rg = 5 ohms, Rs = 0.05 ohms, Ig =?

Rs = IgRg/I-Ig

0.05 = Ig*5/(2-Ig), make Ig subject formular

0.05(2-Ig) = 5Ig

0.1 = 0.05Ig + 5Ig

0.1 = 5.05Ig

Ig = 0.1/5.05

Ig = 0.0198A

= 0.02A = 20mA

focusIQ

1 year ago

IT'S COMPLICATED 😔🤦

indefegable t

10 years ago

not understood to me, pls guys

whizola

11 years ago

if pd across coil and shunt are d same, then ir = It...i *5=2*0.05....i = (2*0.05)/5...I=2*0.01 or 2*10^-2...Converting to milliamps I.e 10^-3amps=1milliamp, ans = 20mA