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. `2 Omega`
C. `0.2 Omega`
D. `2 Omega`

Answer

B

Explanation

(b) The potential difference across ammter and shunt is the 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
potential difference across ammeter and shunt resistance is the same
That `i_(a) xx R = (i - i_(a)) xx S`
or `S = (i_(a) R)/(i - i_(a))`
Given, `i_(a) = 100 A, i = 750 A, R = 13 Omega`
Hence, `S = (100 xx 13)/(750 - 100) = 2 Omega`