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

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 ) xx S `
or ` S = ( I _ a R ) /( I - I _ a ) " "` ...(i)
Given, ` I _ a = 100 A`,
` I = 750 A, R = 13 Omega `
Hence, ` S = ( 100 xx 13 )/( 750 - 100 ) = 2 Omega `