A galvanometer has a resistance of 5 By using a... - JAMB Physics 1999 Question
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?
A
2mA
B
10mA
C
20mA
D
25mA
correct option: c
Shunt resistance is usually given as
R = igrg/(I - ig)
Where R = 0.05Ω
rg = 5Ω
I = 2M
∴ ig = IR/(rg + R)
= (2 x 0.05)/(5 + 0.05)
= 0.10/5.05
= 0.0198A
= 19.8mA
= 20mA
R = igrg/(I - ig)
Where R = 0.05Ω
rg = 5Ω
I = 2M
∴ ig = IR/(rg + R)
= (2 x 0.05)/(5 + 0.05)
= 0.10/5.05
= 0.0198A
= 19.8mA
= 20mA
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