question_answer

The resistance of an ammeter is 13\[{{i}_{2}}>{{i}_{3}}>{{i}_{1}}\]. and its scale is graduated for a current upto 100 A. After an additional shunt has been connected to this ammeter it becomes possible to measure currents upto 750 A by this meter. The value of shunt resistance is

A)  20\[{{i}_{2}}>{{i}_{1}}>{{i}_{3}}\]         

B)                        2\[{{i}_{1}}>{{i}_{2}}>{{i}_{3}}\]                               

C)        0.2\[{{i}_{1}}>{{i}_{3}}>{{i}_{2}}\]           

D)        2\[\sqrt{30}\]

Correct Answer: B

Solution :

Let \[{{\phi }_{0}}\]be the current flowing through ammeter and i the total current. So, a current \[\frac{T}{4}\] will flow through shunt resistance. Potential difference across ammeter and shunt resistance is same, ie,     \[\frac{T}{8}\] or     \[\frac{T}{12}\]                           ?...(i) Given, \[\frac{T}{2}\] Hence, \[\lambda \]