Q.
A moving coil galvanometer converted into an ammeter reads up to 0.03 A by connecting a shunt of resistance 4r across it and ammeter reads up to 0.06 A, when a shunt of resistance r is used. What is the maximum current which can be sent through this galvanometer if no shunt is used?
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a
0.03 A
b
0.04 A
c
0.02 A
d
0.01 A
answer is C.
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Detailed Solution
Let G be the resistance of galvanometer and ig the current which on passing through the galvanometer produces full-scale deflection.Since G and S are in parallel, the potential difference across them will beig × G=i−ig × Sigi= SS+GHence, ig= 4r4r+G × 0.03 ....1ig= rr+G × 0.06 ....2Equating Eqs. (1) and (2), we get4r + G = 2(r + G)⇒ G=2r∴ ig= 4r4r+2r × 0.03=0.02 A from Eq. 1
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