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. `2k Omega`

Answer

B

Explanation

The potential difference across ammeter and shunt is same.
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
ie. `i_(a)xxR=(i-i_(a)) xxS`
or `S=(i_(a)R)/(i-i_(a))`.....(i)
Given, `S=(100xx13)/(750-100)=2 Omega`