A voltmeter coil has resistance `50.0(Omega)`and a resistor of `1.15k(Omega)`is connected in series. It can read potential differences up to 12 volts. If this same coil is used to construct an ammeter which can measure currents up to 2.0A ,What should be the resistance of the shunt used ?

Explanation

`R_eff = (1150 + 50 ) Omega = 1200 Omega `
` i= (12/1200) A = 0.01A `
Let r be the resistence of shunt used, then
the potential across both the registers is
same.
` 0.01 xx 50 = 1.99 xx r `
` rArr r = (0.01 xx 50)/1.99 = 50/199 `
` = 0.251 Omega ` .