A100μA galvanometer with internal resistance 1000Ω is to be converted into an ammeter of 10 A. Find the shunt to be connected
A
0.1Ω
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
1Ω
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
0.01Ω
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
D
0.001Ω
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution
The correct option is D0.01Ω To convert galvanometer into an ammeter a small resistance(shunt) is to be connected in parallel with it. Consider the circuit shown in figure. Since, galvanometer and shunt are parallel to each other voltage across them is same. Hence, (I−Ig)S=IgR ⇒S=IgRI−Ig S=100×10−6×100010−100×10−6