The resistance of an ammeter is `13 Omega` 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

Options

A. `20 Omega`
B. `2Omega`
C. `0.2 Omega`
D. `2kOmega`

Answer

B

Explanation

Let `i_(a)` is the current flowing through ammeter and I is the total current. So a current `i-i_(a)` will flow through shunt resistance.
Potential difference across ammeter and shunt resistance is same.

ie, `i_(a)xxR=(i-i_(a))xxS`
or `S=(i_(a)R)/(i-i_(a))" "......(i)`
Given `i_(a)=100A,i=750A,R=13Omega`
Hence, `S=(100xx13)/(750-100)=2Omega`