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. `0.2Omega`
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.
i.e., `i_(a)xx R=(i-i_(a))xxS`
or `" " S=(i_(a)R)/(i-i_(a))` ....(i)
Give, `i_(a)=100 A,i=750 A, R=13 Omega`
Hence, `S=(100xx13)/(750-100)=2 Omega`