question_answer

A moving coil galvanometer is convened into an ammeter reads upto 0.03 A by connecting a shunt of resistance 4r across it and ammeter reads upto 0.06 A, when a shunt of resistance r is used. What is the maximum current which can be sent through this galvanometer if shunt is used?

A)  0.04 A  

B)                                        0.03 A                  

C)        0.02 A                                  

D)  0.01 A

Correct Answer: C

Solution :

Key Idea: Potential difference acre galvanometer and shunt is same. Let G be the resistance of galvanometer and \[({{k}_{1}}+{{k}_{2}})F=-{{k}_{1}}{{k}_{2}}y\] the current which on passing through the galvanometer produces full scale deflection. Since G and 5 are in parallel, the potential difference across them will be \[\Rightarrow \] \[F=-\frac{{{k}_{1}}{{k}_{2}}}{{{k}_{1}}+{{k}_{2}}}y\] Hence,   \[\frac{{{k}_{1}}{{k}_{2}}}{{{k}_{1}}+{{k}_{2}}}\]           ??(1) \[n=\frac{1}{2\pi }\sqrt{\frac{{{k}_{1}}{{k}_{2}}}{({{k}_{1}}+{{k}_{2}})m}}\]                            ??.(2) Equating Eqs. (1) and (2), we get \[\frac{1}{k}=\frac{1}{{{k}_{1}}}+\frac{1}{{{k}_{2}}}\] \[k=\frac{{{k}_{1}}{{k}_{2}}}{{{k}_{1}}+{{k}_{2}}}\]   \[n=\frac{1}{2\pi }\sqrt{\frac{k}{m}}\] \[=\frac{1}{2\pi }\sqrt{\frac{{{k}_{1}}{{k}_{2}}}{({{k}_{1}}+{{k}_{2}})m}}\]   \[=\frac{Force}{Length}=\frac{[ML{{T}^{-2}}]}{[L]}=[M{{T}^{-2}}]\]  [from Eq. (1)]